To solve the system of equations using the substitution method, we start with the two equations given:
- \( y = x^2 - 4x - 41 \)
- \( y = 4 \)
Since the second equation tells us that \( y \) is equal to 4, we can substitute \( 4 \) for \( y \) in the first equation. This will allow us to solve for \( x \).
Substituting \( y = 4 \) into the first equation gives:
\[ 4 = x^2 - 4x - 41 \]
Now, we'll rearrange this equation to set it to zero:
\[ x^2 - 4x - 41 - 4 = 0 \]
This simplifies to:
\[ x^2 - 4x - 45 = 0 \]
Now we need to factor or use the quadratic formula to solve for \( x \). We will first try to factor:
We need two numbers that multiply to \(-45\) and add to \(-4\). The numbers \(-9\) and \(5\) satisfy this.
So we can factor the equation as:
\[ (x - 9)(x + 5) = 0 \]
Setting each factor to zero gives us:
- \( x - 9 = 0 \) → \( x = 9 \)
- \( x + 5 = 0 \) → \( x = -5 \)
Now we have two \( x \)-values: \( x = 9 \) and \( x = -5 \).
Next, we'll find the corresponding \( y \)-values using the second equation \( y = 4 \).
So no matter what \( x \) is, \( y = 4 \).
Therefore, the solution pairs to the system of equations are:
\[ (9, 4) \quad \text{and} \quad (-5, 4) \]
The final solutions are:
\[ (9, 4) \quad \text{and} \quad (-5, 4) \]