Question

Solve:
$\frac{bx}{a}-\frac{ay}{b}+a+b=0$; $bx-ay+2ab=0$

• A. $x=-a,y=b$
• B. $x=b,y=ab$
• C. $x=-ab,y=a-b$
• D. $x=-b,y=b-a$

Answer 1

The correct option is A $x=-a,y=b$

$\frac{bx}{a}-\frac{ay}{b}+a+b=0$
$b^{2}x-a^{2}y+a^{2}b+a{b}^2=\mathrm{0......}\left(i\right)$
$bx-ay+2ab=\mathrm{0.......}\left(ii\right)$
Multiply equation $\left(ii\right)$ by $b$
$b^{2}x-a^{2}y+a^{2}b+a{b}^2=0$
$b^{2}x-aby+2a{b}^2=0$
--------------------------------------------------------------------
$-a^{2}y+aby+a^{2}b-a{b}^2=0$
$ay(-a+b)=-a^{2}b+a{b}^2$
$ay(-a+b)=ab(b-a)$
$ay=ab$
$y=b$
Substitute $y=b$ in equation $\left(i\right)$
$b^{2}x-a^{2}b+a{b}^2=0$
$b^{2}x=-a^{2}b$
$x=-a$
So, $x=-a,y=b$