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Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
Exercise 3.1
Question 1
Sudha went to market with her friends. They wanted to eat `gol - gappa'
as well as `dahi - bhalla'. The number of plates of gol - gappa taken by
them is half that of dahi - bhalla. The cost of one plate of gol - gappa
was Rs. 10 and cost of one plate of dahi - bhalla was Rs. 5. She spent Rs.
60. Represent the situation algebraically and graphically.
Sol :Let no. of plates of gol - gappa = x
and no. of plates of dhai - bhalla = y
Cost of 1 plate gol - gappa = Rs. 10
Cost of 1 plate dhai - bhalla = Rs. 5
Total money spent = Rs. 20
According to the question,
$\mathrm{x}=\frac{1}{2} \mathrm{y}$ …(1)
10x + 5y = 60 …(2)
From eqn (1), we get
2x – y = 0 …(3)
Now, table for 2x – y = 0
| x | -1 | 0 | 1 | 2 |
| y=2x | -2 | 0 | 2 | 4 |
Now, table for 10x + 5y = 60
| x | -1 | 0 | 6 | 1 |
| $y=\dfrac{60-10x}{5}$ | 14 | 12 | 2 | 10 |
On plotting points on a graph paper and join them to get a straight line representing $x=\frac{1}{2} y$.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 10x + 5y = 60.
Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e. (3,6).
Question 2
Romila went to a stationary shop and purchased 2 pencils and 3 erasers
for Rs. 9. Her friend Sonali saw the new variety of pencils and erasers
with Romila and she also bought 4 pencils and 6 erasers of the same kind
for Rs. 18. Represent this situation algebraically and
graphically.
Sol :Let the cost of one pencil = Rs x
and cost of one eraser = Rs y
Romila spent = Rs. 9
Sonali spent = Rs. 18
According to the question
2x + 3y = 9 …(1)
4x + 6y = 18 …(2)
Now, table for 2x+3y=9
Now, table for 4x+6y=18
On plotting points on a graph paper and join them to get a straight line
representing 2x + 3y = 9.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 4x + 6y = 18.
On plotting points on a graph paper and join them to get a straight line
representing y = 2x + 30.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing y = 3x + 20.
From the graph, it is clear that lines represented by the equations x–y=1 and 2x+3y–12=0 are intersecting at a point A i.e. (3,2)
The given points are at which Person A walks (0,3) and (1,3)
and the points at which person B walks (0,4) and (1,5)
Now, we plot these points on a same graph as shown in the following figure.
(iii) x = 1, y = — 2 (iv) x = – 3,y = 4
(v) x = 1, y = — 2 (vi) x = — 4, y = 2
Sol :
Given equation is 4x-3y+24=0
(i) Justification
On substituting x = 0, y = 8 in LHS of given equation, we get
LHS = 4(0) – 3(8) + 24
(ii) Justification
On substituting x = – 6, y = 0 in LHS of given equation, we get
LHS = 4( – 6) – 3(0) + 24
(iii) Justification
On substituting x = 1, y = – 2 in LHS of given equation, we get
LHS = 4(1) – 3( – 2) + 24
(iv) Justification
On substituting x = – 3, y = 4 in LHS of given equation, we get
LHS = 4( – 3) – 3(4) + 24
(v) Justification
On substituting x = 1, y = – 2 in LHS of given equation, we get
LHS = 4(1) – 3( – 2) + 24
(vi) Justification
On substituting x = – 4, y = 2 in LHS of given equation, we get
(iii) C (3, — 5) (iv) D (4, 2)
(v) E (— 9, 5) (vi) F (— 3, 5)
(vii) G (— 9, — 5)
Sol :
The given equation is 5x – 3y + 30 = 0
(i) Given A ( – 6,0). Here x = – 6 and y = 0
On substituting x = – 6, y = 0 in LHS of given equation, we get
LHS = 5( – 6) – 3(0) + 30
6x + 2y = 1
Sol :
The given pair of linear equations is
3x + y = 2 or 3x + y – 2 = 0
and 6x + 2y = 1 or 6x + 2y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 2
and a2 = 6, b2 = 2 and c2 = – 1
Now, table for 3x-2y+12=0 or $y=\frac{3 x+12}{2}$
Here, the lines intersecting at point B i.e. ( – 2,3)
Hence, the unique solution is x = – 2 and y = 3.
x — 4y = — 14
Sol :
The given pair of linear equations is
3x + 2y = 14 or 3x + 2y – 14 = 0
and x – 4y = – 14 or x – 4y + 14 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 2 and c1 = – 14
and a2 = 1, b2 = – 4 and c2 = 14
Here, the lines intersecting at point B i.e. (2,4)
Hence, the unique solution is x = 2 and y = 4.
3x — 4y = 1
Sol :
The given pair of linear equations is
2x – 3y = 1 or 2x – 3y – 1 = 0
and 3x – 4y = 1 or 3x – 4y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 3 and c1 = – 1
and a2 = 3, b2 = – 4 and c2 = – 1
Here, the lines intersecting at point C i.e. ( – 1, – 1)
Hence, the unique solution is x = – 1 and y = – 1.
5x + 2y = 27
Sol :
The given pair of linear equations is
2x – y = 9 or 2x – y – 9 = 0
and 5x + 2y = 27 or 5x + 2y – 27 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 1 and c1 = – 9
and a2 = 5, b2 = 2 and c2 = – 27
∴ The lines representing the given pair of linear equations will intersect at a point.
Here, the lines intersecting at point C i.e. (5,1)
Hence, the unique solution is x = 5 and y = 1.
2x = y + 3
Sol :
The given pair of linear equations is
x + 3y = 5 or x + 3y – 5 = 0
and 2x – y = 3 or 2x – y – 3 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1, b1 = 3 and c1 = – 5
and a2 = 2, b2 = – 1 and c2 = – 3
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 2x-y=3 or y=2x-3
Here, the lines intersecting at point C i.e. (2,1)
Hence, the unique solution is x = 2 and y = 1.
2x — y = —3
Sol :
The given pair of linear equations is
3x – 5y = – 1 or 3x – 5y + 1 = 0
and 2x – y = – 3 or 2x – y + 3 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = – 5 and c1 = 1
and a2 = 2, b2 = – 1 and c2 = 3
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 2x-y=-3 or y=2x+3
Here, the lines intersecting at point B i.e. ( – 2, – 1)
Hence, the unique solution is x = – 2 and y = – 1.
3x — 9y + 15 = 0
Sol :
The given pair of linear equations is
2x – 6y + 10 = 0
and 3x – 9y + 15 = 0
Now, table for x – y – 1 = 0 or y = x – 1
Here, the lines intersecting at point B i.e. (3,2)
Hence, the unique solution is x = 3 and y = 2.
Does the point (4, 9) lie on any of the lines? Write its equation.
Sol :
The given equation is 3x – 5y = 19
and 3y – 7x + 1 = 0 or 7x – 3y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for 7x-3y=1 or $\mathrm{y}=\frac{7 \mathrm{x}-1}{3}$
From the graph, it is clear that lines represented by the equations 3x – 5y = 19 and 7x – 3y – 1 = 0 are intersecting at a point C i.e. ( – 2, – 5).
Yes, point (4,9) lie on 3y – 7x + 1 = 0.
The given equation is
2x – 3y = 1
and 3x – 4y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for 3x-4y=1 or $\mathrm{y}=\frac{3 \mathrm{x}-1}{4}$
Here, the lines intersecting at point F i.e. ( – 1, – 1)
Yes, point (3,2) lie on the line 3x – 4y = 1
2x + y = 4
Sol :
The given equation is
x – 2y = – 3
and 2x + y = 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for 2x+y=4 or y=4-2x
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect the x axis are B ( – 3,0) and E (2,0)
x — 2y = — 3
Sol :
The given equation is
2x + 3y = 8
and x – 2y = – 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x-2y=-3 or $y=\frac{x+3}{2}$
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect at x axis are B (4,0) and E ( – 3,0).
2x — 3y = — 4
Sol :
The given equation is
x + 2y = 5
and 2x – 3y = – 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x+2y=5 or $y=\frac{5-x}{2}$
Now, table for 2x-3y=-4 or $y=\frac{2 x+4}{3}$
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect the x axis are B (5,0) and E ( – 2,0)
4x + 3y = 24
Sol :
The given equation is
x – y + 1 = 0
and 4x+3y=24 or $x+\frac{3}{4} y=6$
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for 4x+3y=24 or $\mathrm{y}=\frac{4 \mathrm{x}-24}{3}$
Here, the lines intersecting at point C i.e. (3,4)
The points which intersect the x axis are B ( – 1,0) and E (6,0)
x— 2y = 7
Sol :
The given equation is
x + 2y = 1
and x – 2y = 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x-2y=7 or $y=\frac{x-7}{2}$
Here, the lines intersecting at point B i.e. (4, – 1.5)
The points which intersect the x axis are B (1,0) and E (7,0)
x — 2y = —7
Sol :
The given equation is
x + 2y = 1
and x – 2y = – 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x-2y=-7 or $y=\frac{x+7}{2}$
Here, the lines intersecting at point B i.e. ( – 3,2)
The points which intersect the x axis are B (1,0) and E ( – 7,0)
3y — x = 3
Sol :
The given equation is
2x – y = 4
and 3y – x = 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Here, the lines intersecting at point C, i.e. (3,2)
The point which intersects at y axis are A (0, – 4) and D (0,1)
2x — y — 4 = 0
Sol :
The given equation is
2x + 3y = 12
and 2x – y – 4 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for 2x-y-4=0 or y=2x-4
Here, the lines intersecting at point C, i.e. (3,2)
The points which intersects at y axis is A (0,4) and D (0, – 4)
x — y — 3 = 0
Sol :
The given equation is
2x – y = 5
and x – y = 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x – y = 3 or y = x – 3
Here, the lines intersecting at point C, i.e. (2, – 1)
The point which intersects at y axis are A (0, – 5) and D (0, – 3)
x + y + 1 = 0
Sol :
The given equation is
2x – y = 4
and x + y + 1 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x + y + 1 = 0 or y = – (x + 1)
Now, table for 2x – y = 5 or y = 2x + 5
Here, the lines intersecting at point C i.e. (2, – 1)
The point which is intersect at y axis are A (0,5) and D (0, – 5)
2x — 3y — 7 = 0
Shade the region bounded by the lines and the x – axis.
Sol :
The given equation is
3x + 2y = 4
and 2x – 3y = 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
3x — 2y = 12
Sol :
The given equation is
2x + 3y = – 5
and 3x – 2y = 12
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Here, the lines are intersecting at a point C (1,4).
