Question

Solve the equations:
x^2+y^2=13,
xy+6=0

Answer 1

using eqn 1
x=-6/y
then substitute in eqn 2
(-6/y)^2+y^2=13

Answer 2

whats after this ??

Answer 3

Multiply both sides by y^2

Answer 4

So, you'll get:
\[(\frac{-6}{y})^2+y^2=13\]\[(-6)^2+y^4=13y^2\]Quadratic equation..

Answer 5

it has y^4 how is this quadratic?

Answer 6

\[(y^2)^2 - 13y^2 + 36=0\]

Answer 7

Let u = y^2
u^2 - 13u + 36 =0
Solve u, the solve u=y^2

Answer 8

Another thing you could try: \[
x=\sqrt{13}\cos t,\quad y = \sqrt{13}\sin t \\
13\sin t\cos t+6=0 \implies \frac{13}{2}\sin 2t=-6
\]

Answer 9

$$(x^2+y^2)=13$$$$xy=-6\implies 2xy=-12$$
adding those two together,
$$x^2+2xy+y^2=1\implies (x+y)^2=1\implies x=-y\pm1$$
considering plus,
$$x=1-y\\
(1-y)y=-6\implies y^2-y-6=0\implies y=3,\,y=-2$$so, $$x=-2,\, x=3$$

consider minus,
$$x=-(y+1)\\(y+1)y=6\implies y^2+y-6=0\implies y=-3,\,y=2$$
so,
$$x=2,\,x=-3$$
the answers are,
$$(-2,3),\,(3,-2),\,(2,-3),\,(-3,2)$$