KC Sinha: Exercise 3.4 - Mathematics Solution Class 10 Chapter 3 Pair of Linear Equations in Two Variables

[mathjax]

Question 1 A 

Solve the following pair of linear equation by cross - multiplication method:

8x + 5y = 9
3x + 2y = 4
Sol :
Given, pair of equations is
8x + 5y – 9 = 0 and 3x + 2y – 4 = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-20+18}=\frac{y}{-27+32}=\frac{1}{16-15}$
$\Rightarrow \begin{matrix}\frac{x}{-2}&=\frac{y}{5}&=\frac{1}{1}\\\text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{-2}=\frac{1}{1}$
⇒ x = – 2

On taking II and III ratio, we get
$\Rightarrow \frac{y}{5}=\frac{1}{1}$
⇒ y = 5

---

Question 1 B 

Solve the following pair of linear equation by cross - multiplication method:

2x + 3y = 46
3x + 5y = 74
Sol :
Given, pair of equations is
2x + 3y – 46 = 0 and 3x + 5y – 74 = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-222+230}=\frac{y}{-138+148}=\frac{1}{10-9}$
$\Rightarrow \begin{matrix}\frac{x}{8}&=\frac{y}{10}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{8}=\frac{1}{1}$
⇒ x = 8

On taking II and III ratio, we get
$\Rightarrow \frac{y}{10}=\frac{1}{1}$
⇒ y = 10

---

Question 1 C 

Solve the following pair of linear equation by cross - multiplication method:

x + 4y + 9 = 0
5x – 1 = 3y
Sol :
Given, pair of equations is
x + 4y + 9 = 0 and 5x – 3y – 1 = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-4+27}=\frac{y}{45+1}=\frac{1}{-3-20}$

$\Rightarrow \begin{matrix}\frac{x}{23}&=\frac{y}{46}&=\frac{1}{-23}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{23}=\frac{1}{-23}$
⇒ x = – 1

On taking II and III ratio, we get
$\Rightarrow \frac{y}{46}=\frac{1}{-23}$
⇒ y = – 2

---

Question 1 D 

Solve the following pair of linear equation by cross - multiplication method:
2x + 3y – 7 = 0
6x + 5y – 11 = 0
Sol :
Given, pair of equations is
2x + 3y – 7 = 0 and 6x + 5y – 11 = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-33+35}=\frac{y}{-42+22}=\frac{1}{10-18}$
$\Rightarrow \begin{matrix}\frac{x}{2}&=\frac{y}{-20}&=\frac{1}{-8}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{2}=\frac{1}{-8}$
$\Rightarrow x=\frac{1}{-4}$

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-20}=\frac{1}{-8}$
$\Rightarrow \mathrm{y}=\frac{5}{2}$

---

Question 1 E 

Solve the following pair of linear equation by cross - multiplication method:

$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$
Sol :
Given, pair of equations is
$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$

Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$

So, Eq. (1) and (2) reduces to
2u + 3v – 13 = 0
5u – 4v + 2 = 0

By cross - multiplication method, we have

$\Rightarrow \frac{u}{6-52}=\frac{v}{-65-4}=\frac{1}{-8-15}$

$\Rightarrow \begin{matrix}\frac{u}{-46}&=\frac{v}{-69}&=\frac{1}{-23}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{\mathrm{u}}{-46}=\frac{1}{-23}$
⇒ u = 2

On taking II and III ratio, we get
$\Rightarrow \frac{\mathrm{v}}{-69}=\frac{1}{-23}$
⇒ v = 3

So, $u=\frac{1}{x}=2 \Rightarrow x=\frac{1}{2}$
and $v=\frac{1}{y}=3 \Rightarrow y=\frac{1}{3}$

---

Question 1 F 

Solve the following pair of linear equation by cross - multiplication method:

$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
Sol :
Given, pair of equations is
$\frac{x}{6}+\frac{y}{15}=4$

And $\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$

⇒4x-y=57

By cross - multiplication method, we have

$\Rightarrow \frac{x}{-114-120}=\frac{y}{-480+285}=\frac{1}{-5-8}$

$\Rightarrow \begin{matrix}\frac{x}{-234}&=\frac{y}{-195}&=\frac{1}{-13}\

\ \text{I}& \text{II}& \text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{-234}=\frac{1}{-13}$
⇒ x = 18

