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Match the systems of equations with their solution sets.

Linear-Quadratic System of Equations
A .y + 12 = x2 + x

x + y = 3

B .y − 15 = x2 + 4x

x − y = 1

C. y + 5 = x2 − 3x

2x + y = 1

D.y − 6 = x2 − 3x

x + 2y = 2

E. y − 17 = x2 − 9x

-x + y = 1

F. y − 15 = -x2 + 4x

x + y = 1

Solution Set

{(-2, 3), (7, -6)}

{(-5, 8), (3, 0)}

{(-2, 5), (3, -5)}

{(2, 3), (8, 9)}

In general, you can solve the linear equation for y, then substitute that into the quadratic. You can subtract the x-term on the left and complete the square to find the solutions.

A.

(3-x) +12 = x^2 +x

15 = x^2 + 2x

16 = x^2 +2x +1 = (x +1)^2 . . . . add the square of half the x-coefficient to complete the square; next take the square root

±4 -1 = x = {-5, 3) . . . . . identifies the second solution set for system A

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B.

(x -1) -15 = x^2 +4x

-16 = x^2 +3x

-13.75 = x^2 +3x +2.25 = (x +1.5)^2

roots are complex: -1.5 ±i√13.75

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C.

(1-2x) +5 = x^2 -3x

6 = x^2 -x

6.25 = x^2 -x + .25 = (x -.5)^2

±2.5 +.5 = x = {-2, 3} . . . . . identifies the third solution set for system C

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remaining problems are done in a similar way.

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* DNE = does not exist. There is no matching solution set for the complex numbers that are the solutions to this.