Question

Solve the following equations:
$x+y=7+\sqrt{xy}$,
$x^2+y^2=133-xy$

• A. $x=9,y=4$
• B. $x=4,y=8$
• C. $x=2,y-7$
• D. $x=5,y=12$

Answer 1

The correct option is A $x=9,y=4$

$x+y=7+\sqrt{xy}$ ......(i)

$x^2+y^2=133-xy{x}^2+y^2+xy=133$ .......(ii)

using ${(a+b)}^2{=a}^2+b^2+2ab$

${(x+y)}^2-xy=133$ ........(iii)

Substituting (i) in (iii), we get

${(7+\sqrt{xy})}^2-xy=133\Rightarrow 49+xy+14\sqrt{xy}-xy=133\Rightarrow 14\sqrt{xy}=84\Rightarrow \sqrt{xy}=6\Rightarrow xy=36$

$\Rightarrow y=\frac{36}{x}$

Thus $x^2+{\left(\frac{36}{x}\right)}^2+x.\frac{36}{x}=133$ ....(iv)

From (ii), we have

$x^2+\frac{1296}{x^2}=97\Rightarrow x^4-97x^2+1296=0\Rightarrow x^4-16x^2-81x^2+1296=0\Rightarrow x^2(x^2-16)-81(x^2-16)=0\Rightarrow (x^2-16)(x^2-81)=0\Rightarrow x^2=16,81\Rightarrow x=\pm 4,\pm 9$

Negative values are not true for equation (i)
So, $x=4,9$

Now, $y=\frac{36}{x}$

$x=4\Rightarrow y=\frac{36}{4}=9x=9\Rightarrow y=\frac{36}{9}=4$

So, the values of $x$ are $4,9$ and corresponding values of $y$ are $9,4$.