Question

The List II gives the solution to the pair of equations given in the List I, match them correctly.

Answer 1

Let $2x+y=8......\left(1\right)$
$x+6y=15......\left(2\right)$
$\left(1\right)\times 1-\left(2\right)\times 2\Rightarrow -11y=-22\Rightarrow y=2$
Substitute the value of $y$ in $\left(1\right)$
$\Rightarrow 2x+2=8$
$\Rightarrow 2x=6\Rightarrow x=3$
$\therefore x=3$ and $y=2\Rightarrow (3,2)$
Hence $A\to 4$
Let $5x+3y=35......\left(1\right)$
$2x+4y=28......\left(2\right)$
$\left(1\right)\times 4-\left(2\right)\times 3\Rightarrow 14x=56\Rightarrow x=4$
Substitute the value of $x$ in $\left(1\right)$
$\Rightarrow 2\left(4\right)+4y=28$
$\Rightarrow 4y=20$
$\Rightarrow y=5$
$\therefore x=4$ and $y=5\Rightarrow (4,5)$
Hence $B\to 3$
Let $\frac{1}{7x}+\frac{1}{6y}=3......\left(1\right)$
$\frac{1}{2x}-\frac{1}{3y}=5......\left(2\right)$
$\left(1\right)\times 2+\left(2\right)\times 1\Rightarrow \frac{11}{14x}=11$
$\Rightarrow x=\frac{1}{14}$
Substitute the value of $x$ in $\left(2\right)$
$\Rightarrow \frac{1}{2\left(\frac{1}{14}\right)}-\frac{1}{3y}=5$
$\Rightarrow \frac{1}{3y}=2$
$\Rightarrow y=\frac{1}{6}$
$\therefore x=\frac{1}{14}$ and $y=\frac{1}{6}\Rightarrow (\frac{1}{14},\frac{1}{6})$
Hence $C\to 2$
Let $15x+4y=61......\left(1\right)$
$4x=15y=72......\left(2\right)$
$\left(1\right)\Rightarrow 4y=61-15x$
$\Rightarrow y=\frac{61}{4}-\frac{15x}{4}$
Substitute the value of $y$ in $\left(2\right)$
$\Rightarrow 4x+15(\frac{61}{4}-\frac{15x}{4})-72=0$
$\Rightarrow 4x+\frac{915}{4}-\frac{225}{4}-72=0$
$\Rightarrow \frac{-209x}{4}+\frac{627}{4}=0$
$\Rightarrow x=3$
Substitute the value of $x$ in $\left(2\right)$
$\Rightarrow 4\left(3\right)+15y=72$
$\Rightarrow 15y=60$
$\Rightarrow y=4$
$\therefore x=3$ and $y=4\Rightarrow (3,4)$
Hence $D\to 1$