Let:
[tex]\begin{gathered} y=x^2+3x-12\text{ (1)} \\ y=x+3\text{ (2)} \\ \end{gathered}[/tex]Replacing (1) into (2):
[tex]\begin{gathered} x^2+3x-12=x+3 \\ \text{Subtract (x+3) from both sides:} \\ x^2+3x-12-x-3=0 \\ x^2+2x-15=0 \end{gathered}[/tex]The factors of -15 that sum to 2 are -3 and 5, therefore:
[tex]\begin{gathered} x^2+2x-15=(x-3)(x+5)=0 \\ So\colon \\ x=3 \\ or \\ x=-5 \end{gathered}[/tex]For x = 3:
y = 3² + 3*(3) - 12 = 9 + 9 - 12 = 6
y = (3) + 3 = 6
This solution is correct
For x = -5
y = (-5)² + 3(-5) - 12 = 25 - 15 - 12 = -2
y = (-5) + 3 = -2
This solution is correct as well.
Therefore, the solutions are:
(3,6) and (-5,-2)
or
x = 3 and y = 6
x = -5 and y = -2