2x + y = 8
Sol :
The given equation is
x – y = 1
and 2x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x – y = 1 or y = x – 1
Now, table for 2x + y = 8 or y = 2x – 8
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBE are C(3,2), B(0, – 1) and E(0,8)
The coordinates of the vertices of ∆PQR are P( – 4, 2), Q(2, 5)and R(1, 3)
y = x
Y = 2x
x + y = 6
Sol :
The given equation is
y = x
y = 2x
and x + y = 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
The coordinates of the vertices of ∆PQR are P(0, 0), Q(2, 4)and R(3, 3)
3y = x
x + y = 8
Sol :
The given equation is
y = x
3y = x
and x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Now, table for x + y = 8 or y = 8 – x
The coordinates of the vertices of ∆PQR are P(0, 0), Q(4, 4)and R(6, 2)
Sol :
Given, pair of equations
2x + 3y = 7
and (a + b)x + (2a – b)y = 3(a + b + 1)
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 7
and a2 = (a + b), b2 = (2a – b) and c2 = – (a + b + 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{2}{a+b}$,$\frac{b_{1}}{b_{2}}=\frac{3}{2 a-b}$ and $\frac{c_{1}}{c_{2}}=\frac{-7}{-3(a+b+1)}$
⇒ b = 1
Now, substituting the value of b in eqn (2), we get
⇒ a + b = 6
⇒ a + 1 = 6
⇒ a = 5
Sol :
Given, pair of equations
(2a – 1)x – 3y = 5
and 3x + (b – 2)y = 3
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = (2a – 1), b1 = – 3 and c1 = – 5
and a2 = 3, b2 = b – 2 and c2 = 3
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{2 a-1}{3}$,$\frac{b_{1}}{b_{2}}=\frac{-3}{b-2}$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-3}=\frac{5}{3}$
On taking I and III terms, we get
$\frac{2 a-1}{3}=\frac{5}{3}$
⇒ 3(2a – 1) = 15
⇒ 6a – 3 = 15
⇒ 6a = 15 + 3
$\Rightarrow \mathrm{a}=\frac{18}{6}=3$
On taking II and III terms, we get
$\Rightarrow \frac{-3}{b-2}=\frac{5}{3}$
⇒ – 9 = 5(b – 2)
⇒ 5b – 10 = – 9
⇒ 5b = – 9 + 10
$\Rightarrow \mathbf{b}=\frac{1}{5}$
Sol :
Given, pair of equations
kx + 3y – (k – 3) = 0
and 12x + ky – k = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = k, b1 = 3 and c1 = – (k – 3)
and a2 = 12, b2 = k and c2 = – k
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{k}{12}, \frac{b_{1}}{b_{2}}=\frac{3}{k}$ and $\frac{c_{1}}{c_{2}}=\frac{-(k-3)}{-k}=\frac{(k-3)}{k}$
On taking I and II terms, we get
$\frac{k}{12}=\frac{3}{k}$
⇒ k2 = 36
⇒ k = √36
⇒ k = ±6
But k = – 6 not satisfies the last two terms of eqn (1)
On taking II and III terms, we get
$\frac{3}{\mathrm{k}}=\frac{(\mathrm{k}-3)}{\mathrm{k}}$
⇒ 3k = k(k – 3)
⇒ 3k = k2 – 3k
⇒ k2 – 3k – 3k = 0
⇒ k(k – 6) = 0
⇒ k = 0 and 6
Which satisfies the last two terms of eqn (1)
Hence, the required value of k = 0, 6
Sol :
Given, pair of equations
3x + 4y = 12
and (a + b)x + 2(a – b)y = 5a – 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 4 and c1 = – 12
and a2 = (a + b), b2 = 2(a – b) and c2 = – (5a – 1) = 1 + 5a
$a_{1}=3, \quad b_{1}=4, c_{1}=-12$
and $a_{2}=(a+b), b_{2}=2(a-b), c_{2}=-(5 a-1)=1+5 a$
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Solving eqn (1) and (2), we get
Now, substituting the value of b in eqn (2), we get
⇒ a – 4b = 1
⇒ a – 4 = 1
⇒ a = 1 + 4
⇒ a = 5
Sol :
Given, pair of equations
(a – 1)x + 3y = 2
and 6x + (1 – 2b)y = 6
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = (a – 1), b1 = 3 and c1 = – 2
and a2 = 6, b2 = 1 – 2b and c2 = – 6
On taking I and III terms, we get
$\frac{a-1}{6}=\frac{1}{3}$
⇒ 3(a – 1) = 6
⇒ 3a – 3 = 6
⇒ 3a = 6 + 3
$\Rightarrow \mathrm{a}=\frac{9}{3}=3$
On taking II and III terms, we get
$\Rightarrow \frac{3}{1-2 b}=\frac{1}{3}$
⇒ 9 = 1 – 2b
⇒ – 2b = 9 – 1
⇒ – 2b = 8
⇒ b = – 4
Sol :
Given, pair of equations
2x + 3y = 7
and (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 7
and a2 = (a + b + 1), b2 = (a + 2b + 2) and c2 = – {4(a + b) + 1}
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{2}{a+b+1}, \frac{b_{1}}{b_{2}}=\frac{3}{a+2 b+2}$ and $\frac{c_{1}}{c_{2}}=\frac{-7}{-\{4(a+b)+1\}}$
On taking I and II terms, we get
$\frac{2}{a+b+1}=\frac{3}{a+2 b+2}$
⇒ 2(a + 2b + 2) = 3(a + b + 1)
⇒ 2a + 4b + 4 = 3a + 3b + 3
⇒ 2a – 3a – 3b + 4b = 3 – 4
⇒ – a + b = – 1
⇒ a – b = 1 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{2}{a+b+1}=\frac{7}{\{4(a+b)+1\}}$
⇒ 2{4(a + b) + 1)} = 7(a + b + 1)
⇒ 2(4a + 4b + 1) = 7a + 7b + 7
⇒ 8a – 7a + 8b – 7b = – 2 + 7
⇒ a + b = 5 …(2)
Solving eqn (1) and (2), we get
On taking I and II terms, we get
$\frac{a}{12}=\frac{3}{a}$
⇒ a2 = 36
⇒ a = √36
⇒ a = ±6
Sol :
Given, pair of equations
x + 2y = 5
and 3x + ay + 15 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1, b1 = 2 and c1 = – 5
and a2 = 3, b2 = a and c2 = 15
For no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
On taking I and II terms, we get
$\frac{1}{3}=\frac{2}{a}$
$\Rightarrow \mathbf{a}=\mathbf{6}$
Sol :
Given, pair of equations
3x + y = 1
and (2a – 1)x + (a – 1)y = 2a + 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 1
and a2 = (2a – 1), b2 = (a – 1) and c2 = – (2a + 1)
For no solutions,
$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$
On taking I and II terms, we get
$\frac{3}{2 a-1}=\frac{1}{a-1}$
⇒ 3(a – 1) = 2a – 1
⇒ 3a – 3 = 2a – 1
⇒ 3a – 2a = – 1 + 3
⇒ a = 2
Sol :
Given, pair of equations
(3a + 1)x + 3y – 2 = 0
and (a2 + 1)x + (a – 2)y – 5 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3a + 1, b1 = 3 and c1 = – 2
and a2 = a2 + 1, b2 = a – 2 and c2 = – 5
On taking I and II terms, we get
$\frac{3 a+1}{\left(a^{2}+1\right)}=\frac{3}{a-2}$
⇒ (3a + 1)(a – 2) = 3(a2 + 1)
⇒ 3a2 – 6a + a – 2 = 3a2 + 3
⇒ – 5a = 2 + 3
⇒ a = – 1
Sol :
Given, pair of equations
cx + 3y – (c – 3) = 0
and 12x + cy – c = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = c, b1 = 3 and c1 = – (c – 3)
and a2 = 12, b2 = c and c2 = – c
On taking I and III terms, we get
$\frac{c}{12}=\frac{3}{c}$
⇒ c2 = 36
⇒ c = √36
⇒ c = ±6
But c = – 6 not satisfies the eqn (1)
Hence, the required value of c = 6.
Sol :
Given, pair of equations
2x + 3y = 2
and (c + 2)x + (2c + 1)y = 2(c – 1)
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 2
and a2 = c + 2, b2 = 2c + 1 and c2 = – 2(c – 1)
So,
$\begin{matrix}\frac{1}{c+1}&=\frac{c+1}{9}&=\frac{-5}{-8 c+1}\\I&II&III\end{matrix}$
⇒ From (I) and (II)
$\frac{1}{c+1}=\frac{c+1}{9}$
⇒ 9 = (c+1)2
⇒ 9 = c2 + 1 + 2c
⇒ 9 – 1 = c2 + 2c
⇒ 8 = c2 + 2c
⇒ c2 + 2c – 8 = 0
Factorize by splitting the middle term,
c2 + 4c – 2c – 8 = 0
⇒ c ( c + 4 ) – 2 ( c + 4) = 0
⇒ (c+4) (c–2) = 0
⇒ c = –4, c = 2
From (II) and (III)
$\frac{c+1}{9}=\frac{-5}{-8 c+1}$
⇒ (c+1)(–8c=1) = –5 × 9
⇒ –8c2 + c – 8c + 1 = –45
⇒ –8c2 + c – 8c + 1 + 45 = 0
⇒ –8c2 – 7c + 46 = 0
⇒ 8c2 + 7c – 46 = 0
⇒ 8c2 – 16c + 23c –46 = 0
⇒ 8c ( c–2) + 23 ( c–2) = 0
⇒ (8c+23) ( c–2) = 0
⇒ c = –23/8 and c = 2
From (I) and (III)
$\frac{1}{c+1}=\frac{-5}{-8 c+1}$
⇒ –8c+1 = –5(c+1)
⇒ –8c + 1 = –5c – 5
⇒ –8c + 5c = –5 –1
⇒ –3c = –6
⇒ c = 2
So the value of c = 2.
Sol :
The pair of equations are:
(c – 1) x – y = 5
(c + 1) x + (1 – c) y = 3c + 1
These equations can be written as:
(c – 1) x – y – 5 = 0
(c + 1) x + (1 – c) y –( 3c + 1)
On comparing the given equation with standard form i.e.
a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = c – 1 , b1 = –1 , c1 = –5
a2 = c + 1, b2 = 1 – c , c2 = –(3c + 1)
Now, table for 3x+5y=15 or $y=\frac{15-3 x}{5}$
Here, the lines are intersecting at point C (5, 0).
The coordinates of the vertices of ∆ABC are A(0, – 4), B(0, 3)and C(5,0)
Sol :
Given, pair of equations
ax + 2y = 5
and 3x + y = 1
On comparing the given equation with standard form
Here,$\frac{a_{1}}{a_{2}}=\frac{3}{a}$, $\frac{b_{1}}{b_{2}}=-\frac{2}{1}$
Here,$\frac{a_{1}}{a_{2}}=\frac{4}{2}=2$, $\frac{b_{1}}{b_{2}}=\frac{p}{2}$
∴ $\begin{array}{ll}2 \neq &\frac{p}{2}\\\text{I}&\text{I I}\end{array}$
Now, table for x – y + 4 = 0
On plotting points on a graph paper and join them to get a straight line representing x + y – 10 = 0.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – y + 4 = 0.
∴ x = 3, y = 7 is the solution of the pair of linear equations.
Hence, the required number of boys is 3 and girls is 7.
Let the present age of father = x year
and the present age of his son = y year
Two years ago,
Father’s age = (x – 2) year
His son’s age = (y – 2) year
According to the question,
⇒ (x – 2) = 5(y – 2)
⇒ x – 2 = 5y – 10
⇒ x – 5y + 8 = 0 …(1)
After two years,
Father’s age = (x + 2) year
His son’s age = (y + 2) year
According to the question,
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x – 3y – 12 = 0 …(2)
Now, table for x – 5y + 8 = 0
Now, table for x – 3y – 12 = 0
On plotting points on a graph paper and join them to get a straight line representing x – 5y + 8 = 0.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – 3y – 12 = 0.
∴ x = 42, y = 10 is the solution of the pair of linear equations.
Hence, the age of father is 42years and age of his son is 10 years.
Let the number of pants = x
and the number of skirts = y
According to the question
Number of skirts = 2(Number of pants) – 2
y = 2x – 2 …(i)
Also, Number of skirts = 4(Number of pants) – 4
y = 4x – 4 …(ii)
Substituting the value of y = 4x – 4 in eqn (i),we get
4x – 4 = 2x – 2
⇒ 4x – 2x – 4 + 2 = 0
⇒ 2x = 2
⇒ x = 1
Now, substitute the value of x in eqn (ii), we get
y = 4(1) – 4 = 0
∴ x = 1, y = 0 is the solution of the pair of linear equations.
Hence, the number of pants she purchased is 1 and the number of skirts, she purchased is zero i.e., she didn’t buy any skirt.
Let the cost of one pencil = Rs x
and the cost of one eraser = Rs y
According to the question
2x + 3y = 9 …(i)
x + 2y = 5 …(ii)
Now, table for 2x + 3y = 9
On plotting points on a graph paper and join them to get a straight line representing 2x + 3y = 9.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x+2y=5.
∴ x = 3, y = 1 is the solution of the pair of linear equations.
Hence, the cost of one pencil is Rs 3 and cost of one eraser is Rs 1.
| x | 0 | 4.5 |
| $y=\dfrac{9-2x}{3}$ | 3 | 0 |
Now, table for 4x+6y=18
| x | 0 | 3 |
| $y=\dfrac{18-4x}{6}$ | 3 | 1 |
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 4x + 6y = 18.
Here, we can see that both the lines coincide. This is so, because, both the
equations are equivalent, i.e.2(2x + 3y) = 2×9 equation (2) is derived from
the other.
