Exercise 3.12 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.12$
Question $1 .$
Solve by the method of elimination
(i) $2 \mathrm{x}-\mathrm{y}=3 ; 3 \mathrm{x}+\mathrm{y}=7$
(ii) $\mathrm{x}-\mathrm{y}=5 ; 3 \mathrm{x}+2 \mathrm{y}=25$
(iii) $\frac{x}{10}+\frac{y}{5}=14 ; \frac{x}{8}+\frac{y}{6}=15$
(iv) $3(2 x+y)=7 x y ; 3(x+3 y)=11 x y$
(v) $\frac{4}{x}+5 y=7 ; \frac{3}{x}+4 y=5$
(vi) $13 \mathrm{x}+11 \mathrm{y}=70 ; 11 \mathrm{x}+13 \mathrm{y}=74$
Solution:
(i) $2 x-y=3$
$3 x+y=7 \ldots . . \ldots . . .$ (2)
(1) $+(2) \Rightarrow$
$\begin{aligned}
2 x-y &=3 \\
3 x+y &=7 \\
\hline 5 x \quad &=10 \\
x &=\frac{10}{5}=2
\end{aligned}$
Substitute $\mathrm{x}=2$ in (1)
$\begin{aligned}
&2(2)-y=3 \\
&4-y=3
\end{aligned}$
$-y=3-4$
$-\mathrm{y}=-1$
$\therefore$ Solution: $x=2 ; y=1$
Verification:
Substitute $x=2, y=1$ in $(2)$
$3(2)+1=7=$ RHS
$\therefore$ Verified.

Substitute $y=2$ in (1)
$\begin{aligned}
&x-2=5 \\
&x=5+2 \\
&x=7
\end{aligned}$
$\therefore$ Solution: $\mathrm{x}=7, \mathrm{y}=2$
Verification:
Substitute $x=7, y=2$ in $(2)$
$\begin{aligned}
&3(7)+2(2)=21+4=25=\text { RHS } \\
&\therefore \text { Verified. }
\end{aligned}$

Substitute $y=30$ in (1)
$x+2(30)=140$ $x+60=140$ $x=140-60$ $x=80$ $\therefore$ Solution: $x=80 ; y=30$ Verification: Substitute $x=80, y=30$ in $(2)$ $3(80)+4(30)=240+120=360=$ RHS $\therefore$ Verified.
Substitute $x=80, y=30$ in (2)
$3(80)+4(30)=240+120=360=$ RHS
$\therefore$ Verified.
(iv) $3(2 x+y)=7 x y \Rightarrow 6 x+3 y=7 x y$ (1)

3(x + 3y) = 11xy ⇒ 3x + 9y = 11xy ………….. (2)

Substitute a $=1$ in (5)
$6 b+3(1)=7$
$6 \mathrm{~b}+3=7$
$6 \mathrm{~b}=7-3$
$\mathrm{b}=\frac{4}{6}=\frac{2}{3}$
$\therefore \mathrm{a}=\frac{1}{x}=1 \Rightarrow \mathrm{x}=1$
$\mathrm{b}=\frac{1}{y}=\frac{2}{3} \Rightarrow \mathrm{y}=\frac{3}{2}$
$\therefore$ Solution: $\mathrm{x}=1 ; \mathrm{y}=\frac{3}{2}$

Substitute $y=4$ in (1)
$\begin{aligned}
&13 x+11(4)=70 \\
&13 x+44=70 \\
&13 x=70-44=26 \\
&x=\frac{26}{13}=2
\end{aligned}$
$\therefore$ Solution: $\mathrm{x}=2 ; \mathrm{y}=4$
 

Question 2.
The monthly income of $A$ and $B$ are in the ratio $3: 4$ and their monthly expenditures are in the ratio $5: 7$. If each saves $₹ 5,000$ per month, find the monthly income of each.
Solution:
Let the monthly income of $A$ and $B$ be $3 x$ and $4 x$ respectively.
Let the monthly expenditure of $\mathrm{A}$ and $\mathrm{B}$ be $5 \mathrm{y}$ and $7 \mathrm{y}$ respectively.
$\therefore 3 \mathrm{x}-5 \mathrm{y}=5000 \ldots \ldots . . .$ (1)
$4 x-7 y=5000 \ldots \ldots \ldots . .(2)$

Substitute $y=5000$ in (1)
$\begin{aligned}
&3 x-5(5000)=5000 \\
&3 x-25000=5000 \\
&3 x=5000+25000 \\
&3 x=30000 \\
&x=10000
\end{aligned}$
$\therefore$ Monthly income of $\mathrm{A}$ is $3 \mathrm{x}=3 \times 10000=₹ 30000$
Monthly income of B is $4 x=4 \times 10000=₹ 40000$
 

Question $3 .$
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age.
Solution:
Let the man's present age $=x$
Five years ago his age is $=x-5$
Let his son's age be $=\mathrm{y}$
5 years ago his son's age $=\mathrm{y}-5$
$\begin{aligned}
&\therefore x-5=7(y-5) \\
&x-5=7 y-35 \\
&x-7 y=-35+5 \\
&x-7 y=-30 \ldots \ldots . . .(1)
\end{aligned}$
After 5 years, man's age will be $=x+5$
His son's age will be $=\mathrm{y}+5$
$\begin{aligned}
&\therefore x+5=4(y+5) \\
&x+5=4 y+20 \\
&x-4 y=20-5 \\
&\Rightarrow x-4 y=15 \ldots .
\end{aligned}$

Substitute $\mathrm{y}=15$ in (1)
$\begin{aligned}
&x-7(15)=-30 \\
&x-105-30 \\
&x=-30+105 \\
&x=75 \\
&\therefore \text { Man's Age }=75, \text { His son's Age }=15
\end{aligned}$