Question

If $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$,then the value of $x^2+y^2$ is equal to:

• A. $0$
• B. $5$
• C. $12$
• D. $40$

Answer 1

The correct option is D $40$

$\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$
$\Rightarrow \frac{x+y-8}{2}=\frac{x+2y-14}{3}$
$\Rightarrow 3x+3y-24=2x+4y-28$
$\Rightarrow x-y=-4\cdots \left(i\right)$
and$\frac{x+2y-14}{3}=\frac{3x-y}{4}$
$\Rightarrow 4x+8y-56=9x-3y$
$\Rightarrow 5x-11y=-56\cdots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$
$D=\left[\begin{array}{cc}1& -1\\ 5& -11\end{array}\right]=-11\times 1-(-1)\times 5=-11+5=-6D_x=\left[\begin{array}{cc}-4& -1\\ -56& -11\end{array}\right]=-11\times (-4)-(-1)\times (-56)=44-56=-12D_y=\left[\begin{array}{cc}1& -4\\ 5& -56\end{array}\right]=-56\times 1-(-4)\times 5=-56+20=-36\therefore x=\frac{D_x}{D}=\frac{-12}{-6}=2y=\frac{D_y}{D}=\frac{-36}{-6}=6$
So, we have: $(x,y)=(2,6)\Rightarrow x^2+y^2=40$