Let the present age of son = x year
and the age of his father = y year
According to the question
y = 2x + 30
or, 2x – y = – 30 …(1)
After 10 years,
Age of son = (x + 10)year
Age of father = (y + 10)year
So, According to the question
y + 10 = 3(x + 10)
y + 10 = 3x + 30
y = 3x + 20
or, 3x – y = – 20 …(2)
Question 3
Present age of father is 30 years more than twice that of his son. After
10 years, the age of father will be thrice the age of his son. Represent
this situation algebraically and geometrically.
Sol :Let the present age of son = x year
and the age of his father = y year
According to the question
y = 2x + 30
or, 2x – y = – 30 …(1)
After 10 years,
Age of son = (x + 10)year
Age of father = (y + 10)year
So, According to the question
y + 10 = 3(x + 10)
y + 10 = 3x + 30
y = 3x + 20
or, 3x – y = – 20 …(2)
Now, table for y = 2x + 30
| x | 0 | -15 | 5 |
| y=2x+30 | 30 | 0 | 40 |
Now, table for y = 3x + 20
| x | 0 | 5 | 10 |
| y=3x+20 | 20 | 35 | 50 |
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing y = 3x + 20.
Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e.
(10,50).
So, the age of son is 10years and age of his father is 50years.
The given equation is
x + 2y – 4 = 0
and
2x + 4y – 12 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
So, the age of son is 10years and age of his father is 50years.
Question 4
The path of a wheel of train A is given by the equation x + 2y — 4 = 0
and the path of a wheel of another train B is given by the equation 2x +
4y — 12 = O. Represent this situation geometrically.
Sol :The given equation is
x + 2y – 4 = 0
and
2x + 4y – 12 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x + 2y – 4 = 0 or $y=\frac{4-x}{2}$
| x | 0 | 4 | 2 |
| $y=\frac{4-x}{2}$ | 2 | 0 | 1 |
Now, table for 2x + 4y – 12 = 0 or $y=\frac{12-2 x}{4}$
| x | 0 | 6 | 6 |
| $y=\frac{12-2 x}{4}$ | 3 | 0 | 2 |
From the graph, it is clear that lines represented by the equations x + 2y – 4
= 0 and 2x + 4y – 12 = 0 are parallel.
The given equation is
x – y = 1
and
2x + 3y = 12
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Question 5
The path of highway number 1 and 2 are given by the equations x — y = 1
and 2x + 3y = 12 respectively. Represent these equations
geometrically.
Sol :The given equation is
x – y = 1
and
2x + 3y = 12
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x – y = 1 or y = x – 1
| x | 0 | 1 | 3 | 4 |
| y=x-1 | -1 | 0 | 2 | 3 |
Now, table for 2x + 3y = 12 or $y=\frac{12-2 x}{3}$
| x | 0 | 6 | 3 |
| $y=\frac{12-2x}{3}$ | 4 | 0 | 2 |
From the graph, it is clear that lines represented by the equations x–y=1 and 2x+3y–12=0 are intersecting at a point A i.e. (3,2)
Question 6
Person A walks along the path joining points (0, 3) and (1, 3) and person
B walks along the path joining points (0, 4) and (1, 5). Represent this
situation geometrically.
Sol :The given points are at which Person A walks (0,3) and (1,3)
and the points at which person B walks (0,4) and (1,5)
Now, we plot these points on a same graph as shown in the following figure.
Question 7
Examine which of the following pair of values of x and y is a solution of
equation 4x — 3y + 24 = 0.
(i) x = 0, y = 8 (ii) x = — 6, y = 0(iii) x = 1, y = — 2 (iv) x = – 3,y = 4
(v) x = 1, y = — 2 (vi) x = — 4, y = 2
Sol :
Given equation is 4x-3y+24=0
(i) Justification
On substituting x = 0, y = 8 in LHS of given equation, we get
LHS = 4(0) – 3(8) + 24
= 0 – 24 + 24 = 0 = RHS
Hence, x = 0, y = 8 is a solution of the equation 4x – 3y + 24 = 0
Hence, x = 0, y = 8 is a solution of the equation 4x – 3y + 24 = 0
(ii) Justification
On substituting x = – 6, y = 0 in LHS of given equation, we get
LHS = 4( – 6) – 3(0) + 24
= – 24 + 24 = 0 = RHS
Hence, x = – 6, y = 0 is a solution of the equation 4x – 3y + 24 = 0
Hence, x = – 6, y = 0 is a solution of the equation 4x – 3y + 24 = 0
(iii) Justification
On substituting x = 1, y = – 2 in LHS of given equation, we get
LHS = 4(1) – 3( – 2) + 24
= 4 + 6 + 24 = 34 ≠ RHS
Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0
Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0
(iv) Justification
On substituting x = – 3, y = 4 in LHS of given equation, we get
LHS = 4( – 3) – 3(4) + 24
= – 12 – 12 + 24 = 0 = RHS
Hence, x = – 3, y = 4 is a solution of the equation 4x – 3y + 24 = 0
Hence, x = – 3, y = 4 is a solution of the equation 4x – 3y + 24 = 0
(v) Justification
On substituting x = 1, y = – 2 in LHS of given equation, we get
LHS = 4(1) – 3( – 2) + 24
= 4 + 6 + 24 = 34 ≠ RHS
Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0
Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0
(vi) Justification
On substituting x = – 4, y = 2 in LHS of given equation, we get
LHS = 4( – 4) – 3(2) + 24
= – 16 – 6 + 24
= – 22 + 24 = 2 ≠ RHS
Hence, x = – 4, y = 2 is not a solution of the equation 4x – 3y + 24 = 0
Question 8
Examine which of the following points lie on the graph of the linear
equation 5x — 3y + 30 = 0.
(i) A (— 6, 0) (ii) B (0, 10)(iii) C (3, — 5) (iv) D (4, 2)
(v) E (— 9, 5) (vi) F (— 3, 5)
(vii) G (— 9, — 5)
Sol :
The given equation is 5x – 3y + 30 = 0
(i) Given A ( – 6,0). Here x = – 6 and y = 0
On substituting x = – 6, y = 0 in LHS of given equation, we get
LHS = 5( – 6) – 3(0) + 30
=– 30 + 30 = 0 = RHS
So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.
Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.
So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.
Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.
(ii) Given B (0,10). Here x = 0 and y = 10
On substituting x = 0, y = 10 in LHS of given equation, we get
LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS
So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0
Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.
On substituting x = 0, y = 10 in LHS of given equation, we get
LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS
So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0
Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.
(iii) Given C (3, – 5). Here x = 3 and y = –
5
On substituting x = 3, y = – 5 in LHS of given equation, we get
LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS
So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
On substituting x = 3, y = – 5 in LHS of given equation, we get
LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS
So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
(iv) Given D (4,2). Here x = 4 and y = 2
On substituting x = 4, y = 2 in LHS of given equation, we get
LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS
So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
On substituting x = 4, y = 2 in LHS of given equation, we get
LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS
So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
(v) Given E ( – 9,5). Here x = – 9 and y = 5
On substituting x = – 9, y = 5 in LHS of given equation, we get
LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS
So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
On substituting x = – 9, y = 5 in LHS of given equation, we get
LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS
So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0
Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.
(vi) Given F ( – 3,5). Here x = – 3 and y =
5
On substituting x = – 3, y = 5 in LHS of given equation, we get
LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS
So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0
Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.
On substituting x = – 3, y = 5 in LHS of given equation, we get
LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS
So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0
Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.
(vii) Given G ( – 9, – 5). Here x = 3 and y = –
5
On substituting x = – 9, y = – 5 in LHS of given equation, we get
LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS
So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0
Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.
Or Graphically
On substituting x = – 9, y = – 5 in LHS of given equation, we get
LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS
So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0
Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.
Or Graphically
Here, we can see through the graph also that Point A, B, F and G lie on the
graph of the linear equation 5x – 3y + 30 = 0
Question 9 A
Solve graphically the following system of linear equations if it has
unique solution:
3x + y = 26x + 2y = 1
Sol :
The given pair of linear equations is
3x + y = 2 or 3x + y – 2 = 0
and 6x + 2y = 1 or 6x + 2y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 2
and a2 = 6, b2 = 2 and c2 = – 1
$\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}$, $\frac{b_{1}}{b_{2}}=\frac{1}{2}=1$ and $\frac{c_{1}}{c_{2}}=\frac{-2}{-1}=2$
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The lines representing the given pair of linear equations are parallel.
3x — 2y + 12 = 0
Sol :
The given pair of linear equations is
2x – 3y + 13 = 0
and 3x – 2y + 12 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 3 and c1 = 13
and a2 = 3, b2 = – 2 and c2 = 12
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The lines representing the given pair of linear equations are parallel.
Question 9 B
Solve graphically the following system of linear equations if it has
unique solution:
2x — 3y + 13 = 03x — 2y + 12 = 0
Sol :
The given pair of linear equations is
2x – 3y + 13 = 0
and 3x – 2y + 12 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 3 and c1 = 13
and a2 = 3, b2 = – 2 and c2 = 12
$\frac{a_{1}}{a_{2}}=\frac{2}{3}$,$\frac{b_{1}}{b_{2}}=\frac{-3}{-2}=\frac{3}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{13}{12}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will
intersect at a point.
Now, table for 2x-3y+13=0 or $y=\frac{2 x+13}{3}$
| x | -6.5 | -2 | 0 |
| $y=\frac{2 x+13}{3}$ | 0 | 3$\frac{13}{3}$ |
| x | 0 | 4 | -2 |
| $y=\frac{3 x-12}{2}$ | 6 | 0 | 3 |
Here, the lines intersecting at point B i.e. ( – 2,3)
Hence, the unique solution is x = – 2 and y = 3.
Question 9 C
Solve graphically the following system of linear equations if it has
unique solution:
3x + 2y = 14x — 4y = — 14
Sol :
The given pair of linear equations is
3x + 2y = 14 or 3x + 2y – 14 = 0
and x – 4y = – 14 or x – 4y + 14 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 2 and c1 = – 14
and a2 = 1, b2 = – 4 and c2 = 14
$\frac{a_{1}}{a_{2}}=\frac{3}{1}$, $\frac{b_{1}}{b_{2}}=\frac{2}{-4}=-\frac{1}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{-14}{14}=-1$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 3x+2y=14 or $y=\frac{14-3 x}{2}$
| x | 0 | 2 | 4 |
| $y=\frac{14-3 x}{2}$ | 7 | 4 | 1 |
Now, table for x-4y+14=0 or $y=\frac{x+14}{4}$
| x | -2 | 14 | 2 |
| $y=\frac{x+14}{4}$ | 3 | 0 | 4 |
Here, the lines intersecting at point B i.e. (2,4)
Hence, the unique solution is x = 2 and y = 4.
Question 9 D
Solve graphically the following system of linear equations if it has
unique solution:
2x — 3y = 13x — 4y = 1
Sol :
The given pair of linear equations is
2x – 3y = 1 or 2x – 3y – 1 = 0
and 3x – 4y = 1 or 3x – 4y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 3 and c1 = – 1
and a2 = 3, b2 = – 4 and c2 = – 1
$\frac{a_{1}}{a_{2}}=\frac{2}{3}$, $\frac{b_{1}}{b_{2}}=\frac{-3}{-4}=\frac{3}{4}$ and $\frac{c_{1}}{c_{2}}=\frac{-1}{-1}=1$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 2x-3y=1 or $y=\frac{2 x-1}{3}$
| x | 0.5 | 0 | -1 |
| $y=\frac{2 x-1}{3}$ | 0 | -0.33 | -1 |
Now, table for 3x-4y=1 or $y=\frac{3 x-1}{4}$
| x | 0 | 0.33 | -1 |
| $y=\frac{3 x-1}{4}$ | -0.25 | 0 | -1 |
Here, the lines intersecting at point C i.e. ( – 1, – 1)
Hence, the unique solution is x = – 1 and y = – 1.
Question 9 E
Solve graphically the following system of linear equations if it has
unique solution:
2x — y = 95x + 2y = 27
Sol :
The given pair of linear equations is
2x – y = 9 or 2x – y – 9 = 0
and 5x + 2y = 27 or 5x + 2y – 27 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 1 and c1 = – 9
and a2 = 5, b2 = 2 and c2 = – 27
$\frac{a_{1}}{a_{2}}=\frac{2}{5}$ ,$\frac{b_{1}}{b_{2}}=\frac{-1}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{-9}{-27}=\frac{1}{3}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for $2 x-y=9$ or y=2 x-9
| x | 0 | 4.5 | 5 |
| y=2x-9 | -9 | 0 | 1 |
Now, table for 5 x+2 y=27 or $y=\frac{27-5 x}{2}$
| x | 5.4 | 5 | 3 |
| $y=\frac{27-5 x}{2}$ | 0 | 1 | 6 |
Here, the lines intersecting at point C i.e. (5,1)
Hence, the unique solution is x = 5 and y = 1.
