To solve the linear-quadratic system using the elimination method, we start by substituting the expression for \( y \) from the linear equation into the quadratic equation.
The equations we have are:
- \( y = x^2 + 10 \) (quadratic equation)
- \( y = -7x - 2 \) (linear equation)
Step 1: Set the equations equal to each other
Since both equations equal \( y \), we can set them equal to each other:
\[ x^2 + 10 = -7x - 2 \]
Step 2: Rearrange the equation
Move all terms to one side to set the equation to zero:
\[ x^2 + 7x + 12 = 0 \]
Step 3: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ (x + 3)(x + 4) = 0 \]
Step 4: Solve for \( x \)
Setting each factor equal to zero gives us:
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x + 4 = 0 \) → \( x = -4 \)
Step 5: Find the corresponding \( y \) values
Now we substitute these \( x \) values back into either of the original equations to find the corresponding \( y \) values. We'll use the linear equation \( y = -7x - 2 \):
- For \( x = -3 \):
\[ y = -7(-3) - 2 = 21 - 2 = 19 \]
Thus, one solution is \( (-3, 19) \).
- For \( x = -4 \):
\[ y = -7(-4) - 2 = 28 - 2 = 26 \]
Thus, the second solution is \( (-4, 26) \).
Final solutions
The solutions to the system are \( (-3, 19) \) and \( (-4, 26) \).
From the options provided, the correct answer is:
(-3, 19) and (-4, 26).