Question

Solve the following system of equations by elimination method.

$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6},\frac{3}{x}+\frac{2}{y}=0,x\ne 0,y\ne 0$

• A. $(6,-4)$
• B. $(6,4)$
• C. $(-6,-4)$
• D. None of these

Answer 1

The correct option is A $(6,-4)$

Consider the given equations.

$\frac{2}{x}+\frac{2}{3y}=\frac{1}{6}$

$4x+12y=xy$ $-------\left(1\right)$

$\frac{3}{x}+\frac{2}{y}=0$

$2x+3y=0$ $-------\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

Multiply by $2$ in $\left(2\right)$ and by $1$ in $\left(1\right)$ and subtract

$4x+12y=xy$

$4x+6y=0$

—————————

$6y=xy$

$x=6$ $[$ put in $\left(2\right)]$

$2\times 6+3y=0$

$12+3y=0$

$3y=-12$

$y=-4$

Hence, the value of $x$ is $6$ and the value of $y$ is $-4$.