| Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
Exercise 3.2
Question 1 A
Solve the following pair of linear equations by substitution method:
7x— 15y = 2x + 2y = 3
Sol :
Given equations are
7x – 15y = 2 …(i)
x + 2y = 3 …(ii)
From eqn (ii), x = 3 – 2y …(iii)
On substituting x = 3 – 2y in eqn (i), we get
⇒ 7(3 – 2y) – 15y = 2
⇒ 21 – 14y – 15y = 2
⇒ 21 – 29y = 2
⇒ – 29y = – 19
$\Rightarrow \mathrm{y}=\frac{19}{29}$
Now, on putting $y=\frac{19}{29}$ in eqn (iii), we get
$\Rightarrow x=3-2\left(\frac{19}{29}\right)$
$\Rightarrow x=3-\frac{38}{29}$
$\Rightarrow x=\frac{87-38}{29}$
$\Rightarrow x=\frac{49}{29}$
Thus, $x=\frac{49}{29}$ and$y=\frac{19}{29}$ is the required solution.
Question 1 B
Solve the following pair of linear equations by substitution method:
x + y = 14x – y = 4
Sol :
Given equations are
x + y = 14 …(i)
x – y = 4 …(ii)
From eqn (ii), x = 4 + y …(iii)
On substituting x = 4 + y in eqn (i), we get
⇒ 4 + y + y = 14
⇒ 4 + 2y = 14
⇒ 2y = 14 – 4
⇒ 2y = 10
$\Rightarrow \mathrm{y}=\frac{10}{2}=5$
Now, on putting y = 5 in eqn (iii), we get
⇒ x = 4 + 5
⇒ x = 9
Thus, x = 9 and
y = 5 is the required solution.Question 1 C
Solve the following pair of linear equations by substitution method:
3x – y = 39x — 3y = 9
Sol :
Given equations are
3x – y = 3 …(i)
9x – 3y = 9 …(ii)
From eqn (i), y = 3x – 3 …(iii)
On substituting y = 3x – 3 in eqn (ii), we get
⇒ 9x – 3(3x – 3) = 9
⇒ 9x – 9x + 9 = 9
⇒ 9 = 9
This equality is true for all values of x, therefore given pair of equations have infinitely many solutions.
Question 1 D
Solve the following pair of linear equations by substitution method:
0.5x + 0.8y = 3.40.6x — 0.3y = 0.3
Sol :
Given equations are
0.5x + 0.8y = 3.4 …(i)
0.6x – 0.3y = 0.3 …(ii)
From eqn (ii), 2x – y = 1
y = 2x – 1 …(iii)
On substituting y = 0.2x – 1 in eqn (i), we get
⇒ 0.5x + 0.8(2x – 1) = 3.4
⇒ 0.5x + 1.6x – 0.8 = 3.4
⇒ 2.1x = 3.4 + 0.8
⇒ 2.1x = 4.2
$\Rightarrow x=\frac{4.2}{2.1}=2$
Now, on putting x = 2 in eqn (iii), we get
⇒ y = 2(2) – 1
⇒ y = 4 – 1
⇒ y = 3
Thus, x = 2 and y = 3 is the required solution.
Question 2 A
Solve the following pair of linear equations by substitution method:
x + y = a—bax —by = a2 + b2
Sol :
Given equations are
x + y = a – b …(i)
ax – by = a2 + b2 …(ii)
From eqn (i), x = a – b – y …(iii)
On substituting x = a – b – y in eqn (ii), we get
⇒ a(a – b – y) – by = a2 + b2
⇒ a2 – ab – ay – by = a2 + b2
⇒ – ab – y(a + b) = b2
⇒ – y(a + b) = b2 + ab
$\Rightarrow \mathrm{y}=\frac{\mathrm{b}^{2}+\mathrm{ab}}{-(\mathrm{a}+\mathrm{b})}$
$\Rightarrow \mathrm{y}=\frac{\mathbf{b}(\mathbf{b}+\mathbf{a})}{-(\mathbf{a}+\mathbf{b})}=-\mathbf{b}$
Now, on putting y = – b in eqn (iii), we get
⇒ x = a – b – ( – b)
⇒ x = a
Thus, x = a and y = – b is the required solution.
