To solve the given linear-quadratic system using the substitution method, we have the equations:
1. \( y = x^2 - 4x - 41 \) (quadratic equation)
2. \( y = 4 \) (linear equation)
We will substitute the value of \( y \) from the second equation into the first equation.
### Step 1: Substitute \( y = 4 \) into the quadratic equation
Substituting \( y = 4 \) into \( y = x^2 - 4x - 41 \):
\[
4 = x^2 - 4x - 41
\]
### Step 2: Rearrange the equation
Rearranging the equation gives:
\[
x^2 - 4x - 41 - 4 = 0
\]
\[
x^2 - 4x - 45 = 0
\]
### Step 3: Factor or use the quadratic formula
We need to solve the quadratic equation \( x^2 - 4x - 45 = 0 \). We can factor it:
\[
(x - 9)(x + 5) = 0
\]
### Step 4: Set each factor to zero
Setting each factor equal to zero gives us:
\[
x - 9 = 0 \quad \Rightarrow \quad x = 9
\]
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]
### Step 5: Find corresponding \( y \) values
Now we substitute \( x = 9 \) and \( x = -5 \) back into the linear equation \( y = 4 \):
- For \( x = 9 \):
\[
y = 4
\]
- For \( x = -5 \):
\[
y = 4
\]
### Step 6: Write the solutions
Thus, the solutions to the linear-quadratic system are:
1. \( (9, 4) \)
2. \( (-5, 4) \)
These are the points where the quadratic graph intersects the line \( y = 4 \).