KC Sinha Mathematics Solution Class 10 Chapter 3 Pairs of Linear Equations in Two Variables Exercise 3.2


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 = 2
x + 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 = 14
x – 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 andy = 5 is the required solution.

Question 1 C

Solve the following pair of linear equations by substitution method:
3x – y = 3
9x — 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 = 3.4
0.6x — 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—b
ax —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 = 2m
mx — 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.

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 andt = 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 ≠ 0
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.

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.noChaptersLinks
1Real numbersExercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
2PolynomialsExercise 2.1
Exercise 2.2
Exercise 2.3
3Pairs of Linear Equations in Two VariablesExercise 3.1
Exercise 3.2
Exercise 3.3
Exercise 3.4
Exercise 3.5
4Trigonometric Ratios and IdentitiesExercise 4.1
Exercise 4.2
Exercise 4.3
Exercise 4.4
5TrianglesExercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5
6StatisticsExercise 6.1
Exercise 6.2
Exercise 6.3
Exercise 6.4
7Quadratic EquationsExercise 7.1
Exercise 7.2
Exercise 7.3
Exercise 7.4
Exercise 7.5
8Arithmetic Progressions (AP)Exercise 8.1
Exercise 8.2
Exercise 8.3
Exercise 8.4
9Some Applications of Trigonometry: Height and DistancesExercise 9.1
10Coordinates GeometryExercise 10.1
Exercise 10.2
Exercise 10.3
Exercise 10.4
11CirclesExercise 11.1
Exercise 11.2
12ConstructionsExercise 12.1
13Area related to CirclesExercise 13.1
14Surface Area and VolumesExercise 14.1
Exercise 14.2
Exercise 14.3
Exercise 14.4
15ProbabilityExercise 15.1

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