Which system of equations does not have a real solution?

A. y = x 2 + 3x - 5 and x + y = - 10

B. y = x 2 + 3x - 5 and 4x + 5y = 20

C. y = x 2 + 3x - 5 and x + y = - 9

Correct answer is A!

It's A.

Step-by-step explanation:

Let's look at option A:

From the second equation y = - 10 - x. Substituting in the first equation:

-10 - x = x^2 + 3x - 5

x^2 + 4x + 5 = 0

Checking the discriminant b^2 - 4ac we get 16 - 4*1*5 = - 4 so there are no real roots. (A negative discriminant means no real roots).

So A has no real solution.

B.

x^2 + 3x - 5 = (20 - 4x) / 5 = 4 - 0.8x

x^2 + 3.8x - 9 = 0

b^2 - 4ac = (3.8) ^2 - 4*1*-9 = 50.44 (positive) so there are real roots.

C.

x^2 + 3x - 5 = - 9 - x

x^2 + 4x + 4 = 0

b^2 - 4ac = 4^2 - 4*1*4 = 0 so there are real roots.