To solve the system of equations given by:
- \( y = 2x^2 - 9 \)
- \( y = x^2 - 10x - 25 \)
we can set the two equations equal to each other:
\[ 2x^2 - 9 = x^2 - 10x - 25 \]
Now, let's rearrange this equation:
\[ 2x^2 - x^2 + 10x - 9 + 25 = 0 \]
This simplifies to:
\[ x^2 + 10x + 16 = 0 \]
Next, we can factor this quadratic equation. We need two numbers that multiply to 16 and add to 10. Those numbers are 2 and 8. Thus, we can factor the equation:
\[ (x + 2)(x + 8) = 0 \]
Setting each factor to zero gives us:
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] \[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \]
Now that we have the \( x \)-values, we can substitute them back into either original equation to find the corresponding \( y \)-values. Let's use the first equation \( y = 2x^2 - 9 \).
- For \( x = -2 \):
\[ y = 2(-2)^2 - 9 = 2(4) - 9 = 8 - 9 = -1 \]
So, one solution is \((-2, -1)\).
- For \( x = -8 \):
\[ y = 2(-8)^2 - 9 = 2(64) - 9 = 128 - 9 = 119 \]
So, the other solution is \((-8, 119)\).
Now, we can list the solutions in the specified format:
The solutions are \((-8, 119)\) and \((-2, -1)\).
Hence, according to the instruction to enter the solution with the least \( x \)-value first, the final answer is:
The solutions are \(-8\), \(119\), and \(-2\), \(-1\).