Question

$3x+4y=0$ , $\frac{2}{3}y-\frac{3}{4}x=20$
Consider the system of equations above. If (x, y) is the solution to the system, then what is the value of $\frac{x}{y}$?

• A. $-\frac{4}{3}$
• B. $-\frac{3}{4}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $-\frac{4}{3}$

Given : $3x+4y=0$ , $\frac{2}{3}y-\frac{3}{4}x=20$

Multiply the second equation by 4, which will help in eliminating the $x$ - term by adding the equations.

$\Rightarrow$ $4(\frac{2}{3}y-\frac{3}{4})=20\times 4$
$\Rightarrow$ $\frac{8}{3}y-3x=80$
$\Rightarrow$ $-3x+\frac{8}{3}y=80$

Add these two equations, we get

$\Rightarrow$ $3x+4y=0$
$\Rightarrow$ $-3x+\frac{8}{3}y=80$
$\Rightarrow$ $\frac{20}{3}y=80$
$\Rightarrow$ $y=3\times 4$
$\Rightarrow$ $y=12$

Now, plug $y=12$ back into the equation, we get

$\frac{2}{3}y-\frac{3}{4}x=20$
$\Rightarrow$ $\frac{2}{3}12-\frac{3}{4}x=20$
$\Rightarrow$ $-\frac{3}{4}x=20-8$
$\Rightarrow$ $-\frac{3}{4}x=12$
$\Rightarrow$ $x=-16$

The solution for the system of equation is $(-16,12)$

The value of $\frac{x}{y}=-\frac{4}{3}$