Drag the tiles to the correct boxes to complete the pairs. not all tiles will be used.

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)}

  A-2, B-DNE*, C-3, D-DNE, E-4, F-1

Step-by-step explanation:

The first attachment shows the solutions to A and C.

The second attachment shows the solutions to E and F.

There are no real number solutions to systems B and D.

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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.

i think it is x = 1.2

marco saved more.   he saved $35 more.

step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35