Question 9 F
Solve graphically the following system of linear equations if it has
unique solution:
3y = 5 — x2x = y + 3
Sol :
The given pair of linear equations is
x + 3y = 5 or x + 3y – 5 = 0
and 2x – y = 3 or 2x – y – 3 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1, b1 = 3 and c1 = – 5
and a2 = 2, b2 = – 1 and c2 = – 3
$\frac{a_{1}}{a_{2}}=\frac{1}{2}$, $\frac{b_{1}}{b_{2}}=\frac{3}{-1}=-3$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-3}=\frac{5}{3}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for x+3y=5 or $y=\frac{5-x}{3}$
| x | 5 | -1 | 2 |
| $y=\frac{5-x}{3}$ | 0 | 2 | 1 |
| x | 0 | 1.5 | 2 |
| y=2x-3 | -3 | 0 | 1 |
Here, the lines intersecting at point C i.e. (2,1)
Hence, the unique solution is x = 2 and y = 1.
Question 9 G
Solve graphically the following system of linear equations if it has
unique solution:
3x — 5y = —12x — y = —3
Sol :
The given pair of linear equations is
3x – 5y = – 1 or 3x – 5y + 1 = 0
and 2x – y = – 3 or 2x – y + 3 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = – 5 and c1 = 1
and a2 = 2, b2 = – 1 and c2 = 3
$\frac{a_{1}}{a_{2}}=\frac{3}{2}$, $\frac{b_{1}}{b_{2}}=\frac{-5}{-1}=5$ and $\frac{c_{1}}{c_{2}}=\frac{1}{3}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
$\therefore \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 3x-5y=-1 or $y=\frac{3 x+1}{5}$
| x | 0 | 3 | -2 |
| $y=\frac{3 x+1}{5}$ | 0.2 | 2 | -1 |
| x | 0 | -1.5 | -2 |
| y=2x+3 | 3 | 0 | -1 |
Here, the lines intersecting at point B i.e. ( – 2, – 1)
Hence, the unique solution is x = – 2 and y = – 1.
Question 9 H
Solve graphically the following system of linear equations if it has
unique solution:
2x — 6y + 10 = 03x — 9y + 15 = 0
Sol :
The given pair of linear equations is
2x – 6y + 10 = 0
and 3x – 9y + 15 = 0
On comparing the given equations with standard form of pair of linear
equations i.e. a1 x + b1y + c1 = 0
and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = – 6 and c1 = 10
and a2 = 3, b2 = – 9 and c2 = 15
a1 = 2, b1 = – 6 and c1 = 10
and a2 = 3, b2 = – 9 and c2 = 15
$\frac{a_{1}}{a_{2}}=\frac{2}{3}$,$\frac{b_{1}}{b_{2}}=\frac{-6}{-9}=\frac{2}{3}$ and $\frac{c_{1}}{c_{2}}=\frac{10}{15}=\frac{2}{3}$
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
∴ The lines representing the given pair of linear equations will coincide.
x — y — 1 = 0
Sol :
The given pair of linear equations is
3x + y – 11 = 0
and x – y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 11
and a2 = 1, b2 = – 1 and c2 = – 1
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
∴ The lines representing the given pair of linear equations will coincide.
Question 9 I
Solve graphically the following system of linear equations if it has
unique solution:
3x + y — 11 = 0x — y — 1 = 0
Sol :
The given pair of linear equations is
3x + y – 11 = 0
and x – y – 1 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 11
and a2 = 1, b2 = – 1 and c2 = – 1
$\frac{a_{1}}{a_{2}}=\frac{3}{1}=3$, $\frac{b_{1}}{b_{2}}=\frac{1}{-1}=-1$ and $\frac{c_{1}}{c_{2}}=\frac{-11}{-1}=11$
$\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
$\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ The lines representing the given pair of linear equations will intersect at a point.
Now, table for 3x + y – 11 = 0 or y = 11 – 3x
| x | 0 | 1 | 3 |
| y=11-3x | 11 | 8 | 2 |
| x | 0 | 1 | 3 |
| y=x-1 | -1 | 0 | 2 |
Here, the lines intersecting at point B i.e. (3,2)
Hence, the unique solution is x = 3 and y = 2.
Question 10
Solve the following system of linear equations graphically:
3x — 5y = 19, 3y — 7x + 1 = 0Does the point (4, 9) lie on any of the lines? Write its equation.
Sol :
The given equation is 3x – 5y = 19
and 3y – 7x + 1 = 0 or 7x – 3y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x-5y=19 or $y=\frac{3 x-19}{5}$
| x | 0 | 6.33 | -2 |
| $y=\frac{3 x-19}{5}$ | -3.8 | 0 | -5 |
| x | 1 | 0 | 4 |
| $y=\frac{7 x-1}{3}$ | 2 | -0.33 | 9 |
From the graph, it is clear that lines represented by the equations 3x – 5y = 19 and 7x – 3y – 1 = 0 are intersecting at a point C i.e. ( – 2, – 5).
Yes, point (4,9) lie on 3y – 7x + 1 = 0.
Question 11
Solve the following system of linear equations graphically: 2x — 3y = 1,
3x — 4y = 1 Does the point (3, 2) lie on any of the lines? Write its
equation.
Sol :The given equation is
2x – 3y = 1
and 3x – 4y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x-3y=1 or $y=\frac{2 x-1}{3}$
| x | 0.5 | 0 | 3 |
| $y=\frac{2 x-1}{3}$ | 0 | -0.33 | 2 |
| x | 0 | 0.33 | -1 |
| $y=\frac{3 x-1}{4}$ | -0.25 | 0 | -1 |
Here, the lines intersecting at point F i.e. ( – 1, – 1)
Yes, point (3,2) lie on the line 3x – 4y = 1
Question 12 A
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
x — 2y = – 32x + y = 4
Sol :
The given equation is
x – 2y = – 3
and 2x + y = 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x-2y=-3 or $y=\frac{x+3}{2}$
| x | 0 | -3 | 1 |
| $y=\frac{x+3}{2}$ | 1.5 | 0 | 2 |
| x | 2 | 0 | 1 |
| y=4-2x | 0 | 4 | 2 |
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect the x axis are B ( – 3,0) and E (2,0)
Question 12 B
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
2x + 3y = 8x — 2y = — 3
Sol :
The given equation is
2x + 3y = 8
and x – 2y = – 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x+3y=8 or $y=\frac{8-2 x}{3}$
| x | 0 | 4 | 1 |
| $y=\frac{x+3}{2}$ | $\frac{8}{3}$ | 0 | 2 |
| x | -1 | -3 | 1 |
| y=4-2x | -1 | 0 | 2 |
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect at x axis are B (4,0) and E ( – 3,0).
Question 12 C
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
x + 2y = 52x — 3y = — 4
Sol :
The given equation is
x + 2y = 5
and 2x – 3y = – 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x+2y=5 or $y=\frac{5-x}{2}$
| x | 0 | 5 | 1 |
| $y=\frac{5-x}{2}$ | $\frac{5}{2}$ | 0 | 2 |
| x | 0 | -2 | 1 |
| $y=\frac{2 x+4}{3}$ | $\frac{4}{3}$ | 0 | 2 |
Here, the lines intersecting at point C i.e. (1,2)
The points which intersect the x axis are B (5,0) and E ( – 2,0)
Question 12 D
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
x — y + 1 = 04x + 3y = 24
Sol :
The given equation is
x – y + 1 = 0
and 4x+3y=24 or $x+\frac{3}{4} y=6$
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x-y+1=0 or y=x+1
| x | 0 | -1 | 3 |
| y=x+1 | 1 | 0 | 4 |
| x | 0 | 6 | 3 |
| $y=\frac{4 x-24}{3}$ | 8 | 0 | 4 |
Here, the lines intersecting at point C i.e. (3,4)
The points which intersect the x axis are B ( – 1,0) and E (6,0)
Question 12 E
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
x + 2y = 1x— 2y = 7
Sol :
The given equation is
x + 2y = 1
and x – 2y = 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x+2y=1 or $y=\frac{1-x}{2}$
| x | 0 | 1 | 4 |
| $y=\frac{1-x}{2}$ | $\frac{1}{2}$ | 0 | $\frac{-3}{2}$ |
| x | 0 | 7 | 4 |
| $y=\frac{x-7}{2}$ | $\frac{-7}{2}$ | 0 | $\frac{-3}{2}$ |
Here, the lines intersecting at point B i.e. (4, – 1.5)
The points which intersect the x axis are B (1,0) and E (7,0)
Question 12 F
Solve the following system of equations graphically. Also find the points
where the lines intersect x – axis.
x + 2y = 1x — 2y = —7
Sol :
The given equation is
x + 2y = 1
and x – 2y = – 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x+2y=1 or $y=\frac{1-x}{2}$
| x | 0 | 1 | -3 |
| $y=\frac{1-x}{2}$ | $\frac{1}{2}$ | 0 | 2 |
| x | 0 | -7 | -3 |
| $y=\frac{x+7}{2}$ | $\frac{7}{2}$ | 0 | -2 |
Here, the lines intersecting at point B i.e. ( – 3,2)
The points which intersect the x axis are B (1,0) and E ( – 7,0)
Question 13 A
Solve the following system of equations graphically. Also find the points
where the lines meet the y – axis.
2x — y = 43y — x = 3
Sol :
The given equation is
2x – y = 4
and 3y – x = 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x – y = 4 or y = 2x – 4
| x | 0 | 2 | 3 |
| y=2x-4 | -4 | 0 | 2 |
Now, table for 3y-x=3 or $y=\frac{x+3}{3}$
| x | 0 | -3 | 3 |
| $y=\frac{x+3}{3}$ | 1 | 0 | 2 |
Here, the lines intersecting at point C, i.e. (3,2)
The point which intersects at y axis are A (0, – 4) and D (0,1)
Question 13 B
Solve the following system of equations graphically. Also find the points
where the lines meet the y – axis.
2x + 3y — 12 = 02x — y — 4 = 0
Sol :
The given equation is
2x + 3y = 12
and 2x – y – 4 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x+3y=12 or $y=\frac{12-x}{3}$
| x | 0 | 6 | 3 |
| $y=\frac{12-x}{3}$ | 4 | 0 | 2 |
| x | 0 | 2 | 3 |
| y=2x-4 | -4 | 0 | -2 |
Here, the lines intersecting at point C, i.e. (3,2)
The points which intersects at y axis is A (0,4) and D (0, – 4)
Question 13 C
Solve the following system of equations graphically. Also find the points
where the lines meet the y – axis.
2x — y — 5 = 0x — y — 3 = 0
Sol :
The given equation is
2x – y = 5
and x – y = 3
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x – y = 5 or y = 2x – 5
| x | 0 | 5/2 | 2 |
| y=2x-5 | -5 | 0 | -1 |
| x | 0 | 3 | 2 |
| y=x-3 | -3 | 0 | -1 |
Here, the lines intersecting at point C, i.e. (2, – 1)
The point which intersects at y axis are A (0, – 5) and D (0, – 3)
Question 13 D
Solve the following system of equations graphically. Also find the points
where the lines meet the y – axis.
2x — y — 4 = 0x + y + 1 = 0
Sol :
The given equation is
2x – y = 4
and x + y + 1 = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x – y = 4 or y = 2x – 4
| x | 0 | 2 | 1 |
| y=2x-4 | -4 | 0 | -2 |
| x | 0 | -1 | 1 |
| y=-(x+1) | -1 | 0 | -2 |
Here, the lines intersecting at point C, i.e. (1, – 2)
The point which intersects at y axis are A (0, – 4) and D (0, – 1)
2x — y — 5 = 0
Sol :
The given equation is
3x + y = 5
and 2x – y = 5
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
The point which intersects at y axis are A (0, – 4) and D (0, – 1)
Question 13 E
Solve the following system of equations graphically. Also find the points
where the lines meet the y – axis.
