Question

Solve the following equations:
$x^2{y}^2+400=41xy,y^2=5xy-4x^2$.

• A. $(\pm 2,\pm 8)$
• B. $(\pm 4,\pm 4)$
• C. $(\pm 8,\pm 4)$
• D. $(\pm 2,\pm 3)$

Answer 1

Answer: (A), (B)

$(\pm 2,\pm 8)$
$(\pm 4,\pm 4)$

Let $y^2=5xy-4x^2$ ..........(i)

Given, $x^2{y}^2+400=41xy$

$\Rightarrow x^2{y}^2-41xy+400=0\Rightarrow x^2{y}^2-16xy-25xy+400=0\Rightarrow xy(xy-16)-25(xy-6)=0\Rightarrow (xy-25)(xy-6)=0\Rightarrow xy=25,6\Rightarrow y=\frac{25}{x},\frac{16}{x}$

Substituting $y$ in (i)

(1) $y=\frac{25}{x}$

Therefore, ${\left(\frac{25}{x}\right)}^2=5x\left(\frac{25}{x}\right)-4x^2$

$\Rightarrow \frac{625}{x^2}+4x^2=125\Rightarrow {4x}^4-125x^2+625=0\Rightarrow {4x}^4-100x^2-25x^2+625=0\Rightarrow 4x^2(x^2-25)-25(x^2-25)=0\Rightarrow (4x^2-25)(x^2-25)=0\Rightarrow x^2=25,\frac{25}{4}\Rightarrow x=\pm 5,\pm \frac{5}{2}$

Thus $y=\frac{25}{x}$

$\Rightarrow y=\pm 5,\pm 10$

(2) $y=\frac{16}{x}$

Therefore, ${\left(\frac{16}{x}\right)}^2=5x\left(\frac{16}{x}\right)-4x^2$

$\Rightarrow \frac{256}{x^2}+4x^2=80\Rightarrow x^4-20x^2+64=0\Rightarrow x^4-4x^2-16x^2+64=0\Rightarrow x^2(x^2-4)-16(x^2-4)=0\Rightarrow (x^2-16)(x^2-4)=0\Rightarrow x^2=4,16$

$\Rightarrow x=\pm 2,\pm 4$

Thus $y=\frac{16}{x}$

$\Rightarrow y=\pm 8,\pm 4$

So, the values of $x$ are $\pm 5,\pm \frac{5}{2},\pm 2,\pm 4$ and corresponding values of $y$ are $\pm 5,\pm 10,\pm 8,\pm 4$.