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

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 

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 C 

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
⇒ (b^{2 –} 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.