The system of equations from the image is:

1. $-10x^2 + y = -80x + 155$
2. $5x^2 + y = 40x - 85$

Let's solve this system of equations step by step. To eliminate $y$, we can subtract the second equation from the first.

1. Subtract the second equation from the first: $(-10x^2 + y) - (5x^2 + y) = (-80x + 155) - (40x - 85)$

Simplifying both sides:

$-10x^2 - 5x^2 + y - y = -80x - 40x + 155 + 85$ $-15x^2 = -120x + 240$

Now, divide the entire equation by $-15$:

$x^2 = 8x - 16$

Rearrange this into a standard quadratic form:

$x^2 - 8x + 16 = 0$

This is a perfect square trinomial:

$(x - 4)^2 = 0$

Thus, $x = 4$.

2. Substitute $x = 4$ into either of the original equations to find $y$. Let's use the second equation: $5(4)^2 + y = 40(4) - 85$ $5(16) + y = 160 - 85$ $80 + y = 75$ $y = 75 - 80 = -5$

So, the solution to the system is $x = 4$ and $y = -5$.

Would you like more details on this process or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you solve a system of equations using substitution?
2. What is the difference between elimination and substitution methods in solving systems of equations?
3. How would the solution change if one of the quadratic terms had a higher degree?
4. Can a system of equations involving quadratics have multiple solutions? Why or why not?
5. What are some real-world problems where solving systems of equations with quadratics is necessary?

Tip: Always double-check your solution by substituting the values back into both original equations to verify correctness.