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-195}=\frac{1}{-13}$
⇒ y = 15

---

Question 2 A 

Solve the following pair of equations by cross - multiplication method.

ax + by = a – b
bx – ay = a + b
Sol :
Given, pair of equations is
ax + by = a – b ⇒ ax + by –(a – b) = 0
bx – ay = a + b ⇒ bx –ay – (a + b) = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^{2}-b^{2}}$
$\Rightarrow \frac{x}{-b a-b^{2}-a^{2}+a b}=\frac{y}{-b a+b^{2}+a^{2}+a b}=\frac{1}{-a^{2}-b^{2}}$
$\Rightarrow \begin{matrix}\frac{x}{-b^{2}-a^{2}}&=\frac{y}{b^{2}+a^{2}}&=\frac{1}{-a^{2}-b^{2}}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{-b^{2}-a^{2}}=\frac{1}{-a^{2}-b^{2}}$
⇒ x = 1

On taking II and III ratio, we get
$\Rightarrow \frac{y}{b^{2}+a^{2}}=\frac{1}{-a^{2}-b^{2}}$
⇒ y = – 1

---

Question 2 B 

Solve the following pair of equations by cross - multiplication method.

$a^{2}-b^{2}$
$\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$
$a \neq 0, b \neq 0$
Sol :
Given, pair of equations is

$\frac{x}{a}+\frac{y}{b}=a+b$

$\Rightarrow \frac{x}{a}+\frac{y}{b}-(a+b)=0$

And $\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$

$\Rightarrow \frac{x}{a^{2}}+\frac{y}{b^{2}}-2=0$

By cross - multiplication method, we have

$\Rightarrow \frac{x}{\frac{-2}{b}+\frac{a+b}{b^{2}}}=\frac{y}{-\frac{a+b}{a^{2}}+\frac{2}{a}}=\frac{1}{\frac{1}{a b^{2}}-\frac{1}{a^{2} b}}$
$\Rightarrow \frac{b^{2} x}{-2 b+a+b}=\frac{a^{2} y}{-a-b+2 a}=\frac{a^{2} b^{2}}{a-b}$
$\Rightarrow \begin{matrix}\frac{b^{2} x}{a-b}&=\frac{a^{2} y}{a-b}&=\frac{a^{2} b^{2}}{a-b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{b^{2} x}{a-b}=\frac{a^{2} b^{2}}{a-b}$
$\Rightarrow \frac{x}{1}=\frac{a^{2} b^{2}}{b^{2}}$
⇒x = a²

On taking II and III ratio, we get
$\Rightarrow \frac{a^{2} y}{a-b}=\frac{a^{2} b^{2}}{a-b}$
⇒ y = b²

---

Question 2 C 

Solve the following pair of equations by cross - multiplication method.

x – y = a + b
ax + by = a² – b²
Sol :
Given, pair of equations is
x – y = a + b ⇒ x – y –(a + b) = 0
ax + by = a² –b²⇒ ax + by – (a² –b² ) = 0

By cross – multiplication method, we have

$\Rightarrow \frac{x}{a^{2}-b^{2}+b(a+b)}=\frac{y}{-a(a+b)+a^{2}-b^{2}}=\frac{1}{b+a}$
$\Rightarrow \frac{x}{a^{2}+a b}=\frac{y}{-b a-b^{2}}=\frac{1}{b+a}$

$\Rightarrow \begin{matrix}\frac{x}{a(a+b)}&=\frac{y}{-b(a+b)}&=\frac{1}{a+b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{a(a+b)}=\frac{1}{a+b}$
⇒ x = a

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-b(a+b)}=\frac{1}{a+b}$
⇒ y = – b

---

Question 2 D 

Solve the following pair of equations by cross - multiplication method.

