Solve the simultaneous equations involving x, y, and their squares

[chwala]

Homework Statement

See attached

Relevant Equations

simultaneous equations

In my approach,

$y=6-x$

$x^2(6-x)^2+2x(6-x)-35=0$

...

$x^4-12x^3+34x^2+12x-35=0$

using factor theorem, i noted that $f(1)=0$ therefore $(x-1)$ is a factor.

on long division we also have other factors:$(x+1), (x-5)$ and $(x-7)$ from these we can find $y$ values by using $y=6-x$. This may be the only approach ....quite easy this one.

 

[Steve4Physics]

Let $w = xy$, then $x^2y^2 + 2xy - 35 = 0$ becomes $w^2 +2w - 35 = 0$.

$(w-5)(w+7) = 0$. I.e $xy = 5$ or $xy=-7$. And take it from there.

 

[Mark44]

Similar to @Steve4Physics's approach, but factoring the quadratic in x and y first:
$x^2y^2 + 2xy - 35 = 0 \Rightarrow (xy + 7)(xy - 5) = 0$
$\Rightarrow xy = -7 \text{ or } xy = 5$

Solve $x + y = 6$ for y (or x) and substitute this expression into the equations above.