Exercise 3.11 - Chapter 3 Algebra 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.11$
Question $1 .$
Solve, using the method of substitution
(i) $2 \mathrm{x}-3 \mathrm{y}=7 ; 5 \mathrm{x}+\mathrm{y}=9$
(ii) $1.5 \mathrm{x}+0.1 \mathrm{y}=6.2 ; 3 \mathrm{x}-0.4 \mathrm{y}=11.2$
(iii) $10 \%$ of $x+20 \%$ of $y=24 ; 3 x-y=20$
(iv) $\sqrt{2} x-\sqrt{3} y=1 ; \sqrt{3} x-\sqrt{8} y=0$
Solution:
(i) $2 \mathrm{x}-3 \mathrm{y}=7$
$5 x+y=9$
Step (1)
From the equation (2)
$5 x+y=9$
$y=-5 x+9$
Step (2)
substitute (3) in (1)
$2 x-3(-5 x+9)=7$
$2 x+15 x-27=7$
$17 x=7+27$
$17 x=34$
$x=\frac{34}{17}=2 ; x=2$
Step (3)
substitute $x=2$ in (3)
$y=-5(2)+9=-10+9=-1$
Solution: $x=2 ; y=-1$
(ii) $1.5 \mathrm{x}+0.1 \mathrm{y}=6.2$
$3 x-0.1 y=11.2 \ldots \ldots \ldots \ldots \ldots$ (2)
Multiply (1) $x 1015 x+y=62 \ldots \ldots \ldots . .(1)$
(2) $\times 10 \Rightarrow 30 \mathrm{x}-4 \mathrm{y}=112$
Step (1)
From equation (3)
$15 x+y=62$
$y=-15 x+62$
Step (2)
substitute (5) in (4)
$30 x-4(-15 x+62)=112$

$\begin{aligned}
&15 x+y=62 \\
&y=-15 x+62 \ldots \ldots \ldots \ldots(5) \\
&\text { Step (2) } \\
&\text { substitute }(5) \text { in }(4) \\
&30 x-4(-15 x+62)=112 \\
&30 x+60 x-248=112 \\
&90 x=112+248 \\
&90 x=360 \\
&\text { Step }(3) \\
&\text { substitute } x=4 \text { in }(5) \\
&y=-15(4)+62 \\
&=-60+62 \\
&y=2
\end{aligned}$

Solution: $x=4 ; y=2$
(iii) $\Rightarrow \frac{10}{100} x+\frac{20}{100} \frac{y}{5}=24$
$\frac{x}{10}+\frac{y}{5}=24$
$\frac{\frac{x+2 y}{10}}{10}=24$
Step (1)
From equation (2)
$3 \mathrm{x}-\mathrm{y}=20$
$-y=20-3 x$
$y=3 x-20-(3)$
Step (2)
substitute (3) in (1)
$x+2(3 x-20)=240$
$x+6 x-40=240$
$7 x=240+40$
$x=\frac{280}{7}$
$x=40$
Step (3)
substitute $x=40$ in (3)
$y=3(40)-20$
$=120-20=100$
Solution : $x=40$ and $y=100$
(iv) $\sqrt{2} x-\sqrt{3} y=1$
$\sqrt{3} x-\sqrt{8} y=0$
Step (1)
From the equation (2)

$\begin{aligned}
&\sqrt{3} x-\sqrt{8} y=0 \\
&\not \sqrt{8} y=\frac{\sqrt{3}}{\sqrt{8}} x \\
&y=\quad
\end{aligned}$
Step (2)
substitute (3) in (1)
$\begin{aligned}
\sqrt{2} x-\sqrt{3} y &=1 \\
\sqrt{2} x-\sqrt{3}\left(\frac{\sqrt{3}}{\sqrt{8}}\right) x &=1 \\
\frac{\sqrt{2} x}{1}-\frac{3}{\sqrt{8}} x &=1 \\
\frac{\sqrt{16} x-3 x}{\sqrt{8}} &=\frac{4 x-3 x}{\sqrt{8}}=1 \quad x=\sqrt{8}
\end{aligned}$
Step (3)
$\text { substitute } x=\sqrt{8} \text { in (1) }$
$\begin{aligned}
\sqrt{2}(\sqrt{8})-\sqrt{3} y &=1 \\
\sqrt{16}-\sqrt{3} y &=1 \\
4-\sqrt{3} y &=1 \\
-\sqrt{3} y &=1-4 \\
f \sqrt{3} y &=\not 3 y=\sqrt{3}
\end{aligned}$
Solution: $x=\sqrt{8}$ and $y=\sqrt{3}$

Question $2 .$
Raman's age is three times the sum of the ages of his two sons. After 5 years his age will be twice the sum of the ages of his two sons. Find the age of Raman.
Solution:
Let Raman's age $=\mathrm{x}$
Let the sum of his two sons age $=y$
now $x=3 y \Rightarrow x-3 y=0$......... (1)
After 5 years,
Step (3)
$\begin{aligned}
&x+5=2(y+10) \\
&x+5=2 y+20 \\
&x-2 y=20-5 \\
&x-2 y=15
\end{aligned}$
Step (1)
From equation (1) $\mathrm{x}=3 \mathrm{y}$
Step (2)
Substitute $x=3 y$ in (2)
$\begin{aligned}
&3 y-2 y=15 \\
&y=15
\end{aligned}$
Step (3)
Substitute $y=15$ in (1)
$\begin{aligned}
&x=3 y=3 \times 15 \\
&x=45
\end{aligned}$
$\therefore$ Raman's age is 45 years.

Question 3 .
The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13 . If the digits are reversed, the number so formed exceeds the original number by 495 . Find the number.
Solution:
Let the number be $\mathrm{x} 0 \mathrm{y}$
If the digits are reversed the number so formed is $y 0 x$
$\begin{aligned}
&x 0 y=100 x+10 \times 0+1 \times y \\
&y 0 x=100 y+10 \times 0+1 \times x \\
&100 y+x-(100 x+y)=495 \\
&100 y+x-100 x-y=495 \\
&-99 x+99 y=495 \ldots \ldots \ldots \ldots . \ldots(2)
\end{aligned}$
(1) $\times-99 \Rightarrow-99 x-99 y=-1287$
(2)
$\begin{gathered}
\Rightarrow \frac{-99 x+99 y=49}{+198 x=+792} \\
x=\frac{792}{198}=4
\end{gathered}$
Substitute $x=4$ in (1)
$4+y=13=13-4=9$
The number is 409 .