Question

What are the solutions to the following system of equations?

Answer 1

Answer:

  D.  No real solutions

Step-by-step explanation:

Any of the usual methods of finding solutions to the system of equations will show there are none,

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graphing

The graphs of the two equations do not intersect. This means any of the answer choices that identifies specific solution points must be incorrect. There are no real solutions.

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check the answers

You can check to see if any of the proposed solution points satisfy the equations.

  A: (x, y) = (-4, -2)   ⇒   8(-4) -(-2) = -30 ≠ 10 . . . not a solution

  B: (x, y) = (-2, -4)   ⇒   8(-2) -(-4) = -12 ≠ 10 . . . not a solution

  C: same (x, y) as B.

So, none of these answer choices describes a solution to the system of equations. We conclude there are ...

  No real solutions

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solve the system

Using the first equation, you can substitute for y in the second equation:

  8x -(x² +12x +30) = 10

  x² +4x +40 = 0 . . . . . . . put in standard form

  (x +2)² +36 = 0 . . . . . . . write in vertex form

There are no real values of x that will make (x+2)² be negative, so there are no real solutions.

Answer 2

Answer:

D. no real answer

Step-by-step explanation:

8x-y=10 then y=8x-10

$y = {x}^{2} + 12x + 30 = 8x - 10$

${x}^{2} + 4x + 40 = 0$

it's a grade 2 equation, it has no real answer

because

${4}^{2} - 4(1)(40) = - 144 < 0$