$\frac{2 x}{a}+\frac{y}{b}=2$
$\frac{x}{a}-\frac{y}{b}=4$
$a \neq 0, b \neq 0$
Sol :
$\frac{2 x}{a}+\frac{y}{b}=2$
$\frac{x}{a}-\frac{y}{b}=4$

Given, pair of equations is

$\frac{2 x}{a}+\frac{y}{b}=2$

$\Rightarrow \frac{2 \mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-2=0$

And $\frac{x}{a}-\frac{y}{b}=4$

$\Rightarrow \frac{x}{a}-\frac{y}{b}-4=0$

By cross - multiplication method, we have

$\Rightarrow \frac{x}{\frac{-4}{b}-\frac{2}{b}}=\frac{y}{\frac{-2}{a}+\frac{8}{a}}=\frac{1}{-\frac{2}{a b}-\frac{1}{a b}}$
$\Rightarrow \begin{matrix}\frac{b x}{-6}&=\frac{a y}{6}&=\frac{a b}{-3}\\\text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{b x}{-6}=\frac{a b}{-3}$
⇒x = 2a

On taking II and III ratio, we get
$\Rightarrow \frac{\mathrm{ay}}{6}=\frac{\mathrm{ab}}{-3}$
⇒ y = – 2b

---

Question 2 E 

Solve the following pair of equations by cross - multiplication method.
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Sol :
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Given, pair of equations is
2ax + 3by = a + 2b ⇒ 2ax + 3by –(a + 2b) = 0
3ax + 2by = 2a + b ⇒ 3ax + 2by –(2a + b) = 0
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-3 b(2 a+b)+2 b(a+2 b)}=\frac{y}{-3 a(a+2 b)+2 a(2 a+b)}=\frac{1}{4ab-9ab}$
$\Rightarrow \frac{x}{-4 b a-3 b^{2}+4 a b+4 b^{2}}=\frac{y}{-3 a^{2}-6 b a+4 a^{2}+2 a b}=\frac{1}{-5 a b}$
$\Rightarrow \begin{matrix}\frac{x}{b^{2}-4 a b}&=\frac{y}{-4 a b+a^{2}}&=\frac{1}{-5 a b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{b^{2}-4 a b}=\frac{1}{-5 a b}$
$\Rightarrow x=\frac{b(b-4 a)}{-5 a b}$
$\Rightarrow x=\frac{4 a-b}{5 a}$

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-4 a b+a^{2}}=\frac{1}{-5 a b}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{a}(\mathrm{a}-4 \mathrm{b})}{-5 \mathrm{ab}}$
$\Rightarrow \mathrm{y}=\frac{4 \mathrm{b}-\mathrm{a}}{5 \mathrm{b}}$

---

Question 2 F 

Solve the following pair of equations by cross - multiplication method.

$\frac{x}{a}+\frac{y}{b}=2$
ax + by = a²–b²
Sol :
$\frac{x}{a}+\frac{y}{b}=2$
ax + by = a²–b²

Given, pair of equations is

$\frac{x}{a}+\frac{y}{b}=2$
$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-2=0$

ax + by = a² –b²⇒ ax + by – (a² –b² ) = 0

By cross - multiplication method, we have

$\Rightarrow \frac{x}{\frac{-\left(a^{2}+b^{2}\right)}{b}+2 b}=\frac{y}{-2 a+\frac{\left(a^{2}+b^{2}\right)}{a}}=\frac{1}{\frac{b}{a}-\frac{a}{b}}$
$\Rightarrow \frac{b x}{-a^{2}-b^{2}+2 b^{2}}=\frac{a y}{-2 a^{2}+a^{2}+b^{2}}=\frac{a b}{b^{2}-a^{2}}$
$\Rightarrow \frac{b x}{-a^{2}+b^{2}}=\frac{a y}{-a^{2}+b^{2}}=\frac{a b}{b^{2}-a^{2}}$
$\Rightarrow \begin{matrix}\frac{b x}{b^{2}-a^{2}}&=\frac{a y}{b^{2}-a^{2}}&=\frac{a b}{b^{2}-a^{2}}\\ \text{I}& \text{II}& \text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{b x}{b^{2}-a^{2}}=\frac{a b}{b^{2}-a^{2}}$
⇒ x = a

On taking II and III ratio, we get
$\Rightarrow \frac{a y}{b^{2}-a^{2}}=\frac{a b}{b^{2}-a^{2}}$
⇒ y = b

---

Question 3 

Solve the following system of equations by cross - multiplication method.