3x + y — 5 = 02x — y — 5 = 0
Sol :
The given equation is
3x + y = 5
and 2x – y = 5
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x + y = 5 or y = 5 – 3x
| x | 0 | 1 | 2 |
| y=5-3x | 5 | 2 | -1 |
| x | 0 | 1 | 2 |
| y=2x+5 | -5 | -3 | -1 |
Here, the lines intersecting at point C i.e. (2, – 1)
The point which is intersect at y axis are A (0,5) and D (0, – 5)
Question 14 A
Solve the following system of linear equations graphically.
3x + 2y — 4 = 02x — 3y — 7 = 0
Shade the region bounded by the lines and the x – axis.
Sol :
The given equation is
3x + 2y = 4
and 2x – 3y = 7
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x+2y=4 or $y=\frac{4-3 x}{2}$
| x | 0 | $\frac{4}{3}=1.33$ | 2 |
| $y=\frac{4-3 x}{2}$ | 2 | 0 | -1 |
Now, table for 2x-3y=7 or $y=\frac{2 x-7}{3}$
Here, the lines intersecting at a point C i.e. (2, – 1).
2x — 3y + 6 = 0
Shade the region bounded by the lines and the x – axis.
Sol :
The given equation is
3x – 2y = 1
and 2x – 3y = – 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
| x | 0 | $\frac{7}{2}$ | 2 |
| $y=\frac{2 x-7}{3}$ | $\frac{-7}{3}$ | 0 | -1 |
Here, the lines intersecting at a point C i.e. (2, – 1).
Question 14 B
Solve the following system of linear equations graphically.
3x — 2y – 1 = 02x — 3y + 6 = 0
Shade the region bounded by the lines and the x – axis.
Sol :
The given equation is
3x – 2y = 1
and 2x – 3y = – 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x-2y=1 or $y=\frac{3 x-1}{2}$
| x | 0 | $\frac{1}{3}=0.33$ | 3 |
| $y=\frac{3 x-1}{2}$ | -0.5 | 0 | 4 |
Now, table for 2x-3y=-6 or $y=\frac{2 x+6}{3}$
Here, the lines intersecting at a point C, i.e. (3,4).
2x — y = 0
Sol :
The given equation is
2x + y = 6
and 2x – y = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x + y = 6 or y = 2x – 6
| x | 0 | -3 | 3 |
| $y=\frac{2 x+6}{3}$ | 2 | 0 | 4 |
Here, the lines intersecting at a point C, i.e. (3,4).
Question 15 A1
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and x – axis; also find the area of the
shaded region.
2x + y = 62x — y = 0
Sol :
The given equation is
2x + y = 6
and 2x – y = 0
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x + y = 6 or y = 2x – 6
| x | 0 | 3 | 1.5 |
| y=2x-6 | 6 | 0 | 3 |
Now, table for 2x – y = 0 or y = 2x
Here, the lines are intersecting at a point $C\left(\frac{3}{2}, 3\right)$
The coordinates of the vertices of Δ COB are $C\left(\frac{3}{2}, 3\right)$,O(0,0) and B(3,0).
| x | 1 | 2 | 1.5 |
| y=2x | 2 | 4 | 3 |
Here, the lines are intersecting at a point $C\left(\frac{3}{2}, 3\right)$
The coordinates of the vertices of Δ COB are $C\left(\frac{3}{2}, 3\right)$,O(0,0) and B(3,0).
Area $\mathrm{A}=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 3 \times 3=\frac{9}{2}$
=4.5 sq. unit
Question 15 A2
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and x – axis; also find the area of the
shaded region.
2x + 3y = —53x — 2y = 12
Sol :
The given equation is
2x + 3y = – 5
and 3x – 2y = 12
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x+3y=-5 or $y=\frac{-(2 x+5)}{3}$
| x | -1 | $-\frac{1}{2}$ | 2 |
| $y=\frac{-(2 x+5)}{3}$ | -1 | 0 | -3 |
Now, table for 3x-2y=12 or $\mathrm{y}=\frac{3 \mathrm{x}-12}{2}$
Here, the lines are intersecting at a point C(2,-3)
The coordinates of the vertices of ΔCBD are C(2,-3),$\mathrm{B}\left({-\frac{5}{2},0}\right)$ and D(4,0)
| x | 4 | 0 | 2 |
| $y=\frac{3 x-12}{2}$ | 0 | -6 | -3 |
Here, the lines are intersecting at a point C(2,-3)
The coordinates of the vertices of ΔCBD are C(2,-3),$\mathrm{B}\left({-\frac{5}{2},0}\right)$ and D(4,0)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times \frac{13}{2} \times 3$ $\because$ base $\left.=\left(4-\left(\frac{5}{2}\right)\right)=\frac{13}{2}\right.$
$=\frac{39}{4}$ sq. units
4x + 3y — 20 = 0
Sol :
The given equation is
4x – 3y = – 4
and 4x + 3y = 20
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Question 15 A3
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and x – axis; also find the area of the
shaded region.
4x — 3y + 4 = 04x + 3y — 20 = 0
Sol :
The given equation is
4x – 3y = – 4
and 4x + 3y = 20
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 4x-3y=-4 or $y=\frac{4 x+4}{3}$
| x | -1 | 0 | 2 |
| $y=\frac{4 x+4}{3}$ | 0 | $\frac{4}{3}$ | 4 |
Now, table for 4x+3y=20 or $y=\frac{20-4 x}{3}$
Here, the lines are intersecting at a point C(2,4)
| x | 5 | 0 | 2 |
| $y=\frac{20-4 x}{3}$ | 0 | $\frac{20}{3}$ | 4 |
The coordinates of the vertices of ΔCAD
are C(2,4), A(-1,0) and D(5,0)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 6 \times 4$ (∵, base = (5–(–1)) = 6)
= 12 sq.units
2x — y + 2 = 0
Sol :
The given equation is
2x + y = 6
and 2x – y = – 2
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
$=\frac{1}{2} \times 6 \times 4$ (∵, base = (5–(–1)) = 6)
= 12 sq.units
Question 15 A4
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and x – axis; also find the area of the
shaded region.
2x + y = 62x — y + 2 = 0
Sol :
The given equation is
2x + y = 6
and 2x – y = – 2
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 2x + y = 6 or y = 2x – 6
| x | 3 | 0 | 4 |
| y=2x-6 | 0 | 6 | -1.5 |
Now, table for 2x – y = – 2 or y = 2x + 2
| x | -1 | 0 | 1 |
| y=2x+2 | 0 | 2 | 4 |
Here, the lines are intersecting at a point C (1,4).
The coordinates of the vertices of ΔCAD
are C(1,4), A(3,0) and D( – 1,0)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2}\times 4 \times 4$ (base=[3-(-1)]=4)
= 8 sq.unitsQuestion 15 B1
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and y – axis. Also find the area of the
shaded region.
x — y = 12x + y = 8
Sol :
The given equation is
x – y = 1
and 2x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x – y = 1 or y = x – 1
| x | 1 | 0 | 3 |
| y=x-1 | 0 | -1 | 2 |
| x | 4 | 0 | 3 |
| y=2x-8 | 0 | 8 | 2 |
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBE are C(3,2), B(0, – 1) and E(0,8)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 9 \times 3$ (base=8-(-1)=9)
$=\frac{27}{2}$ sq. units
x — y — 1 = 0
Sol :
The given equation is
3x + y = 11
and x – y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x + y = 11 or y = 3x – 11
Now, table for x – y = 1or y = x – 1
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBE are C(3,2), B(0, 11) and E(0, – 1)
$=\frac{27}{2}$ sq. units
Question 15 B2
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and y – axis. Also find the area of the
shaded region.
3x + y — 11 = 0x — y — 1 = 0
Sol :
The given equation is
3x + y = 11
and x – y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x + y = 11 or y = 3x – 11
| x | $\frac{11}{3}$ | 0 | 3 |
| $y=\frac{1-x}{2}$ | 0 | 11 | 2 |
| x | 1 | 0 | 3 |
| y=x-1 | 0 | -1 | 2 |
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBE are C(3,2), B(0, 11) and E(0, – 1)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 12 \times 3$ (base=[11-(-1)]=12)
= 18 sq.units
2x — y — 4 = 0
Sol :
The given equation is
x + 2y = 7
and 2x – y = 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
= 18 sq.units
Question 15 B3
Solve the following pair of linear equations graphically and shade the
region bounded by these lines and y – axis. Also find the area of the
shaded region.
x + 2y — 7 = 02x — y — 4 = 0
Sol :
The given equation is
x + 2y = 7
and 2x – y = 4
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x+2y=7 or $y=\frac{7-x}{2}$
Now, table for 2x-y=4 or y=2x-4
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBD are C(3,2),B(0, $\frac{7}{2}$) and E(0, – 4)
Area $=\frac{1}{2} \times$ base $\times$ height
| x | 7 | 0 | 4 |
| $y=\frac{7-x}{2}$ | 0 | $\frac{7}{2}$ | $\frac{-3}{2}$ |
| x | 2 | 0 | 3 |
| y=2x-4 | 0 | -4 | 2 |
Here, the lines are intersecting at a point C (3,2).
The coordinates of the vertices of ΔCBD are C(3,2),B(0, $\frac{7}{2}$) and E(0, – 4)
Area $=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times \frac{15}{2} \times 3$ (base=$\left(\frac{7}{2}\right)-(-4)=\frac{15}{2}$)
$=\frac{45}{4} \mathrm{sq} \cdot$ units
3x + 2y = 14
Sol :
The given equation is
4x – y = 4
and 3x + 2y = 14
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 4x – y = 4 or y = 4x – 4
Now, table for 3x+2y=14 or $y=\frac{14-3 x}{2}$
Here, the lines are intersecting at point C(2,4).
2x + y = 8
Sol :
The given equation is
x – y = 1
and 2x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x – y = 1 or y = x – 1
Now, table for 2x + y = 8 or y = 8 – 2x
Here, the lines are intersecting at point C(3, 2).
Also find the vertices of the triangle formed by the two lines and x – axis.
Sol :
The given equation is
5x – 6y = – 30
and 5x + 4y = 20
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
$=\frac{45}{4} \mathrm{sq} \cdot$ units
Question 16 A
Solve the following system of linear equations graphically. Also shade
the region bounded by the lines and y – axis.
4x — y = 43x + 2y = 14
Sol :
The given equation is
4x – y = 4
and 3x + 2y = 14
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 4x – y = 4 or y = 4x – 4
| x | 0 | 1 | 2 |
| y=4x-4 | -4 | 0 | 4 |
| x | 0 | $\frac{14}{3}$ | 2 |
| $y=\frac{14-3 x}{2}$ | 7 | 0 | 4 |
Here, the lines are intersecting at point C(2,4).
Question 16 B
Solve the following system of linear equations graphically. Also shade
the region bounded by the lines and y – axis.
x — y = 12x + y = 8
Sol :
The given equation is
x – y = 1
and 2x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for x – y = 1 or y = x – 1
| x | 1 | 0 | 3 |
| y=x-1 | 0 | -1 | 2 |
| x | 4 | 0 | 3 |
| y=8-2x | 0 | 8 | 2 |
Here, the lines are intersecting at point C(3, 2).
Question 17
Solve the following system of linear equations graphically:
5x — 6y + 30 = 0; 5x + 4y — 20 = 0Also find the vertices of the triangle formed by the two lines and x – axis.
Sol :
The given equation is
5x – 6y = – 30
and 5x + 4y = 20
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 5x-6y=-3 or $\mathrm{y}=\frac{5 \mathrm{x}+3}{6}$
Now, table for 5x+4y=20 or $y=\frac{20-5 x}{4}$
Here, the lines are intersecting at point C (0,5).
The coordinates of the vertices of ∆ACD are A( – 6,0), C(0,5)and D(4,0)
Sol :
The given equation is
3x – y = – 9
and 3x + 4y = 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x – y = – 9 or y = 3x + 9
Now, table for 3x+4y=6 or $y=\frac{6-3 x}{4}$
Here, the lines are intersecting at point C ( – 2, 3).
