Marimzoey0pa3rky
Answered

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

Verified Answer :

HomertheGenius

y = 8 x - 10
x² + 12 x + 30 = 8 x - 10
x² + 4 x + 40 = 0
and because the discriminant is: D = b² - 4 ac = 16 - 160 = - 144 < 0
answer is: D ) No real solutions

heidischardt

Answer:

no real solutions

Step-by-step explanation:

y = x^2 + 12x + 30

8x − y = 10 are out 2 equations.

1. -8x from both sides in this 8x − y = 10 equation.

we are left with -y=-8x+10. multiply both sides by -1: y=8x-10.

2. Plug in one of the two equations for the other's y variable:

8x-10=x^2 +12x +30.

3. solve: +10 to each side, and -8x from each side:

x^2 +4x + 40 = 0

4. use discriminant formula: b^2-4(a)(c)

(4)^2 - (4)(1)(40): 16-160=-144.

5. Because the answer is negative, there is two complex solutions. But bc in every complex solution there is an imaginary number, and bc imaginary numbers cannot be graphed, the answer is "NO REAL SOLUTIONS."

hope this helps :)

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