a(x + y) + b(x – y) = a² – ab + b²
a(x + y) – b(x – y) = a² + ab + b²
Sol :
The given system of equations can be re - written as
ax + ay + bx – by – a² + ab – b² = 0
⇒(a + b)x + (a – b)y – (a² – ab + b² ) = 0 …(1)
and ax + ay – bx + by – a² – ab – b² = 0
⇒ (a – b)x + (a + b)y – (a² + ab + b² ) = 0 …(2)
Now, by cross – multiplication method, we have
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-(a-b)\left(a^{2}+a b+b^{2}\right)+(a+b)\left(a^{2}-a b+b^{2}\right)}$

$=\frac{y}{-\left(a^{2}-a b+b^{2}\right)(a-b)+\left(a^{2}+a b+b^{2}\right)(a+b)}=\frac{1}{(a+b)(a+b)-(a-b)(a-b)}$
$\Rightarrow \frac{x}{-\left(a^{3}-b^{3}\right)+\left(a^{3}+b^{3}\right)}$
$=\frac{y}{-a^{3}+a^{2} b-b^{2} a+b a^{2}-a b^{2}+b^{3}+a^{3}+a^{2} b+b^{2} a+b a^{2}+a b^{2}+b^{3}}$
$=\frac{1}{a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2}}$
$\Rightarrow \begin{matrix}\frac{x}{2 b^{2}}&=\frac{y}{4 a^{2} b+2 b^{2}}&=\frac{1}{4 a b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow x=\frac{b^{2}}{2 a}$

On taking II and III ratio, we get
$\Rightarrow \frac{y}{4 a^{2} b+2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow \mathrm{y}=\frac{2 \mathrm{a}^{2}+\mathrm{b}^{2}}{2 \mathrm{a}}$

---

Question 4 A 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x – 3y – 7 = 0
3x – 3y – 15 = 0
Sol :
Given pair of linear equations
x – 3y – 7 = 0
3x – 3y – 15 = 0
⇒ x – y – 5 = 0 …(ii)
As we can see that a₁ = 1, b₁ = – 3 and c₁ = – 7
and a_{2 =} 1, b₂ = – 1 and c₂ = – 5

$\frac{a_{1}}{a_{2}}=\frac{1}{1}=1$, $\frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3$ and $\frac{c_{1}}{c_{2}}=\frac{-7}{-5}=\frac{7}{5}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

$\Rightarrow \frac{x}{15-7}=\frac{y}{-7+5}=\frac{1}{-1+3}$
$\Rightarrow \begin{matrix}\frac{x}{8}&=\frac{y}{-2}&=\frac{1}{2}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{8}=\frac{1}{2}$
⇒ x = 4

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-2}=\frac{1}{2}$
⇒ y = – 1

---

Question 4 B 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

2x + y = 5
3x + 2y = 8
Sol :
Given pair of linear equations
2x + y = 5
3x + 2y = 8
As we can see that a₁ = 2, b₁ = 1 and c₁ = – 5
and a_{2 =} 3, b₂ = 2 and c₂ = – 8

$\frac{a_{1}}{a_{2}}=\frac{2}{3}$ , $\frac{b_{1}}{b_{2}}=\frac{1}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-8}=\frac{5}{8}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$

∴ Given pair of equations has unique solution

By cross - multiplication method, we have

$\Rightarrow \frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}$

$\Rightarrow \begin{matrix}\frac{x}{2}&=\frac{y}{1}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{2}=1$
⇒ x = 2

On taking II and III ratio, we get
$\Rightarrow \frac{y}{1}=1$
⇒ y = 1

---

Question 4 C 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

3x – 5y = 20
6x – 10y = 40
Sol :
Given pair of linear equations
3x – 5y = 20
6x – 10y = 40
⇒ 3x – 5y = 20 …(ii)
As we can see that a₁ = 3, b₁ = – 5 and c₁ = – 20
and a_{2 =} 3, b₂ = – 5 and c₂ = – 20

$\frac{a_{1}}{a_{2}}=\frac{3}{3}=1$, $\frac{b_{1}}{b_{2}}=\frac{-5}{-5}=1$ and $\frac{c_{1}}{c_{2}}=\frac{-20}{-20}=1$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

∴ Given pair of equations has infinitely many solutions.

