Question

Solve the following equations:
$x+\sqrt{xy}+y=65$,
$x^2+xy+y^2=2275$.

• A. $x=45,y=5$
• B. $x=40,y=15$
• C. $x=20,y=5$
• D. $x=5,y=45$

Answer 1

Answer: (A), (D)

$x=45,y=5$
$x=5,y=45$

$x+\sqrt{xy}+y=65x+y=65-\sqrt{xy}$ ......(i)

$x^2+y^2+xy=2275$ .......(ii)

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

${(x+y)}^2-xy=2275$ ........(iii)

Substituting (i) in (ii), we have

${(65-\sqrt{xy})}^2-xy=2275$

$\Rightarrow 4225+xy-130\sqrt{xy}-xy=2275$

$\Rightarrow -130\sqrt{xy}=-1950$

$\Rightarrow \sqrt{xy}=15$

$\Rightarrow xy=225$

$\Rightarrow y=\frac{225}{x}$

$\Rightarrow x^2+{\left(\frac{225}{x}\right)}^2+x.\frac{225}{x}=2275$

$\Rightarrow x^2+\frac{50652}{x^2}=2050$

$\Rightarrow x^4-2050x^2+50652=0$

$\Rightarrow x^4-25x^2-2025x^2+50625=0$

$\Rightarrow x^2(x^2-25)-2025(x^2-25)=0$

$\Rightarrow (x^2-2025)(x^2-25)=0$

$\Rightarrow x^2=25,2025$

$\Rightarrow x=\pm 5,\pm 45$

Negative values of $x$ are not satisfied by (i)
So, $x=5,45$

Now, $y=\frac{225}{x}$

When $x=5\Rightarrow y=\frac{225}{5}=45$

When $x=45\Rightarrow y=\frac{225}{45}=5$

So, the values of $x$ are $5,45$ and corresponding values of $y$ are $45,5$.