Question

Solve for $x$: $\{\begin{array}{c}{x}^2+2xy+y^2=64\\ y-x=12\end{array}$

• A. $-2$
• B. $-4$
• C. $2$
• D. $10$

Answer 1

The correct option is A $-2$

Given, $x^2+2xy+y^2=64$

$\therefore (x+y{)}^2=64$ .......... $\because (x+y{)}^2=x^2+2xy+y^2$

$\Rightarrow x+y=8$....(1)

$\Rightarrow -x+y=12$....(2)

Add the (1) and (2), we get

$\Rightarrow 2y=20$

$\Rightarrow y=\frac{20}{2}=10$

Substitute the value of $y=10$ in (1)

$\Rightarrow x+10=8$

$\Rightarrow x=8-10$

$\Rightarrow x=-2$