Answer:
There are two pairs of solutions: (2,7) and (-1,4)
Step-by-step explanation:
We will use substitution.
y = x^2 + 3
y = x +5
Since the second equation is equal to y, replace y in the first equation with the second equation.
y = x^2 + 3
x + 5 = x^2 + 3
Rearrange so that one side is equal to 0.
5 - 3 = x^2 - x
2 = x^2 - x
0 = x^2 - x - 2
You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).
a = 1, b = -1, c = -2
Zeros = [tex]\frac{-b + \sqrt{b^{2}-4ac } }{2a}[/tex] and [tex]\frac{-b - \sqrt{b^{2}-4ac } }{2a}[/tex]
Zeros = 2 and -1
Now that you have your x values, plug them into the equations to find their corresponding y values.
y = x^2 + 3
y = (2)^2 + 3
y = 7
Pair #1: (2,7)
y = x^2 + 3
y = (-1)^2 + 3
y = 4
Pair #2: (-1,4)
Therefore, there are two pairs of solutions: (2,7) and (-1,4).
