Question

Solve the given equations simultaneously:

$\frac{x}{a}+\frac{y}{b}=2$ and $ax-by=a^2-b^2$.

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

Answer 1

The correct option is C $x=a,y=b$

The given system of equations may be written as
$bx+ay-2ab=0$
$ax-by-(a^2-b^2)=0$

By cross-multiplication, we have
$\frac{x}{-a(a^2-b^2)-(-b)(-2ab)}=\frac{-y}{-b(a^2-b^2)-a(-2ab)}=\frac{1}{b\times -b-a\times a}$

$\Rightarrow \frac{x}{-a(a^2-b^2)-2a{b}^2}=\frac{-y}{-b(a^2-b^2)+2a^{2}b}=\frac{1}{-b^2-a^2}$

$\Rightarrow \frac{x}{-a(a^2-b^2+2b^2)}=\frac{-y}{-b(a^2-b^2-2a^2)}=\frac{1}{-(a^2+b^2)}$

$\Rightarrow \frac{x}{-a(a^2+b^2)}=\frac{-y}{-b(-a^2-b^2)}=\frac{1}{-(a^2+b^2)}$

$\Rightarrow x=\frac{-a(a^2+b^2)}{-(a^2+b^2)}=a$ and $y=\frac{-b(a^2+b^2)}{-(a^2+b^2)}=b$

Hence, the solution of the given system of equations is $x=a,y=b$.