Alright, let's solve the system of equations step by step.
We are given:
[tex]\[ 3x + y = 11 \quad \text{(Equation 1)} \][/tex]
[tex]\[ x^2 - 2y = 18 \quad \text{(Equation 2)} \][/tex]
### Step 1: Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from Equation 1
Rearrange Equation 1:
[tex]\[ y = 11 - 3x \][/tex]
### Step 2: Substitute [tex]\( y \)[/tex] into Equation 2
Replace [tex]\( y \)[/tex] in Equation 2 with [tex]\( 11 - 3x \)[/tex]:
[tex]\[ x^2 - 2(11 - 3x) = 18 \][/tex]
### Step 3: Simplify the resulting equation
Distribute the -2 inside the parentheses:
[tex]\[ x^2 - 22 + 6x = 18 \][/tex]
Move all terms to one side to form a standard quadratic equation:
[tex]\[ x^2 + 6x - 40 = 0 \][/tex]
### Step 4: Solve the quadratic equation
We can factor the quadratic equation:
[tex]\[ (x + 10)(x - 4) = 0 \][/tex]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = -10 \quad \text{or} \quad x = 4 \][/tex]
### Step 5: Solve for [tex]\( y \)[/tex] using the values of [tex]\( x \)[/tex]
Case 1: If [tex]\( x = -10 \)[/tex]
[tex]\[ y = 11 - 3(-10) \][/tex]
[tex]\[ y = 11 + 30 \][/tex]
[tex]\[ y = 41 \][/tex]
Case 2: If [tex]\( x = 4 \)[/tex]
[tex]\[ y = 11 - 3(4) \][/tex]
[tex]\[ y = 11 - 12 \][/tex]
[tex]\[ y = -1 \][/tex]
### Step 6: Write the solutions as ordered pairs [tex]\( (x, y) \)[/tex]
The solutions to the system of equations are:
[tex]\[ (-10, 41) \quad \text{and} \quad (4, -1) \][/tex]
Thus, the solutions are:
[tex]\[
\boxed{(-10, 41) \text{ and } (4, -1)}
\][/tex]