The coordinates of the vertices of ∆ACE are A( – 3,0), C( – 2,3)and E(2,0)
Sol :
The given equation is
3x – 4y = – 9
and 3x + y = 9
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
| x | -6 | 6 | 0 |
| $y=\frac{5+3 x}{6}$ | 0 | 10 | 5 |
| x | 4 | -4 | 0 |
| $y=\frac{20-5 x}{4}$ | 0 | 10 | 5 |
Here, the lines are intersecting at point C (0,5).
The coordinates of the vertices of ∆ACD are A( – 6,0), C(0,5)and D(4,0)
Question 18
Draw the graphs of the equations 3x — y + 9 = 0 and 3x + 4y — 6 =
0.
Also determine the vertices of the triangle formed by the lines and the x –
axis.Sol :
The given equation is
3x – y = – 9
and 3x + 4y = 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x – y = – 9 or y = 3x + 9
| x | -3 | 0 | -2 |
| y=3x+9 | 0 | 9 | 3 |
| x | 0 | 2 | -2 |
| $y=\frac{6-3 x}{4}$ | $\frac{3}{2}$ | 0 | 3 |
Here, the lines are intersecting at point C ( – 2, 3).
The coordinates of the vertices of ∆ACE are A( – 3,0), C( – 2,3)and E(2,0)
Question 19
Draw the graphs of the following equations 3x — 4y + 6 = 0; 3x + y — 9 =
0.
Also, determine the coordinates of the vertices of the triangle formed
by these lines and the x – axis.Sol :
The given equation is
3x – 4y = – 9
and 3x + y = 9
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 3x-4y=-9 or $y=\frac{3 x+9}{4}$
Now, table for 3x+y=9 or y=9-3x
Here, the lines are intersecting at point C (2, 3).
The coordinates of the vertices of ∆ACD are A( – 2,0), C(2,3)and D(3,0)
2y – x = 8
5y – x = 14
y – 2x = 1
Sol :
The given equation is
– x + 2y = 8
– x + 5y = 14
and – 2x + y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
| x | -2 | 0 | 4 |
| $y=\frac{3 x+9}{4}$ | 0 | 1.5 | 3 |
| x | 3 | 0 | 2 |
| y=9-3x | 0 | 9 | 3 |
Here, the lines are intersecting at point C (2, 3).
The coordinates of the vertices of ∆ACD are A( – 2,0), C(2,3)and D(3,0)
Question 20 A
Use a single graph paper and draw the graph of the following equations. Obtain the vertices of the triangle so formed:2y – x = 8
5y – x = 14
y – 2x = 1
Sol :
The given equation is
– x + 2y = 8
– x + 5y = 14
and – 2x + y = 1
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for -x+2y=8 or $y=\frac{8+x}{2}$
| x | 0 | -8 |
| $\mathrm{y}=\frac{3x+9}{4}$ | 4 | 0 |
Now, table for -x+5y=14 or $\mathrm{y}=\frac{14+\mathrm{x}}{5}$
| x | 0 | -14 |
| $\mathrm{y}=\frac{14+\mathrm{x}}{5}$ | $\frac{14}{5}$ | 0 |
Now, table for -2x+y=1 or y=1+2x
| x | 0 | $\frac{-1}{2}$ |
| y=1+2x | 1 | 0 |
The coordinates of the vertices of ∆PQR are P( – 4, 2), Q(2, 5)and R(1, 3)
Question 20 B
Use a single graph paper and draw the graph of the following equations. Obtain the vertices of the triangle so formed:y = x
Y = 2x
x + y = 6
Sol :
The given equation is
y = x
y = 2x
and x + y = 6
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for y = x
| x | 1 | -1 |
| y=x | 1 | -1 |
Now, table for y = 2x
| x | 1 | -1 |
| y=9-3x | 2 | -2 |
Now, table for x + y = 6 or y = 6 – x
| x | 0 | 6 |
| y=6-x | 6 | 0 |
The coordinates of the vertices of ∆PQR are P(0, 0), Q(2, 4)and R(3, 3)
Question 20 C
Use a single graph paper and draw the graph of the following equations.
Obtain the vertices of the triangle so formed:
y = x3y = x
x + y = 8
Sol :
The given equation is
y = x
3y = x
and x + y = 8
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for y = x
| x | 1 | -1 |
| y=x | 1 | -1 |
Now, table for 3y = x
| x | 3 | -3 |
| $\mathrm{y}=\frac{\mathrm{x}}{3}$ | 1 | -1 |
Now, table for x + y = 8 or y = 8 – x
| x | 0 | 8 |
| y=8-x | 8 | 0 |
The coordinates of the vertices of ∆PQR are P(0, 0), Q(4, 4)and R(6, 2)
Question 21 A
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
2x + 3y = 7, (a + b) x + (2a – b) y = 3(a + b + 1)Sol :
Given, pair of equations
2x + 3y = 7
and (a + b)x + (2a – b)y = 3(a + b + 1)
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 7
and a2 = (a + b), b2 = (2a – b) and c2 = – (a + b + 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{2}{a+b}$,$\frac{b_{1}}{b_{2}}=\frac{3}{2 a-b}$ and $\frac{c_{1}}{c_{2}}=\frac{-7}{-3(a+b+1)}$
$\begin{matrix}\frac{2}{a+b}&=\frac{3}{2 a-b}&=\frac{-7}{-3(a+b+1)}\\\text{I}&\text{II}&\text{III}\end{matrix}$
On taking I and II terms, we get
$\frac{2}{a+b}=\frac{3}{2 a-b}$
⇒ 2(2a – b) = 3(a + b)
⇒ 4a – 2b = 3a + 3b
⇒ 4a – 3a – 3b – 2b = 0
⇒ a – 5b = 0 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{2}{a+b}=\frac{7}{3(a+b+1)}$
⇒ 6(a + b + 1) = 7(a + b)
⇒ 6a + 6b + 6 = 7a + 7b
⇒ 6a – 7a + 6b – 7b = – 6
⇒ – a – b = – 6
⇒ a + b = 6 …(2)
Solving eqn (1) and (2), we get
$\frac{2}{a+b}=\frac{3}{2 a-b}$
⇒ 2(2a – b) = 3(a + b)
⇒ 4a – 2b = 3a + 3b
⇒ 4a – 3a – 3b – 2b = 0
⇒ a – 5b = 0 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{2}{a+b}=\frac{7}{3(a+b+1)}$
⇒ 6(a + b + 1) = 7(a + b)
⇒ 6a + 6b + 6 = 7a + 7b
⇒ 6a – 7a + 6b – 7b = – 6
⇒ – a – b = – 6
⇒ a + b = 6 …(2)
Solving eqn (1) and (2), we get
$\begin{array}a-5b&=0\\a+b&=6\\-\phantom{a}-\phantom{b}&\phantom{=}-\phantom{6}\\ \hline -6b&=-6\end{array}$
⇒ b = 1
Now, substituting the value of b in eqn (2), we get
⇒ a + b = 6
⇒ a + 1 = 6
⇒ a = 5
Question 21 B
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
(2a – 1) x – 3y = 5, 3x + (b – 2) y = 3Sol :
Given, pair of equations
(2a – 1)x – 3y = 5
and 3x + (b – 2)y = 3
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = (2a – 1), b1 = – 3 and c1 = – 5
and a2 = 3, b2 = b – 2 and c2 = 3
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{2 a-1}{3}$,$\frac{b_{1}}{b_{2}}=\frac{-3}{b-2}$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-3}=\frac{5}{3}$
∴$\begin{matrix}\frac{2 a-1}{3}&=\frac{-3}{b-2}&=\frac{5}{3}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{2 a-1}{3}=\frac{5}{3}$
⇒ 3(2a – 1) = 15
⇒ 6a – 3 = 15
⇒ 6a = 15 + 3
$\Rightarrow \mathrm{a}=\frac{18}{6}=3$
On taking II and III terms, we get
$\Rightarrow \frac{-3}{b-2}=\frac{5}{3}$
⇒ – 9 = 5(b – 2)
⇒ 5b – 10 = – 9
⇒ 5b = – 9 + 10
$\Rightarrow \mathbf{b}=\frac{1}{5}$
Question 21 C
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
kx + 3y – (k – 3) = 0, 12x + ky – k = 0Sol :
Given, pair of equations
kx + 3y – (k – 3) = 0
and 12x + ky – k = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = k, b1 = 3 and c1 = – (k – 3)
and a2 = 12, b2 = k and c2 = – k
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{k}{12}, \frac{b_{1}}{b_{2}}=\frac{3}{k}$ and $\frac{c_{1}}{c_{2}}=\frac{-(k-3)}{-k}=\frac{(k-3)}{k}$
∴$\begin{matrix}\frac{k}{12}&=\frac{3}{k}&=\frac{(k-3)}{k}\\\text{I}&\text{II}&\text{III}\end{matrix}$…(1)
$\frac{k}{12}=\frac{3}{k}$
⇒ k2 = 36
⇒ k = √36
⇒ k = ±6
But k = – 6 not satisfies the last two terms of eqn (1)
On taking II and III terms, we get
$\frac{3}{\mathrm{k}}=\frac{(\mathrm{k}-3)}{\mathrm{k}}$
⇒ 3k = k(k – 3)
⇒ 3k = k2 – 3k
⇒ k2 – 3k – 3k = 0
⇒ k(k – 6) = 0
⇒ k = 0 and 6
Which satisfies the last two terms of eqn (1)
Hence, the required value of k = 0, 6
Question 21 D
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
3x + 4y = 12, (a + b) x + 2 (a – b) y = 5a – 1Sol :
Given, pair of equations
3x + 4y = 12
and (a + b)x + 2(a – b)y = 5a – 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 4 and c1 = – 12
and a2 = (a + b), b2 = 2(a – b) and c2 = – (5a – 1) = 1 + 5a
$a_{1}=3, \quad b_{1}=4, c_{1}=-12$
and $a_{2}=(a+b), b_{2}=2(a-b), c_{2}=-(5 a-1)=1+5 a$
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{3}{a+b}$, $\frac{b_{1}}{b_{2}}=\frac{4}{2(a-b)}$ and $\frac{c_{1}}{c_{2}}=\frac{-12}{-(5 a-1)}=\frac{12}{(5 a-1)}$
∴ $\begin{matrix}\frac{3}{a+b}&=\frac{4}{2(a-b)}&=\frac{-12}{-(5 a-1)}\\\text{I}&\text{II}&\text{III}\end{matrix}$
On taking I and II terms, we get
$\frac{3}{a+b}=\frac{4}{2(a-b)}$
⇒ 6(a – b) = 4(a + b)
⇒ 6a – 6b = 4a + 4b
⇒ 6a – 4a – 6b – 4b = 0
⇒ 2a – 10b = 0
⇒ a – 5b = 0 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{3}{a+b}=\frac{12}{5 a-1}$
⇒ 3(5a – 1) = 12(a + b)
⇒ 15a – 3 = 12a + 12b
⇒ 15a – 12a – 12b = 3
⇒ 3a – 12b = 3
⇒ a – 4b = 1 …(2)
$\frac{3}{a+b}=\frac{4}{2(a-b)}$
⇒ 6(a – b) = 4(a + b)
⇒ 6a – 6b = 4a + 4b
⇒ 6a – 4a – 6b – 4b = 0
⇒ 2a – 10b = 0
⇒ a – 5b = 0 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{3}{a+b}=\frac{12}{5 a-1}$
⇒ 3(5a – 1) = 12(a + b)
⇒ 15a – 3 = 12a + 12b
⇒ 15a – 12a – 12b = 3
⇒ 3a – 12b = 3
⇒ a – 4b = 1 …(2)
Solving eqn (1) and (2), we get
$\begin{aligned}a-5 b&=0\\a-4 b&=1\\-\phantom{a}+\phantom{4 b}&\phantom{=}-\phantom{1}\\ \hline -b&=-1\end{aligned}$
⇒ b = 1Now, substituting the value of b in eqn (2), we get
⇒ a – 4b = 1
