Question

What are the solutions to the following system?

Answer 1

The solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$.

Solution:

Given equation:

$-2 x^{2}+y=-5$

Add $2 x^{2}$ on both sides.

$-2 x^{2}+y+2 x^{2}=-5+2 x^{2}$

$y=-5+2 x^{2}$ -------- (1)

$y=-3 x^{2}+5$ (given)  -------- (2)

Equate (1) and (2).

$-5+2 x^{2}=-3 x^{2}+5$

Add $3 x^{2}$ on both sides.

$-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}$

$-5+5x^{2}=5$

Add 5 on both sides.

$-5+5x^{2}+5=5+5$

$5x^{2}=10$

Divide by 5 on both sides, we get

$x^{2}=2$

Taking square root on both sides, we get

$x=\sqrt{2}, x=-\sqrt{2}$

Substitute $x=\sqrt{2}$ in (1).

$y=-5+2(\sqrt{2})^2$

$y=-5+4$

$y=-1$

Substitute $x=-\sqrt{2}$ in (1).

$y=-5+2(-\sqrt{2})^2$

$y=-5+4$

$y=-1$

Therefore the solutions are $(\sqrt{2},-1) \text { and }(-\sqrt{2},-1)$.

Option C is the correct answer.