To solve the system of equations using the elimination method, we'll set the two equations equal to each other since both are equal to \( y \).
The two equations are:
- \( y = x^2 + 10 \)
- \( y = -7x - 2 \)
Setting them equal to each other gives:
\[ x^2 + 10 = -7x - 2 \]
To solve for \( x \), we can rearrange the equation by moving all terms to one side:
\[ x^2 + 7x + 10 + 2 = 0 \]
This simplifies to:
\[ x^2 + 7x + 12 = 0 \]
Next, we will factor the quadratic equation:
\[ (x + 3)(x + 4) = 0 \]
Setting each factor to zero gives us the solutions for \( x \):
- \( x + 3 = 0 \) → \( x = -3 \)
- \( x + 4 = 0 \) → \( x = -4 \)
Now that we have the \( x \)-values, we can substitute these back into either original equation to find the corresponding \( y \)-values. We'll use the second equation, \( y = -7x - 2 \).
For \( x = -3 \):
\[ y = -7(-3) - 2 = 21 - 2 = 19 \]
The first solution is \( (-3, 19) \).
For \( x = -4 \):
\[ y = -7(-4) - 2 = 28 - 2 = 26 \]
The second solution is \( (-4, 26) \).
Putting it all together, the solutions to the system of equations are:
\[ \boxed{(-3, 19)} \quad \text{and} \quad \boxed{(-4, 26)} \]
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