| Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
Exercise 3.4
Question 1 A
Solve the following pair of linear equation by cross - multiplication method:
8x + 5y = 93x + 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
$\Rightarrow \frac{x}{-2}=\frac{1}{1}$
⇒ x = – 2
On taking II and III ratio, we get
$\Rightarrow \frac{y}{5}=\frac{1}{1}$
⇒ y = 5
3x + 5y = 74
Sol :
Given, pair of equations is
2x + 3y – 46 = 0 and 3x + 5y – 74 = 0
By cross - multiplication method, we have
$\Rightarrow \frac{y}{5}=\frac{1}{1}$
⇒ y = 5
Question 1 B
Solve the following pair of linear equation by cross - multiplication method:
2x + 3y = 463x + 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
$\Rightarrow \frac{x}{8}=\frac{1}{1}$
⇒ x = 8
On taking II and III ratio, we get
$\Rightarrow \frac{y}{10}=\frac{1}{1}$
⇒ y = 10
5x – 1 = 3y
Sol :
Given, pair of equations is
x + 4y + 9 = 0 and 5x – 3y – 1 = 0
By cross - multiplication method, we have
$\Rightarrow \frac{y}{10}=\frac{1}{1}$
⇒ y = 10
Question 1 C
Solve the following pair of linear equation by cross - multiplication method:
x + 4y + 9 = 05x – 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
$\Rightarrow \frac{x}{23}=\frac{1}{-23}$
⇒ x = – 1
On taking II and III ratio, we get
$\Rightarrow \frac{y}{46}=\frac{1}{-23}$
⇒ y = – 2
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
$\Rightarrow \frac{y}{46}=\frac{1}{-23}$
⇒ 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$$\frac{5}{x}-\frac{4}{y}=-2$
Sol :
Given, pair of equations is
$\frac{2}{x}+\frac{3}{y}=13$
$\frac{5}{x}-\frac{4}{y}=-2$
Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$
So, Eq. (1) and (2) reduces to
2u + 3v – 13 = 0
5u – 4v + 2 = 0
2u + 3v – 13 = 0
5u – 4v + 2 = 0
By cross - multiplication method, we have
$\Rightarrow \frac{u}{6-52}=\frac{v}{-65-4}=\frac{1}{-8-15}$
$\Rightarrow \begin{matrix}\frac{u}{-46}&=\frac{v}{-69}&=\frac{1}{-23}\\ \text{I}&\text{II}&\text{III}\end{matrix}$
On taking I and III ratio, we get
$\Rightarrow \frac{\mathrm{u}}{-46}=\frac{1}{-23}$
⇒ u = 2
$\Rightarrow \frac{\mathrm{u}}{-46}=\frac{1}{-23}$
⇒ u = 2
On taking II and III ratio, we get
$\Rightarrow \frac{\mathrm{v}}{-69}=\frac{1}{-23}$
⇒ v = 3
$\Rightarrow \frac{\mathrm{v}}{-69}=\frac{1}{-23}$
⇒ v = 3
So, $u=\frac{1}{x}=2 \Rightarrow x=\frac{1}{2}$
and $v=\frac{1}{y}=3 \Rightarrow y=\frac{1}{3}$
and $v=\frac{1}{y}=3 \Rightarrow y=\frac{1}{3}$
Question 1 F
Solve the following pair of linear equation by cross - multiplication method:
$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$$\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
Sol :
Given, pair of equations is
$\frac{x}{6}+\frac{y}{15}=4$
⇒5x+2y=4×30
⇒5x+2y-120=0
⇒5x+2y-120=0
And $\frac{x}{3}-\frac{y}{12}=\frac{19}{4}$
⇒4x-y=57
By cross - multiplication method, we have
$\Rightarrow \frac{x}{-114-120}=\frac{y}{-480+285}=\frac{1}{-5-8}$
$\Rightarrow \begin{matrix}\frac{x}{-234}&=\frac{y}{-195}&=\frac{1}{-13}\
\ \text{I}& \text{II}& \text{III}\end{matrix}$
$\Rightarrow \frac{x}{-234}=\frac{1}{-13}$
⇒ x = 18
On taking II and III ratio, we get
$\Rightarrow \frac{y}{-195}=\frac{1}{-13}$
⇒ y = 15
$\Rightarrow \frac{y}{-195}=\frac{1}{-13}$
⇒ y = 15
Question 2 A
Solve the following pair of equations by cross - multiplication method.
