Q2 Solve the system.

Question

y=x2+2x-2
y=x+10

Answer 1

To solve the system, we can set the two equations equal to each other.

x^2 + 2x - 2 = x + 10

Now, we can rearrange the equation to set it equal to zero:

x^2 + 2x - x - 12 = 0
x^2 + x - 12 = 0

Next, we can factor the equation:

(x + 4)(x - 3) = 0

Setting each factor equal to zero:

x + 4 = 0 -> x = -4
x - 3 = 0 -> x = 3

Now that we have the possible values of x, we can substitute them back into one of the equations to solve for y. Let's use the second equation:

y = x + 10

For x = -4:
y = -4 + 10 = 6
So, one solution is (-4, 6).

For x = 3:
y = 3 + 10 = 13
So, the other solution is (3, 13).

Therefore, the solutions to the system are (-4, 6) and (3, 13).