⇒ a – 4 = 1
⇒ a = 1 + 4
⇒ a = 5
Question 21 E
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
(a – 1) x + 3y = 2, 6x + (1 – 2b) y = 6Sol :
Given, pair of equations
(a – 1)x + 3y = 2
and 6x + (1 – 2b)y = 6
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = (a – 1), b1 = 3 and c1 = – 2
and a2 = 6, b2 = 1 – 2b and c2 = – 6
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{a-1}{6}$,$\frac{b_{1}}{b_{2}}=\frac{3}{1-2 b}$ and $\frac{c_{1}}{c_{2}}=\frac{-2}{-6}=\frac{1}{3}$
∴$\begin{matrix}\frac{a-1}{6}&=\frac{3}{1-2 b}&=\frac{1}{3}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{a-1}{6}=\frac{1}{3}$
⇒ 3(a – 1) = 6
⇒ 3a – 3 = 6
⇒ 3a = 6 + 3
$\Rightarrow \mathrm{a}=\frac{9}{3}=3$
On taking II and III terms, we get
$\Rightarrow \frac{3}{1-2 b}=\frac{1}{3}$
⇒ 9 = 1 – 2b
⇒ – 2b = 9 – 1
⇒ – 2b = 8
⇒ b = – 4
Question 21 F
Find the values of a and b for which the following system of linear
equations has infinitely many solutions:
2x + 3y = 7, (a + b + 1) x + (a + 2b + 2) y = 4 (a + b) + 1Sol :
Given, pair of equations
2x + 3y = 7
and (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 7
and a2 = (a + b + 1), b2 = (a + 2b + 2) and c2 = – {4(a + b) + 1}
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{2}{a+b+1}, \frac{b_{1}}{b_{2}}=\frac{3}{a+2 b+2}$ and $\frac{c_{1}}{c_{2}}=\frac{-7}{-\{4(a+b)+1\}}$
∴$\begin{matrix}\frac{2}{a+b+1}&=\frac{3}{a+2 b+2}&=\frac{7}{\{4(a+b)+1\}}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{2}{a+b+1}=\frac{3}{a+2 b+2}$
⇒ 2(a + 2b + 2) = 3(a + b + 1)
⇒ 2a + 4b + 4 = 3a + 3b + 3
⇒ 2a – 3a – 3b + 4b = 3 – 4
⇒ – a + b = – 1
⇒ a – b = 1 …(1)
On taking I and III terms, we get
$\Rightarrow \frac{2}{a+b+1}=\frac{7}{\{4(a+b)+1\}}$
⇒ 2{4(a + b) + 1)} = 7(a + b + 1)
⇒ 2(4a + 4b + 1) = 7a + 7b + 7
⇒ 8a – 7a + 8b – 7b = – 2 + 7
⇒ a + b = 5 …(2)
Solving eqn (1) and (2), we get
$\begin{aligned}a-b&=1\\a+b&=5\\ \hline 2a&=6\end{aligned}$
⇒ a = 3
Now, substituting the value of a in eqn (1), we get
⇒ a – b = 1
⇒ 3 – b = 1
⇒ b = 2
Sol :
Given, pair of equations
ax + 3y = a – 2
and 12x + ay = a
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = a, b1 = 3 and c1 = – (a – 2)
and a2 = 12, b2 = a and c2 = – a
For no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
⇒ a = 3
Now, substituting the value of a in eqn (1), we get
⇒ a – b = 1
⇒ 3 – b = 1
⇒ b = 2
Question 22 A
For what value of a, the following system of linear equations has no
solutions:
ax + 3y = a – 2, 12x + ay = aSol :
Given, pair of equations
ax + 3y = a – 2
and 12x + ay = a
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = a, b1 = 3 and c1 = – (a – 2)
and a2 = 12, b2 = a and c2 = – a
For no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴$\begin{matrix}\frac{\mathrm{a}}{12}&=\frac{3}{\mathrm{a}} &\neq \frac{-(\mathrm{a}-2)}{-\mathrm{a}}\\\text{I}&\text{II}&\text{III}\end{matrix}$
On taking I and II terms, we get
$\frac{a}{12}=\frac{3}{a}$
⇒ a2 = 36
⇒ a = √36
⇒ a = ±6
Question 22 B
For what value of a, the following system of linear equations has no
solutions:
x + 2y = 5, 3x + ay + 15 = 0Sol :
Given, pair of equations
x + 2y = 5
and 3x + ay + 15 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1, b1 = 2 and c1 = – 5
and a2 = 3, b2 = a and c2 = 15
For no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴$\begin{matrix}\frac{1}{3}&=\frac{2}{a} &\neq \frac{-5}{15}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{1}{3}=\frac{2}{a}$
$\Rightarrow \mathbf{a}=\mathbf{6}$
Question 22 C
For what value of a, the following system of linear equations has no
solutions:
3x + y = 1, (2a – 1) x + (a – 1) y = 2a + 1Sol :
Given, pair of equations
3x + y = 1
and (2a – 1)x + (a – 1)y = 2a + 1
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = 1 and c1 = – 1
and a2 = (2a – 1), b2 = (a – 1) and c2 = – (2a + 1)
For no solutions,
$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$
∴$\begin{matrix}\frac{3}{2 a-1}&=\frac{1}{a-1}& \neq \frac{-1}{-(2 a+1)}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{3}{2 a-1}=\frac{1}{a-1}$
⇒ 3(a – 1) = 2a – 1
⇒ 3a – 3 = 2a – 1
⇒ 3a – 2a = – 1 + 3
⇒ a = 2
Question 22 D
For what value of a, the following system of linear equations has no
solutions:
(3a + 1) x + 3y – 2 = 0, (a2 + 1) x + (a – 2) y – 5 =
0Sol :
Given, pair of equations
(3a + 1)x + 3y – 2 = 0
and (a2 + 1)x + (a – 2)y – 5 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3a + 1, b1 = 3 and c1 = – 2
and a2 = a2 + 1, b2 = a – 2 and c2 = – 5
For no solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴$\begin{matrix}\frac{3 a+1}{\left(a^{2}+1\right)}&=\frac{3}{a-2}& \neq \frac{-2}{-5}\\\text{I}&\text{II}&\text{III}\end{matrix}$
$\frac{3 a+1}{\left(a^{2}+1\right)}=\frac{3}{a-2}$
⇒ (3a + 1)(a – 2) = 3(a2 + 1)
⇒ 3a2 – 6a + a – 2 = 3a2 + 3
⇒ – 5a = 2 + 3
⇒ a = – 1
Question 23 A
For what value of c, the following system of linear equations has
infinite number of solutions:
cx + 3y – (c – 3) = 0, 12x + cy – c = 0Sol :
Given, pair of equations
cx + 3y – (c – 3) = 0
and 12x + cy – c = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = c, b1 = 3 and c1 = – (c – 3)
and a2 = 12, b2 = c and c2 = – c
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{c}{12}$,$\frac{b_{1}}{b_{2}}=\frac{3}{c}$ and $\frac{c_{1}}{c_{2}}=\frac{-(c-3)}{-c}=\frac{c-3}{c}$
Here,$\frac{a_{1}}{a_{2}}=\frac{c}{12}$,$\frac{b_{1}}{b_{2}}=\frac{3}{c}$ and $\frac{c_{1}}{c_{2}}=\frac{-(c-3)}{-c}=\frac{c-3}{c}$
∴$\begin{matrix}\frac{c}{12}&=\frac{3}{c}&=\frac{c-3}{c}\\\text{I}&\text{II}&\text{III}\end{matrix}$…(1)
$\frac{c}{12}=\frac{3}{c}$
⇒ c2 = 36
⇒ c = √36
⇒ c = ±6
But c = – 6 not satisfies the eqn (1)
Hence, the required value of c = 6.
Question 23 B
For what value of c, the following system of linear equations has
infinite number of solutions:
2x + 3y = 2, (c + 2) x + (2c + 1) y = 2 (c – 1)Sol :
Given, pair of equations
2x + 3y = 2
and (c + 2)x + (2c + 1)y = 2(c – 1)
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 2, b1 = 3 and c1 = – 2
and a2 = c + 2, b2 = 2c + 1 and c2 = – 2(c – 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, $\frac{a_{1}}{a_{2}}=\frac{2}{c+2}$ ,$\frac{b_{1}}{b_{2}}=\frac{3}{2 c+1}$ and
$\frac{c_{1}}{c_{2}}=\frac{-2}{-2(c-1)}=\frac{1}{c-1}$
On taking I and III terms, we get
$\frac{2}{c+2}=\frac{3}{2 c+1}$
⇒ 2(2c + 1) = 3(c + 2)
⇒ 4c + 2 = 3c + 6
⇒ 4c – 3c = 6 – 2
⇒ c = 4
Hence, the required value of c = 4.
Sol :
The pair of equations are:
x + (c + 1) y = 5
(c + 1) x + 9y = 8c – 1
These equations can be written as:
x + (c + 1) y – 5 = 0
(c + 1) x + 9y – (8c – 1) = 0
On comparing the given equation with standard form i.e.
a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1 , b1 = c + 1 , c1 = –5
a2 = c + 1, b2 = 9 , c2 = –(8c – 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\begin{matrix}\frac{2}{c+2}&=\frac{3}{2 c+1}&=\frac{1}{c-1}\\I&II&III\end{matrix}$…(1)
$\frac{2}{c+2}=\frac{3}{2 c+1}$
⇒ 2(2c + 1) = 3(c + 2)
⇒ 4c + 2 = 3c + 6
⇒ 4c – 3c = 6 – 2
⇒ c = 4
Hence, the required value of c = 4.
Question 23 C
For what value of c, the following system of linear equations has
infinite number of solutions:
x + (c + 1) y = 5, (c + 1) x + 9y = 8c – 1Sol :
The pair of equations are:
x + (c + 1) y = 5
(c + 1) x + 9y = 8c – 1
These equations can be written as:
x + (c + 1) y – 5 = 0
(c + 1) x + 9y – (8c – 1) = 0
On comparing the given equation with standard form i.e.
a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 1 , b1 = c + 1 , c1 = –5
a2 = c + 1, b2 = 9 , c2 = –(8c – 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
So,
$\frac{a_{1}}{a_{2}}=\frac{1}{c+1}$ , $\frac{b_{1}}{b_{2}}=\frac{c+1}{9}$, $\frac{c_{1}}{c_{2}}=\frac{-5}{-8 c+1}$
So,
$\begin{matrix}\frac{1}{c+1}&=\frac{c+1}{9}&=\frac{-5}{-8 c+1}\\I&II&III\end{matrix}$
⇒ From (I) and (II)
$\frac{1}{c+1}=\frac{c+1}{9}$
⇒ 9 = (c+1)2
⇒ 9 = c2 + 1 + 2c
⇒ 9 – 1 = c2 + 2c
⇒ 8 = c2 + 2c
⇒ c2 + 2c – 8 = 0
Factorize by splitting the middle term,
c2 + 4c – 2c – 8 = 0
⇒ c ( c + 4 ) – 2 ( c + 4) = 0
⇒ (c+4) (c–2) = 0
⇒ c = –4, c = 2
From (II) and (III)
$\frac{c+1}{9}=\frac{-5}{-8 c+1}$
⇒ (c+1)(–8c=1) = –5 × 9
⇒ –8c2 + c – 8c + 1 = –45
⇒ –8c2 + c – 8c + 1 + 45 = 0
⇒ –8c2 – 7c + 46 = 0
⇒ 8c2 + 7c – 46 = 0
⇒ 8c2 – 16c + 23c –46 = 0
⇒ 8c ( c–2) + 23 ( c–2) = 0
⇒ (8c+23) ( c–2) = 0
⇒ c = –23/8 and c = 2
From (I) and (III)
$\frac{1}{c+1}=\frac{-5}{-8 c+1}$
⇒ –8c+1 = –5(c+1)
⇒ –8c + 1 = –5c – 5
⇒ –8c + 5c = –5 –1
⇒ –3c = –6
⇒ c = 2
So the value of c = 2.
