The system of equations in the image is:

$1) \quad 2x^2 - 3x - y = -5$ $-x + y = 5$

To solve this system, we can use substitution or elimination. I'll start by solving the second equation for $y$ and substituting it into the first equation.

From the second equation: $-x + y = 5 \quad \Rightarrow \quad y = x + 5$

Now substitute this expression for $y$ into the first equation: $2x^2 - 3x - (x + 5) = -5$

Simplifying: $2x^2 - 3x - x - 5 = -5$ $2x^2 - 4x - 5 = -5$ $2x^2 - 4x = 0$ Factor the equation: $2x(x - 2) = 0$

Thus, $x = 0$ or $x = 2$.

Case 1: $x = 0$

Substitute $x = 0$ into $y = x + 5$: $y = 0 + 5 = 5$

So one solution is $(x, y) = (0, 5)$.

Case 2: $x = 2$

Substitute $x = 2$ into $y = x + 5$: $y = 2 + 5 = 7$

So the second solution is $(x, y) = (2, 7)$.

Solutions:

The system has two solutions: $(0, 5) \quad \text{and} \quad (2, 7)$

Would you like more details or have any questions?

Related Questions:

1. How would you solve this system using the elimination method?
2. What if the system included $x^3$ instead of $x^2$? How would that change the solution?
3. Can this system be solved graphically, and how would the graph look?
4. How does the degree of the equation affect the number of possible solutions?
5. What happens if we change the constant on the right side of the first equation?

Tip:

When solving a system of equations, substitution works best if one of the equations can be easily solved for one variable.