Question

Please Help c:

What are the solutions to the following system of equations?
y = x2 + 12x + 30
8x − y = 10

(−4, −2) and (2, 5)

(−2, −4) and (2, 5)

(−2, −4) and (5, 2)

No Real Solutions

Answer 1

Hello,

y=x²+12x+30
y=8x-10

==>x²+12x+30=8x-10
==>x²+4x+40=0
Δ=16-4*40<0 ==> no real solutions.

Answer 2

2nd equation:
8 x-y=10 ⇒ y= 8 x-10
We will switch that in 1st equation:
8 x-10=x²+12 x+30, or:
x² + 4 x +40 = 0, a=1,B=4, c=40
Discriminant : D= b² - 4 ac= 16 - 160 = -144 and because it is negative there are no real solutions.
Answer: D)