To solve the given system of equations, we have:
[tex]\[
\left\{
\begin{array}{r}
x^2 + y^2 = 25 \\
2x + y = -5
\end{array}
\right.
\][/tex]
Let's solve it step-by-step:
1. First Equation: [tex]\(x^2 + y^2 = 25\)[/tex]
2. Second Equation: [tex]\(2x + y = -5\)[/tex]
We will solve the second equation for [tex]\(y\)[/tex]:
[tex]\[
y = -5 - 2x
\][/tex]
Now substitute this expression for [tex]\(y\)[/tex] into the first equation:
[tex]\[
x^2 + (-5 - 2x)^2 = 25
\][/tex]
Simplify the equation:
[tex]\[
x^2 + (25 + 20x + 4x^2) = 25
\][/tex]
[tex]\[
x^2 + 4x^2 + 20x + 25 = 25
\][/tex]
[tex]\[
5x^2 + 20x + 25 = 25
\][/tex]
Subtract 25 from both sides:
[tex]\[
5x^2 + 20x = 0
\][/tex]
Factor out the common factor:
[tex]\[
5x(x + 4) = 0
\][/tex]
This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = 0 \quad \text{or} \quad x = -4
\][/tex]
Now we will find the corresponding [tex]\(y\)[/tex] values for each [tex]\(x\)[/tex]:
- For [tex]\(x = 0\)[/tex]:
[tex]\[
y = -5 - 2(0) = -5
\][/tex]
So one solution is [tex]\((0, -5)\)[/tex].
- For [tex]\(x = -4\)[/tex]:
[tex]\[
y = -5 - 2(-4) = -5 + 8 = 3
\][/tex]
So the other solution is [tex]\((-4, 3)\)[/tex].
Therefore, the solutions to the system of equations are [tex]\((0, -5)\)[/tex] and [tex]\((-4, 3)\)[/tex].
Among the given options, the correct one is:
[tex]\[
(0, -5) \text{ and } (-4, 3)
\][/tex]
