To solve the linear-quadratic system using the elimination method, we have the following equations:
- \( y = x^2 + 4x + 5 \)
- \( y = 12x - 7 \)
We can set the two equations equal to each other by substituting for \( y \):
\[ x^2 + 4x + 5 = 12x - 7 \]
Next, we move all terms to one side of the equation:
\[ x^2 + 4x + 5 - 12x + 7 = 0 \]
Simplifying this gives:
\[ x^2 - 8x + 12 = 0 \]
Now we can factor the quadratic:
\[ (x - 6)(x - 2) = 0 \]
So, the solutions for \( x \) are:
\[ x = 6 \quad \text{or} \quad x = 2 \]
Now we will find the corresponding \( y \)-values by substituting each \( x \) back into one of the original equations. Let’s use the linear equation \( y = 12x - 7 \).
- For \( x = 6 \):
\[ y = 12(6) - 7 = 72 - 7 = 65 \] So, one solution is \( (6, 65) \).
- For \( x = 2 \):
\[ y = 12(2) - 7 = 24 - 7 = 17 \] So, the other solution is \( (2, 17) \).
The solutions to the system are \( (6, 65) \) and \( (2, 17) \).
Thus, the final answer is:
(6,65) and (2,17).