---

Question 4 D 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x – 3y – 3 = 0
3x – 9y – 2 = 0
Sol :
Given pair of linear equations
x – 3y – 3 = 0
3x – 9y – 2 = 0
As we can see that a₁ = 1, b₁ = – 3 and c₁ = – 3
and a_{2 =} 3, b₂ = – 9 and c₂ = – 2

$\frac{a_{1}}{a_{2}}=\frac{1}{3}$ , $\frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}$ and $\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

∴ Given pair of equations has no solution

---

Question 4 E 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x + y = 2
2x + 2y = 4
Sol :
Given pair of linear equations
x + y = 2
2x + 2y = 4
⇒ x + y – 2 = 0
As we can see that a₁ = 1, b₁ = 1 and c₁ = – 2
and a_{2 =} 1, b₂ = 1 and c₂ = – 2

$\frac{a_{1}}{a_{2}}=\frac{1}{1}=1$, $\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{1}=3$ and $\frac{c_{1}}{c_{2}}=\frac{-2}{-2}=1$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

∴ Given pair of equations has infinitely many solution

---

Question 4 F 

Which of the following pair of linear equations has unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it by using cross - multiplication method:

x + y = 2
2x + 2y = 6
Sol :
Given pair of linear equations
x + y = 2
2x + 2y = 6
⇒ x + y – 3 = 0
As we can see that a₁ = 1, b₁ = 1 and c₁ = – 2
and a_{2 =} 1, b₂ = 1 and c₂ = – 3

$\frac{a_{1}}{a_{2}}=\frac{1}{1}=1$ , $\frac{b_{1}}{b_{2}}=\frac{1}{1}=1$ and $\frac{c_{1}}{c_{2}}=\frac{-2}{-3}=\frac{2}{3}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

∴ Given pair of equations has no solution

---

Question 5 A 

Solve the following system of linear equations by cross - multiplication method.

$\frac{15}{x+y}+\frac{7}{x-y}-10=0$
$\frac{15}{x+y}+\frac{7}{x-y}-10=0$
[Hint: Let $u=\frac{1}{x-y}$ and $v=\frac{1}{x-y}$]

Sol :
Given, pair of equations is
$\frac{5}{x+y}+\frac{2}{x-y}+1=0$ …(1)

$\frac{15}{x+y}+\frac{7}{x-y}-10=0$ …(2)

Let $u=\frac{1}{x+y}$ and $v=\frac{1}{x-y}$

Now, the Eq. (1) and (2) reduces to
5u + 2v + 1 = 0
15u + 7v – 10 = 0

By cross - multiplication method, we have

$\Rightarrow \frac{\mathrm{u}}{-20-7}=\frac{\mathrm{v}}{15+50}=\frac{1}{35-30}$
$\Rightarrow \begin{matrix}\frac{u}{-27}&=\frac{v}{65}&=\frac{1}{5}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{\mathrm{u}}{-27}=\frac{1}{5}$

$\Rightarrow \mathrm{u}=\frac{-27}{5}$

On taking II and III ratio, we get
$\Rightarrow \frac{v}{65}=\frac{1}{5}$
⇒ v = 13

So, $u=\frac{1}{x+y}=\frac{-27}{5}$

$\Rightarrow \mathrm{x}+\mathrm{y}=-\frac{5}{27}$ …(a)

and $v=\frac{1}{x-y}=13$ 

$\Rightarrow \mathrm{x}-\mathrm{y}=\frac{1}{13}$ …(b)

On adding Eq. (a) and (b), we get
$2 x=-\frac{5}{27}+\frac{1}{13}$
$\Rightarrow 2 \mathrm{x}=\frac{-65+27}{27 \times 13}$
$\Rightarrow 2 \mathrm{x}=\frac{-38}{351}$
$\Rightarrow x=\frac{-19}{351}$

On putting the value of $x=\frac{-19}{351}$ in Eq. (a), we get
$\frac{-19}{351}+y=-\frac{5}{27}$
$\Rightarrow \mathrm{y}=-\frac{5}{27}+\frac{19}{351}$ 
$\Rightarrow \mathrm{y}=\frac{-65+19}{351}$
$\Rightarrow \mathrm{y}=-\frac{46}{351}$

---

Question 5 B 

Solve the following system of linear equations by cross - multiplication method.

ax – ay = 2
(a – 1)x + (a + 1)y = 2(a² + 1)
[Hint: Let $u = \frac{1}{x-y}$ and $v = \frac{1}{x-y}$]

