Use the substitution method to solve this linear�quadratic system. 𝑦 = 𝑥

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 \( y = 4 \).