Question 2 B
Solve the following pair of linear equations by substitution method:
x + y = 2mmx — ny = m2 + n2
Sol :
Given equations are
x + y = 2m …(i)
mx – ny = m2 + n2 …(ii)
From eqn (i), x = 2m – y …(iii)
On substituting x = 2m – y in eqn (ii), we get
⇒ m(2m – y) – ny = m2 + n2
⇒ 2m2 – my – ny = m2 + n2
⇒ – y(m + n) = m2 – 2m2 + n2
⇒ – y(m + n) = – m2 + n2
$\Rightarrow y =\frac{\mathrm{n}^{2}-\mathrm{m}^{2}}{-(\mathrm{a}+\mathrm{b})}\left(\because\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right)=(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})\right)$
$\Rightarrow \mathrm{y}=\frac{(\mathrm{n}-\mathrm{m})(\mathrm{n}+\mathrm{m})}{-(\mathrm{n}+\mathrm{m})}$
⇒ y = – (n – m) = m – n
Now, on putting y = m – n in eqn (iii), we get
⇒ x = 2m – (m – n)
⇒ x = 2m – m + n
⇒ x = m + n
Thus, x = m + n and y = m – n is the required solution.
Question 3 A
Solve the following system of equations by substitution method:
$\frac{x}{2}+y=0.8$$x+\frac{y}{2}=\frac{7}{10}$
Sol :
Given equations are
$\frac{x}{2}+y=0.8$ …(i)
$x+\frac{y}{2}=\frac{7}{10}$ …(ii)
eqn (i) can be re – written as,
$\Rightarrow \frac{x}{2}+y=0.8$
$\Rightarrow\left(\frac{x+2 y}{2}\right)=0.8$
⇒ x + 2y = 0.8×2
⇒ x + 2y = 1.6 …(iii)
On substituting x = 1.6 – 2y in eqn (ii), we get
$\Rightarrow 1.6-2 y+\frac{y}{2}=\frac{7}{10}$
$\Rightarrow \frac{3.2-4 y+y}{2}=\frac{7}{10}$
$\Rightarrow 3.2-3 \mathrm{y}=\frac{7}{10} \times 2$
$\Rightarrow-3 y=\frac{7}{10} \times 2-3.2$
$\Rightarrow-3 y=\frac{14}{10}-\frac{32}{10}$
$\Rightarrow-3 y=-\frac{18}{10}$
$\Rightarrow \mathrm{y}=\frac{18}{10} \times \frac{1}{3}$
$\Rightarrow \mathrm{y}=\frac{6}{10}=0.6$
Now, putting the y = 0.6 in eqn (iii), we get
⇒ x + 2y = 1.6
⇒ x + 2(0.6) = 1.6
⇒ x + 1.2 = 1.6
⇒ x = 0.4
Thus, x = 0.4 and y = 0.6 is the required solution.
$\frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6$
Sol :
Given equations are
s – t = 3 …(i)
$\frac{s}{3}+\frac{t}{2}=6$ …(ii)
From eqn (i), we get
⇒ s = 3 + t …(iii)
On substituting s = 3 + t in eqn (ii), we get
$\Rightarrow \frac{3+t}{3}+\frac{t}{2}=6$
$\Rightarrow \frac{2(3+t)+3 t}{6}=6$
⇒ 6 + 2t + 3t = 6×6
⇒ 5t = 36 – 6
$\Rightarrow \mathrm{t}=\frac{30}{5}=6$
Now, putting the t = 6 in eqn (iii), we get
⇒ s = 3 + 6
⇒ s = 9
Thus, s = 9 and
t = 6 is the required solution.