Sol :
Given, pair of equations is
ax – ay = 2
(a – 1)x + (a + 1)y = 2(a² + 1)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{2 a\left(a^{2}+1\right)+2(a+1)}=\frac{y}{-2(a-1)+2 a\left(a^{2}+1\right)}=\frac{1}{a(a+1)+a(a-1)}$

$\Rightarrow \frac{x}{2 a^{3}+2 a+2 a+2}=\frac{y}{-2 a+2+2 a^{3}+2 a}=\frac{1}{a^{2}+a+a^{2}-a}$

$\Rightarrow \begin{matrix}\frac{x}{2a^3+4a+2}&=\frac{y}{2a^3+2}&=\frac{1}{2a^2}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get

$\Rightarrow \frac{x}{2 a^{3}+4 a+2}=\frac{1}{2 a^{2}}$

$\Rightarrow \mathrm{x}=\frac{\mathrm{a}^{3}+2 \mathrm{a}+1}{\mathrm{a}^{2}}$

On taking II and III ratio, we get

$\Rightarrow \frac{\mathrm{y}}{2 \mathrm{a}^{3}+2}=\frac{1}{2 \mathrm{a}^{2}}$

$\Rightarrow \mathrm{y}=\frac{\mathrm{a}^{3}+1}{\mathrm{a}^{2}}$

---

Question 6 

If the cost of 2 pencils and 3 erasers is Rs. 9 and the cost of 4 pencils and 6 erasers is Rs. 18. Find the cost of each pencil and each eraser.

Sol :
Let the cost of one pencil = Rs x
and cost of one eraser = Rs y
According to the question
2x + 3y = 9 …(1)
4x + 6y = 18
⇒2(2x + 3y) = 18
⇒2x + 3y = 9 …(2)
As we can see From Eq. (1) and (2)

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

∴Given pair of linear equations has infinitely many solutions.

---

Question 7 

The paths traced by the wheels of two trains are given by equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the paths cross each other?

Sol :
Given paths traced by the wheel of two trains are
x + 2y – 4 = 0 …(i)
2x + 4y – 12 = 0
⇒ x + 2y – 6 = 0 …(ii)
As we can see that a₁ = 1, b₁ = 2 and c₁ = – 4
and a_{2 =} 1, b₂ = 2 and c₂ = – 6

$\frac{a_{1}}{a_{2}}=\frac{1}{1}=1$, $\frac{b_{1}}{b_{2}}=\frac{2}{2}=1$ and $\frac{c_{1}}{c_{2}}=\frac{-4}{-6}=\frac{2}{3}$

$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ Given pair of equations has no solution
Hence, two paths will not cross each other.

---

Question 8 

The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.

Sol :
Given ratio of incomes = 9:7
And the ratio of their expenditures = 4:3
Saving of each person = Rs. 2000
Let incomes of two persons = 9x and 7x
And their expenditures = 4y and 3y
According to the question,
9x – 4y = 2000
⇒9x – 4y – 2000 = 0 …(i)
7x – 3y = 2000
⇒7x – 3y – 2000 = 0 …(ii)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{8000-6000}=\frac{y}{-14000+18000}=\frac{1}{-27+28}$

$\Rightarrow \begin{matrix}\frac{x}{2000}&=\frac{y}{4000}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratios, we get
$\Rightarrow \frac{x}{2000}=\frac{1}{1}$
⇒ x = 2000

On taking II and III ratios, we get
$\Rightarrow \frac{y}{4000}=\frac{1}{1}$
⇒ y = 4000

Hence, the monthly incomes of two persons are 9(2000) = Rs18000 and 7(2000) = Rs14000

---

Question 9 

The sum of two - digits number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Sol :
Let unit’s digit = y
and the ten’s digit = x
So, the original number = 10x + y
The sum of the number = 10x + y
The sum of the digit = x + y
reversing number = x + 10y
According to the question,
10x + y + x + 10y = 66
⇒11x + 11y = 66
⇒ x + y = 6 …(i)
x – y = 2 …(ii)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-2-6}=\frac{y}{-6+2}=\frac{1}{-1-1}$

$\Rightarrow \begin{matrix}\frac{x}{-8}&=\frac{y}{-4}&=\frac{1}{-2}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{-8}=\frac{1}{-1}$
⇒ x = 4

On taking II and III ratio, we get
$\Rightarrow \frac{y}{-4}=\frac{1}{-2}$
⇒ y = 2

So, the original number = 10x + y
= 10(4) + 2
= 42
Reversing the number = x + 10y
= 24
Hence, the two digit number is 42 and 24. These are two such numbers.

