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

Question

2 − 4𝑥 − 41 & 𝑦 = 4

Answer 1

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 \( y = 4 \).