Question

Solve $31x^2{y}^2-7y^4-112xy+64=0$,
$x^2-7xy+4y^2+8=0$

• A. $x=\pm 3,y=\pm 1$
• B. $x=\pm 1,y=\pm 3$
• C. $x=2,y=3$
• D. $x=3,y=4$

Answer 1

The correct option is A $x=\pm 3,y=\pm 1$

Given, $31x^2{y}^2-7y^4-112xy+64=0$ .....(1)

and $x^2-7xy+4y^2+8=0$ ......(2)

From (2) we have $-8=x^2-7xy+4y^2$; and substituting in (1),
$31x^2{y}^2-7y^4+14xy(x^2-7xy+4y^2)+(x2-7xy+4y^2{)}^2=0$;
$\therefore 31x^2{y}^2-7y^4+(x^2-7xy+4y^2)(14xy+x^2-7xy+4y^2)=0$
$\therefore 31x^2{y}^2-7y^4+(x^2+4y^2{)}^2-(7xy{)}^2=0$;
that is, $x^4-10x^2{y}^2+9y^4=0$ ........ (3)
$\therefore (x^2-y^2)(x^2-9y^2)=0$;
Hence, $x=\pm y$, or $x=\pm 3y$
Taking theses cases in succession and substituting in (2), we obtain
$x=y=\pm 2$;
$x=-y=\pm \sqrt{-\frac{2}{3}}$;
$x=\pm 3,y=\pm 1$;
$x=\pm 3\sqrt{-\frac{4}{17}},y=\mp \sqrt{-\frac{4}{17}}$.