Question

Solve by cross-multiplication method :
$(a-b)x+(a+b)y=2(a^2-b^2)$,
$(a+b)x-(a-b)y=4ab$

• A. $x=(a-b)$ and $y=(2a-b)$
• B. $x=(a+b)$ and $y=(a-b)$
• C. $x=(2a+5b)$ and $y=(a-b^2)$
• D. $x=(2a-3b)$ and $y=(3a-b)$

Answer 1

The correct option is B $x=(a+b)$ and $y=(a-b)$

Writing the equations in the standard form, we get.
$(a-b)x+(a+b)y=2(a^2-b^2)$,
$(a+b)x-(a-b)y=4ab$
Applying the cross-multiplication method, we get
$-4ab(a+b)-2(a-b)(a^2-b2)$
$=-2(a+b)[2ab+(a-b{)}^2]$
$=-2(a+b)(2ab+a^2+b^2-2ab)$
$=-2(a+b)(a^2+b^2)$
Simplification of the expression under $y$ :
$-2(a^2-b^2)(a+b)+4ab(a-b)$
$=-2(a-b)\left[\right(a+b\left)\right(a+b)-2ab]$
$=-2(a-b)(a^2+b^2+2ab-2ab)$
$=-2(a-b)(a^2+b^2)$
Simplification of the expression under $1$ :
$-(a-b{)}^2-(a+b{)}^2$
$=-(a^2+b^2-2ab)-(a^2+b^2+2ab)$
$=-2(a^2+b^2)$
Hence,
$\frac{x}{-2(a+b)(a^2+b^2)}=\frac{y}{-2(a-b)(a^2+b^2)}=\frac{1}{-2(a^2+b^2)}$
$\Rightarrow \frac{x}{a+b}=\frac{y}{a-b}=\frac{1}{1}$
$\Rightarrow x=(a+b)$ and $y=(a-b)$