ax + by = a – bbx – 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
$\Rightarrow \frac{x}{-b^{2}-a^{2}}=\frac{1}{-a^{2}-b^{2}}$
⇒ 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
$\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$
$a \neq 0, b \neq 0$
Sol :
Given, pair of equations is
$\Rightarrow \frac{y}{b^{2}+a^{2}}=\frac{1}{-a^{2}-b^{2}}$
⇒ 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}$
$\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
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
$\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 + bax + 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
$\Rightarrow \frac{\mathrm{ay}}{6}=\frac{\mathrm{ab}}{-3}$
⇒ 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
$\Rightarrow \frac{x}{b^{2}-4 a b}=\frac{1}{-5 a b}$
$\Rightarrow x=\frac{b(b-4 a)}{-5 a b}$
$\Rightarrow x=\frac{4 a-b}{5 a}$
$\Rightarrow \frac{x}{b^{2}-4 a b}=\frac{1}{-5 a b}$
$\Rightarrow x=\frac{b(b-4 a)}{-5 a b}$
$\Rightarrow x=\frac{4 a-b}{5 a}$
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
$\frac{x}{a}+\frac{y}{b}=2$
$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-2=0$
$\Rightarrow \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}-2=0$
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
$\Rightarrow \frac{b x}{b^{2}-a^{2}}=\frac{a b}{b^{2}-a^{2}}$
⇒ 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
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
$\Rightarrow \frac{a y}{b^{2}-a^{2}}=\frac{a b}{b^{2}-a^{2}}$
⇒ y = b
Question 3
Solve the following system of equations by cross - multiplication method.
a(x + y) + b(x – y) = a2 – ab + b2a(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
$\Rightarrow \frac{x}{-(a-b)\left(a^{2}+a b+b^{2}\right)+(a+b)\left(a^{2}-a b+b^{2}\right)}$
$=\frac{y}{-\left(a^{2}-a b+b^{2}\right)(a-b)+\left(a^{2}+a b+b^{2}\right)(a+b)}=\frac{1}{(a+b)(a+b)-(a-b)(a-b)}$
$\Rightarrow \frac{x}{-\left(a^{3}-b^{3}\right)+\left(a^{3}+b^{3}\right)}$
$=\frac{y}{-a^{3}+a^{2} b-b^{2} a+b a^{2}-a b^{2}+b^{3}+a^{3}+a^{2} b+b^{2} a+b a^{2}+a b^{2}+b^{3}}$
$=\frac{1}{a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2}}$
$\Rightarrow \begin{matrix}\frac{x}{2 b^{2}}&=\frac{y}{4 a^{2} b+2 b^{2}}&=\frac{1}{4 a b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$
$\Rightarrow \frac{x}{-\left(a^{3}-b^{3}\right)+\left(a^{3}+b^{3}\right)}$
$=\frac{y}{-a^{3}+a^{2} b-b^{2} a+b a^{2}-a b^{2}+b^{3}+a^{3}+a^{2} b+b^{2} a+b a^{2}+a b^{2}+b^{3}}$
$=\frac{1}{a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2}}$
$\Rightarrow \begin{matrix}\frac{x}{2 b^{2}}&=\frac{y}{4 a^{2} b+2 b^{2}}&=\frac{1}{4 a b}\\ \text{I}&\text{II}&\text{III}\end{matrix}$
On taking I and III ratio, we get
$\Rightarrow \frac{x}{2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow x=\frac{b^{2}}{2 a}$
$\Rightarrow \frac{x}{2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow x=\frac{b^{2}}{2 a}$
On taking II and III ratio, we get
$\Rightarrow \frac{y}{4 a^{2} b+2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow \mathrm{y}=\frac{2 \mathrm{a}^{2}+\mathrm{b}^{2}}{2 \mathrm{a}}$
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 a2 = 1, b2 = – 1 and c2 = – 5
$\Rightarrow \frac{y}{4 a^{2} b+2 b^{3}}=\frac{1}{4 a b}$
$\Rightarrow \mathrm{y}=\frac{2 \mathrm{a}^{2}+\mathrm{b}^{2}}{2 \mathrm{a}}$
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 = 03x – 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 a2 = 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 $\frac{c_{1}}{c_{2}}=\frac{-7}{-5}=\frac{7}{5}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}$
∴ 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
$\Rightarrow \frac{x}{8}=\frac{1}{2}$
⇒ x = 4
On taking II and III ratio, we get
$\Rightarrow \frac{y}{-2}=\frac{1}{2}$
⇒ y = – 1
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 a2 = 3, b2 = 2 and c2 = – 8
$\Rightarrow \frac{y}{-2}=\frac{1}{2}$
⇒ 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 = 53x + 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 a2 = 3, b2 = 2 and c2 = – 8
$\frac{a_{1}}{a_{2}}=\frac{2}{3}$ , $\frac{b_{1}}{b_{2}}=\frac{1}{2}$ and $\frac{c_{1}}{c_{2}}=\frac{-5}{-8}=\frac{5}{8}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ 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
$\Rightarrow \frac{x}{2}=1$
⇒ x = 2
On taking II and III ratio, we get
$\Rightarrow \frac{y}{1}=1$
⇒ y = 1
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 a2 = 3, b2 = – 5 and c2 = – 20
$\Rightarrow \frac{y}{1}=1$
⇒ 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 = 206x – 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 a2 = 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 $\frac{c_{1}}{c_{2}}=\frac{-20}{-20}=1$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
∴ Given pair of equations has infinitely many solutions.