ax – by = a2 – b2
Sol :
Given equations are
$\frac{x}{a}+\frac{y}{b}=2$ …(i)
ax – by = a2 – b2 …(ii)
eqn (i) can be re – written as,
$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$
$\Rightarrow\left(\frac{b x+a y}{a b}\right)=2$
⇒ bx + ay = ab×2
⇒ bx + ay = 2ab
$\Rightarrow x=\frac{2 a b-a y}{b}$ …(iii)
On substituting $\mathrm{X}=\frac{2 \mathrm{ab}-\mathrm{ay}}{\mathrm{b}}$ in eqn (ii), we get
$\Rightarrow \mathrm{a}\left(\frac{2 \mathrm{ab}-\mathrm{ay}}{\mathrm{b}}\right)-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}$
$\Rightarrow \frac{2 a^{2} b-a^{2} y}{b}-b y=a^{2}-b^{2}$
$\Rightarrow \frac{2 a^{2} b-a^{2} y-b^{2} y}{b}=a^{2}-b^{2}$
⇒ 2a2 b – a2 y – b2 y = b(a2 – b2)
⇒ 2a2 b – y(a2 + b2 ) = a2b – b3
⇒ – y(a2 + b2 ) = a2 b – b3 – 2a2b
⇒ – y(a2 + b2 ) = – a2b – b3
⇒ – y(a2 + b2 ) = – b(a2 + b2)
$\Rightarrow y=\frac{-b\left(a^{2}+b^{2}\right)}{-\left(a^{2}+b^{2}\right)}=b$
Now, on putting y = b in eqn (iii), we get
$\Rightarrow x=\frac{2 a b-a b}{b}$
⇒ x = a
Thus, x = a and y = b is the required solution.
x + y = 2ab
Sol :
Given equations are
$\frac{b x}{a}+\frac{a y}{b}=a^{2}+b^{2}$ …(i)
x + y = 2ab …(ii)
eqn (i) can be re - written as,
$\Rightarrow \frac{b x}{a}+\frac{a y}{b}=a^{2}+b^{2}$
$\Rightarrow\left(\frac{\mathrm{b}^{2} \mathrm{x}+\mathrm{a}^{2} \mathrm{y}}{\mathrm{ab}}\right)=\mathrm{a}^{2}+\mathrm{b}^{2}$
⇒ b2 x + a2 y = ab × (a2 + b2)
$\Rightarrow x=\frac{a b \times\left(a^{2}+b^{2}\right)-a^{2} y}{b^{2}}$ …(iii)
Now, from eqn (ii), y = 2ab – x …(iv)
On substituting y = 2ab – x in eqn (iii), we get
$\Rightarrow x=\frac{a b \times\left(a^{2}+b^{2}\right)-a^{2}(2 a b-x)}{b^{2}}$
$\Rightarrow x=\frac{a^{3} b+b^{3} a-2 a^{3} b+a^{2} x}{b^{2}}$
⇒ b2 x = b3 a – a3b + a2x
⇒ b2x – a2x = b3a – a3b
⇒ (b2 – a2) x = ab(b2 – a2)
⇒ x = ab
Now, on putting x = ab in eqn (iv), we get
⇒ y = 2ab – ab
⇒ y = ab
Thus, x = ab and y = ab is the required solution.
⇒ x + 2y = 1.6 …(iii)
On substituting x = 1.6 – 2y in eqn (ii), we get
$\Rightarrow 1.6-2 y+\frac{y}{2}=\frac{7}{10}$
$\Rightarrow \frac{3.2-4 y+y}{2}=\frac{7}{10}$
$\Rightarrow 3.2-3 \mathrm{y}=\frac{7}{10} \times 2$
$\Rightarrow-3 y=\frac{7}{10} \times 2-3.2$
$\Rightarrow-3 y=\frac{14}{10}-\frac{32}{10}$
$\Rightarrow-3 y=-\frac{18}{10}$
$\Rightarrow \mathrm{y}=\frac{18}{10} \times \frac{1}{3}$
$\Rightarrow \mathrm{y}=\frac{6}{10}=0.6$
Now, putting the y = 0.6 in eqn (iii), we get
⇒ x + 2y = 1.6
⇒ x + 2(0.6) = 1.6
⇒ x + 1.2 = 1.6
⇒ x = 0.4
Thus, x = 0.4 and y = 0.6 is the required solution.