---

Question 10 

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we add 1 to the denominator. What is the fraction?

Sol :
Let the numerator = x
and the denominator = y

So, the fraction $=\frac{x}{y}$

According to the question,

Condition I:
$\frac{x+1}{y-1}=1$
⇒ x + 1 = y – 1
⇒ x – y = – 2
⇒ x – y + 2 = 0 …(i)

Condition II:
$\frac{x}{y+1}=\frac{1}{2}$
⇒ 2x = y + 1
⇒ 2x – y = 1
⇒ 2x – y – 1 = 0 …(ii)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{1+2}=\frac{y}{4+1}=\frac{1}{-1+2}$

$\Rightarrow \begin{matrix}\frac{x}{3}&=\frac{y}{5}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{3}=\frac{1}{1}$
⇒ x = 3

On taking II and III ratio, we get

$\Rightarrow \frac{y}{5}=\frac{1}{1}$
⇒ y = 5

So, the numerator is 3 and the denominator is 5
Hence, the fraction is $\frac{3}{5}$

---

Question 11 

The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4 apples is Rs. 28. Find the cost of an orange and an apple.

Sol :
Let the cost of an orange = Rs x
And the cost of an apple = Rs y
According to the question,
5x + 3y = 35
And 2x + 4y = 28
⇒ x + 2y = 14
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-42+70}=\frac{y}{-35+70}=\frac{1}{10-3}$

$\Rightarrow \begin{matrix}\frac{x}{28}&=\frac{y}{35}&=\frac{1}{7}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{28}=\frac{1}{7}$
⇒ x = 4

On taking II and III ratio, we get
$\Rightarrow \frac{y}{35}=\frac{1}{7}$
⇒ y = 5
Hence, the cost of an orange is Rs. 4 and cost of an apple is Rs. 5

---

Question 12 

A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days, she has to pay Rs. 1000 as hostel charges, whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and cost of food per day.

Sol :
Let fixed hostel charge (monthly) = Rs x
and cost of food for one day = Rs y
In case of student A,
x + 20y = 1000
x + 20y – 1000 = 0 …(i)
In case of student B,
x + 26y = 1180
x + 26y – 1180 = 0 …(ii)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{-23600+26000}=\frac{y}{-1000+1180}=\frac{1}{26-20}$

$\Rightarrow \begin{matrix}\frac{x}{2400}&=\frac{y}{180}&=\frac{1}{6}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{2400}=\frac{1}{6}$
⇒ x = 400

On taking II and III ratio, we get
$\Rightarrow \frac{y}{180}=\frac{1}{6}$
⇒ y = 30
Hence, monthly fixed charges is Rs. 400 and cost of food per day is Rs. 30

---

Question 13 

A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Sol :

Let the numerator = x
and the denominator = y

So, the fraction $=\frac{x}{y}$
According to the question,

Condition I:
$\frac{x-1}{y}=\frac{1}{3}$
⇒ 3(x – 1) = y
⇒ 3x – 3 = y
⇒ 3x – y – 3 = 0 …(i)

Condition II:
$\frac{x}{y+8}=\frac{1}{4}$
⇒ 4x = y + 8
⇒ 4x – y = 8
⇒ 4x – y – 8 = 0 …(ii)
By cross - multiplication method, we have

$\Rightarrow \frac{x}{8-3}=\frac{y}{-12+24}=\frac{1}{-3+4}$

$\Rightarrow \begin{matrix}\frac{x}{5}&=\frac{y}{12}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$

On taking I and III ratio, we get
$\Rightarrow \frac{x}{5}=\frac{1}{1}$
⇒ x = 5

On taking II and III ratio, we get
$\Rightarrow \frac{y}{12}=\frac{1}{1}$
⇒ y = 12
So, the numerator is 5 and the denominator is 12
Hence, the fraction is $\frac{3}{5}$