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 a2 = 3, b2 = – 9 and c2 = – 2
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 = 03x – 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 a2 = 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 $\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ 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 = 22x + 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 a2 = 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 $\frac{c_{1}}{c_{2}}=\frac{-2}{-2}=1$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
∴ Given pair of equations has infinitely many solution
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 a2 = 1, b2 = 1 and c2 = – 3
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 = 22x + 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 a2 = 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 $\frac{c_{1}}{c_{2}}=\frac{-2}{-3}=\frac{2}{3}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ 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)
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
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
$\Rightarrow \frac{\mathrm{u}}{-27}=\frac{1}{5}$
$\Rightarrow \frac{\mathrm{u}}{-27}=\frac{1}{5}$
$\Rightarrow \mathrm{u}=\frac{-27}{5}$
On taking II and III ratio, we get
$\Rightarrow \frac{v}{65}=\frac{1}{5}$
⇒ v = 13
$\Rightarrow \frac{v}{65}=\frac{1}{5}$
⇒ 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
$2 x=-\frac{5}{27}+\frac{1}{13}$
$\Rightarrow 2 \mathrm{x}=\frac{-65+27}{27 \times 13}$
$\Rightarrow 2 \mathrm{x}=\frac{-38}{351}$
$\Rightarrow x=\frac{-19}{351}$
$2 x=-\frac{5}{27}+\frac{1}{13}$
$\Rightarrow 2 \mathrm{x}=\frac{-65+27}{27 \times 13}$
$\Rightarrow 2 \mathrm{x}=\frac{-38}{351}$
$\Rightarrow x=\frac{-19}{351}$
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
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
$\Rightarrow \frac{\mathrm{y}}{2 \mathrm{a}^{3}+2}=\frac{1}{2 \mathrm{a}^{2}}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{a}^{3}+1}{\mathrm{a}^{2}}$
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)
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.
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.
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 a2 = 1, b2 = 2 and c2 = – 6
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?
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 a2 = 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.
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
$\Rightarrow \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$
∴ 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.
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
$\Rightarrow \frac{x}{2000}=\frac{1}{1}$
⇒ x = 2000
On taking II and III ratios, we get
$\Rightarrow \frac{y}{4000}=\frac{1}{1}$
⇒ y = 4000
$\Rightarrow \frac{y}{4000}=\frac{1}{1}$
⇒ y = 4000
Hence, the monthly incomes of two persons are 9(2000) = Rs18000 and 7(2000) = Rs14000
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
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?
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
$\Rightarrow \frac{y}{-4}=\frac{1}{-2}$
⇒ 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.
Sol :
Let the numerator = x
and the denominator = 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?
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)
$\frac{x+1}{y-1}=1$
⇒ 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
$\frac{x}{y+1}=\frac{1}{2}$
⇒ 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
⇒ y = 5
So, the numerator is 3 and the denominator is 5
Hence, the fraction is $\frac{3}{5}$
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
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.
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
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
$\Rightarrow \frac{y}{35}=\frac{1}{7}$
⇒ 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.
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
$\Rightarrow \begin{matrix}\frac{x}{2400}&=\frac{y}{180}&=\frac{1}{6}\\ \text{I}&\text{II}&\text{III}\end{matrix}$
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
Sol :
$\Rightarrow \frac{y}{180}=\frac{1}{6}$
⇒ 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.
Let the numerator = x
and the denominator = y
and the denominator = y
So, the fraction $=\frac{x}{y}$
According to the question,
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)
$\frac{x-1}{y}=\frac{1}{3}$
⇒ 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
$\frac{x}{y+8}=\frac{1}{4}$
⇒ 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
$\Rightarrow \begin{matrix}\frac{x}{5}&=\frac{y}{12}&=\frac{1}{1}\\ \text{I}&\text{II}&\text{III}\end{matrix}$
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 $\frac{3}{5}$
$\Rightarrow \frac{y}{12}=\frac{1}{1}$
⇒ y = 12
So, the numerator is 5 and the denominator is 12
Hence, the fraction is $\frac{3}{5}$
| 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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