To solve the system of equations by elimination, we have the following two equations:

1. $2x^2 - 3x - y = -5$
2. $-x + y = 5$

Step 1: Solve the second equation for $y$

From the second equation, $-x + y = 5$, we can isolate $y$: $y = x + 5$

Step 2: Substitute $y = x + 5$ into the first equation

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

Simplify the equation: $2x^2 - 3x - x - 5 = -5$ $2x^2 - 4x - 5 = -5$

Step 3: Solve for $x$

Now, add 5 to both sides: $2x^2 - 4x = 0$

Factor out $2x$: $2x(x - 2) = 0$

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

Step 4: Solve for $y$

• If $x = 0$, substitute into $y = x + 5$: $y = 0 + 5 = 5$ So, one solution is $(0, 5)$.

• If $x = 2$, substitute into $y = x + 5$: $y = 2 + 5 = 7$ So, another solution is $(2, 7)$.

Step 5: Check solutions in the original system

• For $(0, 5)$:

  • First equation: $2(0)^2 - 3(0) - 5 = -5$, which is true.
  • Second equation: $-0 + 5 = 5$, which is true.

• For $(2, 7)$:

  • First equation: $2(2)^2 - 3(2) - 7 = -5$, which simplifies to $8 - 6 - 7 = -5$, true.
  • Second equation: $-2 + 7 = 5$, which is true.

Both solutions are valid. Therefore, the correct answers are:

• $(0, 5)$
• $(2, 7)$

Do you want more details or have any questions?

Related Questions:

1. How does substitution differ from elimination in solving systems of equations?
2. Can this system of equations be solved graphically, and how would the graphs look?
3. What are the advantages of using elimination over substitution?
4. How would this system change if the second equation was altered slightly, like $-x + y = 6$?
5. What methods are used to solve systems of nonlinear equations?

Tip: Always check your solutions in both equations to verify accuracy, especially in systems with quadratic components!