Question 3 B
Solve the following system of equations by substitution method:
s —t = 3$\frac{\mathrm{s}}{3}+\frac{\mathrm{t}}{2}=6$
Sol :
Given equations are
s – t = 3 …(i)
$\frac{s}{3}+\frac{t}{2}=6$ …(ii)
From eqn (i), we get
⇒ s = 3 + t …(iii)
On substituting s = 3 + t in eqn (ii), we get
$\Rightarrow \frac{3+t}{3}+\frac{t}{2}=6$
$\Rightarrow \frac{2(3+t)+3 t}{6}=6$
⇒ 6 + 2t + 3t = 6×6
⇒ 5t = 36 – 6
$\Rightarrow \mathrm{t}=\frac{30}{5}=6$
Now, putting the t = 6 in eqn (iii), we get
⇒ s = 3 + 6
⇒ s = 9
Thus, s = 9 and
t = 6 is the required solution.Question 3 C
Solve the following system of equations by substitution method:
$\frac{x}{a}+\frac{y}{b}=2$, a ≠ 0, b ≠ 0ax – by = a2 – b2
Sol :
Given equations are
$\frac{x}{a}+\frac{y}{b}=2$ …(i)
ax – by = a2 – b2 …(ii)
eqn (i) can be re – written as,
$\Rightarrow \frac{x}{a}+\frac{y}{b}=2$
$\Rightarrow\left(\frac{b x+a y}{a b}\right)=2$
⇒ bx + ay = ab×2
⇒ bx + ay = 2ab
$\Rightarrow x=\frac{2 a b-a y}{b}$ …(iii)
On substituting $\mathrm{X}=\frac{2 \mathrm{ab}-\mathrm{ay}}{\mathrm{b}}$ in eqn (ii), we get
$\Rightarrow \mathrm{a}\left(\frac{2 \mathrm{ab}-\mathrm{ay}}{\mathrm{b}}\right)-\mathrm{by}=\mathrm{a}^{2}-\mathrm{b}^{2}$
$\Rightarrow \frac{2 a^{2} b-a^{2} y}{b}-b y=a^{2}-b^{2}$
$\Rightarrow \frac{2 a^{2} b-a^{2} y-b^{2} y}{b}=a^{2}-b^{2}$
⇒ 2a2 b – a2 y – b2 y = b(a2 – b2)
⇒ 2a2 b – y(a2 + b2 ) = a2b – b3
⇒ – y(a2 + b2 ) = a2 b – b3 – 2a2b
⇒ – y(a2 + b2 ) = – a2b – b3
⇒ – y(a2 + b2 ) = – b(a2 + b2)
$\Rightarrow y=\frac{-b\left(a^{2}+b^{2}\right)}{-\left(a^{2}+b^{2}\right)}=b$
Now, on putting y = b in eqn (iii), we get
$\Rightarrow x=\frac{2 a b-a b}{b}$
⇒ x = a
Thus, x = a and y = b is the required solution.
Question 3 D
Solve the following system of equations by substitution method:
$\frac{b x}{a}+\frac{a y}{b}=a^{2}+b^{2}$x + y = 2ab
Sol :
Given equations are
$\frac{b x}{a}+\frac{a y}{b}=a^{2}+b^{2}$ …(i)
x + y = 2ab …(ii)
eqn (i) can be re - written as,
$\Rightarrow \frac{b x}{a}+\frac{a y}{b}=a^{2}+b^{2}$
$\Rightarrow\left(\frac{\mathrm{b}^{2} \mathrm{x}+\mathrm{a}^{2} \mathrm{y}}{\mathrm{ab}}\right)=\mathrm{a}^{2}+\mathrm{b}^{2}$
⇒ b2 x + a2 y = ab × (a2 + b2)
$\Rightarrow x=\frac{a b \times\left(a^{2}+b^{2}\right)-a^{2} y}{b^{2}}$ …(iii)
Now, from eqn (ii), y = 2ab – x …(iv)
On substituting y = 2ab – x in eqn (iii), we get
$\Rightarrow x=\frac{a b \times\left(a^{2}+b^{2}\right)-a^{2}(2 a b-x)}{b^{2}}$
$\Rightarrow x=\frac{a^{3} b+b^{3} a-2 a^{3} b+a^{2} x}{b^{2}}$
⇒ b2 x = b3 a – a3b + a2x
⇒ b2x – a2x = b3a – a3b
⇒ (b2 – a2) x = ab(b2 – a2)
⇒ x = ab
Now, on putting x = ab in eqn (iv), we get
⇒ y = 2ab – ab
⇒ y = ab
Thus, x = ab and y = ab is the required solution.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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