Certainly! Let's find the solutions for the given system of equations:
[tex]\[
\begin{cases}
3x + y = 11 \\
x^2 - 2y = 18
\end{cases}
\][/tex]
### Step 1: Solve the first equation for [tex]\( y \)[/tex]
First, let's isolate [tex]\( y \)[/tex] in the first equation:
[tex]\[
3x + y = 11
\][/tex]
Subtract [tex]\( 3x \)[/tex] from both sides:
[tex]\[
y = 11 - 3x
\][/tex]
### Step 2: Substitute [tex]\( y \)[/tex] into the second equation
Next, we substitute [tex]\( y = 11 - 3x \)[/tex] into the second equation:
[tex]\[
x^2 - 2y = 18
\][/tex]
Substitute [tex]\( y \)[/tex]:
[tex]\[
x^2 - 2(11 - 3x) = 18
\][/tex]
### Step 3: Simplify and solve for [tex]\( x \)[/tex]
Distribute and simplify the equation:
[tex]\[
x^2 - 22 + 6x = 18
\][/tex]
Combine like terms:
[tex]\[
x^2 + 6x - 40 = 0
\][/tex]
This is a quadratic equation. Now, we solve for [tex]\( x \)[/tex]:
[tex]\[
x^2 + 6x - 40 = 0
\][/tex]
Factorize the quadratic equation:
[tex]\[
(x + 10)(x - 4) = 0
\][/tex]
Set each factor to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x + 10 = 0 \quad \text{or} \quad x - 4 = 0
\][/tex]
[tex]\[
x = -10 \quad \text{or} \quad x = 4
\][/tex]
### Step 4: Solve for [tex]\( y \)[/tex] using the values of [tex]\( x \)[/tex]
Using [tex]\( x = -10 \)[/tex]:
[tex]\[
y = 11 - 3(-10) = 11 + 30 = 41
\][/tex]
Using [tex]\( x = 4 \)[/tex]:
[tex]\[
y = 11 - 3(4) = 11 - 12 = -1
\][/tex]
### Step 5: Present the solutions as ordered pairs
The solutions to the system of equations are:
[tex]\[
(-10, 41) \quad \text{and} \quad (4, -1)
\][/tex]
