To find the solutions to the given system of equations:
[tex]\[
\left\{\begin{aligned}
-2x^2 + y &= -5 \\
y &= -3x^2 + 5
\end{aligned}\right.
\][/tex]
we will follow a step-by-step approach.
### Step 1: Set the Equations Equal to Each Other
Since both equations equal [tex]\( y \)[/tex], we can set the right-hand sides of the equations equal to each other:
[tex]\[
-2x^2 - 5 = -3x^2 + 5
\][/tex]
### Step 2: Solve for [tex]\( x \)[/tex]
Rearranging the terms to isolate [tex]\( x^2 \)[/tex], we get:
[tex]\[
-2x^2 + 3x^2 = 5 + 5
\][/tex]
[tex]\[
x^2 = 10
\][/tex]
Taking the square root of both sides, we find:
[tex]\[
x = \sqrt{10} \quad \text{or} \quad x = -\sqrt{10}
\][/tex]
### Step 3: Substitute [tex]\( x \)[/tex] Back into One of the Original Equations
We can use either of the original equations to find [tex]\( y \)[/tex]. Let's use [tex]\( y = -3x^2 + 5 \)[/tex]:
For [tex]\( x = \sqrt{10} \)[/tex]:
[tex]\[
y = -3(\sqrt{10})^2 + 5
\][/tex]
[tex]\[
y = -3(10) + 5
\][/tex]
[tex]\[
y = -30 + 5
\][/tex]
[tex]\[
y = -25
\][/tex]
So one solution is [tex]\((\sqrt{10}, -25)\)[/tex].
For [tex]\( x = -\sqrt{10} \)[/tex]:
[tex]\[
y = -3(-\sqrt{10})^2 + 5
\][/tex]
[tex]\[
y = -3(10) + 5
\][/tex]
[tex]\[
y = -30 + 5
\][/tex]
[tex]\[
y = -25
\][/tex]
So the other solution is [tex]\((- \sqrt{10}, -25)\)[/tex].
### Conclusions
The solutions to the system of equations are:
[tex]\[
(\sqrt{10}, -25) \quad \text{and} \quad (-\sqrt{10}, -25)
\][/tex]
None of the provided options [tex]\((0,2)\)[/tex], [tex]\((1,-2)\)[/tex], [tex]\((\sqrt{2},-1)\)[/tex] and [tex]\((- \sqrt{2},-1)\)[/tex], [tex]\((\sqrt{5}, -10)\)[/tex] and [tex]\((- \sqrt{5}, -10)\)[/tex] match our solutions [tex]\((\sqrt{10}, -25)\)[/tex] and [tex]\((- \sqrt{10}, -25)\)[/tex].
