To solve the system of equations using the elimination method, we start with the given equations:
- \( y = x^2 + 10 \) (Equation 1)
- \( y = -7x - 2 \) (Equation 2)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 + 10 = -7x - 2 \]
Now, we'll rearrange this equation to one side:
\[ x^2 + 7x + 12 = 0 \]
Next, we can factor this quadratic equation:
\[ (x + 3)(x + 4) = 0 \]
Setting each factor equal to zero gives us the solutions for \( x \):
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x + 4 = 0 \) → \( x = -4 \)
Now we can substitute these values back into either original equation to find the corresponding values of \( y \).
For \( x = -3 \):
Using Equation 2:
\[ y = -7(-3) - 2 = 21 - 2 = 19 \]
So one point is \( (-3, 19) \).
For \( x = -4 \):
Using Equation 2 again:
\[ y = -7(-4) - 2 = 28 - 2 = 26 \]
So another point is \( (-4, 26) \).
Thus, the solutions for the system are the points \( (-3, 19) \) and \( (-4, 26) \).
Therefore, the correct answer is:
D. (-3,19) and (-4,26).