Question 23 D
For what value of c, the following system of linear equations has
infinite number of solutions:
(c – 1) x – y = 5, (c + 1) x + (1 – c) y = 3c + 1Sol :
The pair of equations are:
(c – 1) x – y = 5
(c + 1) x + (1 – c) y = 3c + 1
These equations can be written as:
(c – 1) x – y – 5 = 0
(c + 1) x + (1 – c) y –( 3c + 1)
On comparing the given equation with standard form i.e.
a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = c – 1 , b1 = –1 , c1 = –5
a2 = c + 1, b2 = 1 – c , c2 = –(3c + 1)
For infinitely many solutions,
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
So,
$\frac{a_{1}}{a_{2}}=\frac{c-1}{c+1}$, $\frac{b_{1}}{b_{2}}=\frac{-1}{1-c}$, $\frac{c_{1}}{c_{2}}=\frac{-5}{-3 c-1}$
So,
$\frac{c-1}{c+1}=\frac{-1}{1-c}$
⇒ (c–1)(1–c) = – (c+1)
⇒ c – c2 –1 + c = –c – 1
⇒ c – c2 –1 + c + c + 1 = 0
⇒ 3c – c2 = 0
⇒ c ( 3 – c ) = 0
⇒ c = 0 , c = 3
From (II) and (III)
$\frac{-1}{1-c}=\frac{-5}{-3 c-1}$
⇒ –(–3c–1) = –5(1–c)
⇒ 3c + 1 = –5 + 5c
⇒ 3c + 1 + 5 – 5c =0
⇒ 6 – 2c = 0
⇒ 6 = 2c
⇒ c = 3
From (I) and (III)
$\frac{c-1}{c+1}=\frac{-5}{-3 c-1}$
⇒ (c–1)(–3c–1) = –5 (c+1)
⇒ –3c2 – c + 3c + 1 = –5c – 5
⇒ –3c2 – c + 3c + 1 + 5c + 5 = 0
⇒ –3c2 + 7c + 6 = 0
⇒ 3c2 – 7c – 6 = 0
⇒ 3c2 – 9c + 2c – 6 = 0
⇒ 3c ( c–3) + 2 (c–3) = 0
⇒ (3c+2) ( c–3) = 0
⇒ c = –2/3 and c = 3
Hence the value of c is 3.
Sol :
The given equation is
4x – 5y = 20
and 3x + 5y = 15
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
So,
$\begin{matrix}\frac{c-1}{c+1}&=\frac{-1}{1-c}&=\frac{-5}{-3 c-1}\\I&II&III\end{matrix}$
$\frac{c-1}{c+1}=\frac{-1}{1-c}$
⇒ (c–1)(1–c) = – (c+1)
⇒ c – c2 –1 + c = –c – 1
⇒ c – c2 –1 + c + c + 1 = 0
⇒ 3c – c2 = 0
⇒ c ( 3 – c ) = 0
⇒ c = 0 , c = 3
From (II) and (III)
$\frac{-1}{1-c}=\frac{-5}{-3 c-1}$
⇒ –(–3c–1) = –5(1–c)
⇒ 3c + 1 = –5 + 5c
⇒ 3c + 1 + 5 – 5c =0
⇒ 6 – 2c = 0
⇒ 6 = 2c
⇒ c = 3
From (I) and (III)
$\frac{c-1}{c+1}=\frac{-5}{-3 c-1}$
⇒ (c–1)(–3c–1) = –5 (c+1)
⇒ –3c2 – c + 3c + 1 = –5c – 5
⇒ –3c2 – c + 3c + 1 + 5c + 5 = 0
⇒ –3c2 + 7c + 6 = 0
⇒ 3c2 – 7c – 6 = 0
⇒ 3c2 – 9c + 2c – 6 = 0
⇒ 3c ( c–3) + 2 (c–3) = 0
⇒ (3c+2) ( c–3) = 0
⇒ c = –2/3 and c = 3
Hence the value of c is 3.
Question 24
Solve the following system of equations graphically. Also determine the
vertices of the triangle formed by the lines and y – axis.
4x – 5y – 20 = 0, 3x + 5y – 15 = 0Sol :
The given equation is
4x – 5y = 20
and 3x + 5y = 15
Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.
Table for 4x-5y=20 or $\mathrm{y}=\frac{4 \mathrm{x}-20}{5}$
| x | 0 | 5 |
| $y=\dfrac{4x-20}{5}$ | -4 | 0 |
Now, table for 3x+5y=15 or $y=\frac{15-3 x}{5}$
| x | 0 | 5 |
| $y=\dfrac{15-3x}{5}$ | 3 | 0 |
Here, the lines are intersecting at point C (5, 0).
The coordinates of the vertices of ∆ABC are A(0, – 4), B(0, 3)and C(5,0)
Question 25 A
Find the value of a for which the following system of equations has
unique solution:
ax + 2y = 5, 3x + y = 1Sol :
Given, pair of equations
ax + 2y = 5
and 3x + y = 1
On comparing the given equation with standard form
i.e. a1 x +
b1y + c1 = 0 and a2 x +
b2y + c2 = 0, we get
a1 = a, b1 = 2 and c1 = – 5
and a2 = 3, b2 = 1 and c2 = – 1
and a2 = 3, b2 = 1 and c2 = – 1
For unique solutions,
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{a}{3}$, $\frac{b_{1}}{b_{2}}=\frac{2}{1}$
∴$\begin{array}{ll} \frac{a}{3} \neq &\frac{2}{1}\\\text{I}& \text{II}\end{array}$On taking I and II terms, we get
$\frac{a}{3} \neq \frac{2}{1}$
⇒ a≠6
Thus, given lines have a unique solution for all real values of a, except 6.
Sol :
Sol :
Given, pair of equations
3x + 2y = 4
and ax – y = 3
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = – 2 and c1 = – 4
and a2 = a, b2 = – 1 and c2 = – 3
$\frac{a}{3} \neq \frac{2}{1}$
⇒ a≠6
Thus, given lines have a unique solution for all real values of a, except 6.
Question 25 B
Find the value of a for which the following system of equations has
unique solution:
9x + py – 1 = 0, 3x + 4y – 2 = 0Sol :
Given, pair of equations
9x + py – 1 = 0
and 3x + 4y – 2 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 9, b1 = p and c1 = – 1
and a2 = 3, b2 = 4 and c2 = – 2
9x + py – 1 = 0
and 3x + 4y – 2 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 9, b1 = p and c1 = – 1
and a2 = 3, b2 = 4 and c2 = – 2
For unique solutions,
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{9}{3}=3$, $\frac{b_{1}}{b_{2}}=\frac{p}{4}$
∴ 3 ≠ p/4
I II
Here,$\frac{a_{1}}{a_{2}}=\frac{9}{3}=3$, $\frac{b_{1}}{b_{2}}=\frac{p}{4}$
∴ 3 ≠ p/4
I II
On taking I and II terms, we get
p ≠ 12
Thus, given lines have a unique solution for all real values of p, except 12.
p ≠ 12
Thus, given lines have a unique solution for all real values of p, except 12.
Question 25 C
Find the value of a for which the following system of equations has
unique solution:
3x + 2y = 4, ax – y = 3Sol :
Given, pair of equations
3x + 2y = 4
and ax – y = 3
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 3, b1 = – 2 and c1 = – 4
and a2 = a, b2 = – 1 and c2 = – 3
For unique solutions,
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{3}{a}$, $\frac{b_{1}}{b_{2}}=-\frac{2}{1}$
∴ $\begin{array}{ll}\frac{3}{a} \neq &\frac{2}{-1}\\\text{I}&\text{II}\end{array}$
On taking I and II terms, we get
$\frac{3}{a} \neq \frac{2}{-1}$
$\Rightarrow a \neq-\frac{3}{2}$
Thus, given lines have a unique solution for all real values of a, except $-\frac{3}{2}$.
Sol :
Given, pair of equations
4x + py + 8 = 0
and 2x + 2y + 2 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 4, b1 = p and c1 = 8
and a2 = 2, b2 = 2 and c2 = 2
$\frac{3}{a} \neq \frac{2}{-1}$
$\Rightarrow a \neq-\frac{3}{2}$
Thus, given lines have a unique solution for all real values of a, except $-\frac{3}{2}$.
Question 25 D
Find the value of a for which the following system of equations has
unique solution:
4x + py + 8 = 0, 2x + 2y + 2 = 0Sol :
Given, pair of equations
4x + py + 8 = 0
and 2x + 2y + 2 = 0
On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get
a1 = 4, b1 = p and c1 = 8
and a2 = 2, b2 = 2 and c2 = 2
For unique solutions,
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
Here,$\frac{a_{1}}{a_{2}}=\frac{4}{2}=2$, $\frac{b_{1}}{b_{2}}=\frac{p}{2}$
∴ $\begin{array}{ll}2 \neq &\frac{p}{2}\\\text{I}&\text{I I}\end{array}$
On taking I and II terms, we get
$2 \neq \frac{p}{2}$
⇒ p≠4
Thus, given lines have a unique solution for all real values of p, except 4.
Let the number of boys = x
and the number of girls = y
$2 \neq \frac{p}{2}$
⇒ p≠4
Thus, given lines have a unique solution for all real values of p, except 4.
Question 26
10 students of class X took part in mathematics quiz. If the number of
girls is 4 more than the number of boys, find the number of boys and girls
who took part in the quiz.
Sol :Let the number of boys = x
and the number of girls = y
Now, table for x + y – 10 = 0
| x | 0 | 10 |
| y=10-x | 10 | 0 |
Now, table for x – y + 4 = 0
| x | 0 | -4 |
| y | 4 | 0 |
On plotting points on a graph paper and join them to get a straight line representing x + y – 10 = 0.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – y + 4 = 0.
∴ x = 3, y = 7 is the solution of the pair of linear equations.
Hence, the required number of boys is 3 and girls is 7.
Question 27
Form the pair of linear equations in the following problems and find
their solutions graphically. Two years ago, a father was five times as old
as his son. Two years later, his age will be 8 more than three times the
age of the son.
Sol :Let the present age of father = x year
and the present age of his son = y year
Two years ago,
Father’s age = (x – 2) year
His son’s age = (y – 2) year
According to the question,
⇒ (x – 2) = 5(y – 2)
⇒ x – 2 = 5y – 10
⇒ x – 5y + 8 = 0 …(1)
After two years,
Father’s age = (x + 2) year
His son’s age = (y + 2) year
According to the question,
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x – 3y – 12 = 0 …(2)
Now, table for x – 5y + 8 = 0
| x | 2 | -8 |
| $y=\dfrac{x+8}{5}$ | 2 | 0 |
Now, table for x – 3y – 12 = 0
| x | 0 | 12 |
| $y=\dfrac{x-12}{3}$ | -4 | 0 |
On plotting points on a graph paper and join them to get a straight line representing x – 5y + 8 = 0.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x – 3y – 12 = 0.
∴ x = 42, y = 10 is the solution of the pair of linear equations.
Hence, the age of father is 42years and age of his son is 10 years.
Question 28
Champa went to a 'sale' to purchase some pants and skirts. When her
friends asked her how many of each she had bought, she answered, "The
number of skirts is two less than twice the number of pants purchased.
Also the number of skirts is four less than tbur times the number of pants
purchased". Find how many pants and skirts Champa bought?
Sol :Let the number of pants = x
and the number of skirts = y
According to the question
Number of skirts = 2(Number of pants) – 2
y = 2x – 2 …(i)
Also, Number of skirts = 4(Number of pants) – 4
y = 4x – 4 …(ii)
Substituting the value of y = 4x – 4 in eqn (i),we get
4x – 4 = 2x – 2
⇒ 4x – 2x – 4 + 2 = 0
⇒ 2x = 2
⇒ x = 1
Now, substitute the value of x in eqn (ii), we get
y = 4(1) – 4 = 0
∴ x = 1, y = 0 is the solution of the pair of linear equations.
We can solve this problem through graphically also
Hence, the number of pants she purchased is 1 and the number of skirts, she purchased is zero i.e., she didn’t buy any skirt.
Question 29
Priyanka purchased 2 pencils and 3 erasers for Rs. 9. Sayeeda purchased 1
pencil and two erasers for Rs. 5. Find the cost of one pencil and one
eraser.
Sol :Let the cost of one pencil = Rs x
and the cost of one eraser = Rs y
According to the question
2x + 3y = 9 …(i)
x + 2y = 5 …(ii)
Now, table for 2x + 3y = 9
| x | 0 | 4.5 |
| $y=\dfrac{9-2x}{3}$ | 3 | 0 |
Now, table for x + 2y = 5
| x | 0 | 5 |
| $y=\dfrac{5-x}{5}$ | 2.5 | 0 |
On plotting points on a graph paper and join them to get a straight line representing 2x + 3y = 9.
Similarly, on plotting the points on the same graph paper and join them to get a straight line representing x+2y=5.
∴ x = 3, y = 1 is the solution of the pair of linear equations.
Hence, the cost of one pencil is Rs 3 and cost of one eraser is Rs 1.
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