Question

Find the value of x and y using elimination method:
$\frac{-1}{x}+\frac{2}{y}=0$ and $\frac{x}{2}+\frac{y}{3}=1$

• A. $(\frac{6}{7},\frac{-12}{7})$
• B. $(\frac{-6}{7},\frac{12}{7})$
• C. $(\frac{6}{7},\frac{12}{7})$
• D. $(\frac{-6}{7},\frac{-12}{7})$

Answer 1

Answer: (C)

$(\frac{6}{7},\frac{12}{7})$

First cross multiply the given equation,
$\frac{-1}{x}+\frac{2}{y}=0\Rightarrow 2x-y=0$ ----- (1)
$\frac{x}{2}+\frac{y}{3}=1\Rightarrow 3x+2y=6$ ----- (2)
Multiply equation (1) by 2 and then add the equation (3) and (4)
$4x-2y=0$ ----- (3)
$3x+2y=6$ ----- (4)
_______________
$7x=6$
$x=\frac{6}{7}$
Put the value of $x$ in equation (1), we get
$2x-y=0$
$2\times \frac{6}{7}-y=0$
$y=\frac{12}{7}$
Therefore the solution is $(\frac{6}{7},\frac{12}{7})$