solve the system by the method of substitution

{y = x^3 - x^2 + 2

 y = x^2 + 15x + 2

A. (5, 102), (-3, -34), (0,2)

B. (5, 102), (-1, -12)

C. (5, 102), (3, 56), (1, 18)

D. no real solution

E. )-3, -34), (1, 18)

\(y_1 = x^3 - x^2 + 2\\  y_2 = x^2 + 15x + 2\\\)

Sub in (0,2) becasue it is the easiest.  Yes that makes both statements true

If x=-1

\(y_1=-1-1+2=0 \qquad \mbox{So if x=-1, y=0 So B is not correct}\)

If x=1

\(y_1=1-1+2=2 \qquad \mbox{not 18 so C and E are no good} \)

So that leaves A and D

Check the other two point in A and if they both work then that is the answer otherwise it must be D

{y = x^3 - x^2 + 2

 y = x^2 + 15x + 2

For "y" in the first equation, substitute the second....and we have

x^3- x^2 + 2 = x^2 + 15x + 2     subtract the terms on the right side

x^3 - 2x^2 -15x = 0   factor

x [ x^2 -2x - 15] = 0

x (x - 5) (x + 3)  = 0   and setting each factor to 0, x = 0 or x = 5 or x = -3

Subbing these values back into either equation  gives the values {-3, -34}, {0, 2} and {5, 102}

Answer A is correct