Consider the following system of equations. Determine the value of k if a solution is (-3,2)... Determine a second solution if possible.

Question

x^2+6x+y+k=0

3x+y+k=0

Answer 1

x^2 +6x + y + k = 0

3x + y + k = 0

plug in (-3,2) or x=-3, y=2

(-3)^2 + 6(-3) + (2) + k =0

9-18+2 + k = 0

k=18-11= 7

k = 7

3(-3) +2 + k=0

-9+2 + k=0

k = 7

3x+y = -7

x^2 +6x +y = -7

subtract 1st from 2nd to eliminate y

x^2 +3x = 0

x(x+3)=0

x = 0, -3

plug 0 into the equations to solve for y

y=-7

(0,-7) and (-3,2) are two solutions