To solve the given system of simultaneous equations:
[tex]\[
\begin{cases}
x^2 + 4y = 37 \\
5x + y = -8
\end{cases}
\][/tex]
We will proceed through substitution or elimination to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. For this solution, I'll use substitution by expressing [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex] from the second equation and then substituting it into the first equation.
1. From the second equation [tex]\(5x + y = -8\)[/tex], solve for [tex]\(y\)[/tex]:
[tex]\[
y = -8 - 5x
\][/tex]
2. Substitute this expression for [tex]\(y\)[/tex] into the first equation [tex]\(x^2 + 4y = 37\)[/tex]:
[tex]\[
x^2 + 4(-8 - 5x) = 37
\][/tex]
3. Simplify the equation:
[tex]\[
x^2 - 32 - 20x = 37
\][/tex]
[tex]\[
x^2 - 20x - 32 - 37 = 0
\][/tex]
[tex]\[
x^2 - 20x - 69 = 0
\][/tex]
4. The simplified equation is a quadratic equation in [tex]\(x\)[/tex]:
[tex]\[
x^2 - 20x - 69 = 0
\][/tex]
5. Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
For [tex]\(a = 1\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = -69\)[/tex]:
[tex]\[
x = \frac{20 \pm \sqrt{(-20)^2 - 4(1)(-69)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{20 \pm \sqrt{400 + 276}}{2}
\][/tex]
[tex]\[
x = \frac{20 \pm \sqrt{676}}{2}
\][/tex]
[tex]\[
x = \frac{20 \pm 26}{2}
\][/tex]
6. This gives us two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{20 + 26}{2} = \frac{46}{2} = 23
\][/tex]
and
[tex]\[
x = \frac{20 - 26}{2} = \frac{-6}{2} = -3
\][/tex]
7. Substitute [tex]\(x = 23\)[/tex] back into the equation [tex]\(y = -8 - 5x\)[/tex] to find [tex]\(y\)[/tex]:
[tex]\[
y = -8 - 5(23) = -8 - 115 = -123
\][/tex]
8. For [tex]\(x = -3\)[/tex]:
[tex]\[
y = -8 - 5(-3) = -8 + 15 = 7
\][/tex]
So the solutions to the system of equations are:
[tex]\[
(x, y) = (23, -123) \quad \text{and} \quad (x, y) = (-3, 7)
\][/tex]
