Worksheet on systems of quadratic equations from Kuta Software's Infinite Algebra 2, containing 14 problems with equations and points to verify solutions.

Looking at the worksheet, I need to solve each system of quadratic equations. Since there are 14 problems and no specific instruction on which ones to solve, I'll work through a few representative examples to demonstrate the method, then provide answers for all.

Let me start with problem 5:
5) 3x² + 2y² - 54y - 143 = 0
x - 3y - 3 = 0

From the second equation: x = 3y + 3
Substitute into first equation:
3(3y+3)² + 2y² - 54y - 143 = 0
3(9y² + 18y + 9) + 2y² - 54y - 143 = 0
27y² + 54y + 27 + 2y² - 54y - 143 = 0
29y² - 116 = 0
y² = 4
y = ±2

If y = 2: x = 3(2) + 3 = 9
If y = -2: x = 3(-2) + 3 = -3

Solutions: (9, 2) and (-3, -2)

Let me check problem 6:
6) 2x² + 3y² + 3x - 12y - 42 = 0
x + 3y = 0

From second equation: x = -3y
Substitute:
2(-3y)² + 3y² + 3(-3y) - 12y - 42 = 0
18y² + 3y² - 9y - 12y - 42 = 0
21y² - 21y - 42 = 0
y² - y - 2 = 0
(y-2)(y+1) = 0
y = 2 or y = -1

If y = 2: x = -6
If y = -1: x = 3

Solutions: (-6, 2) and (3, -1)

Let me verify problem 9:
9) x² + y² + x + 3y + 2 = 0
x - y = 0

So x = y
Substitute:
y² + y² + y + 3y + 2 = 0
2y² + 4y + 2 = 0
y² + 2y + 1 = 0
(y+1)² = 0
y = -1
x = -1

Solution: (-1, -1)

Now let me solve several more systematically:

Problem 7:
x² + 2y² - 11x - 3y + 31 = 0
-x + y + 4 = 0 → x = y + 4

(y+4)² + 2y² - 11(y+4) - 3y + 31 = 0
y² + 8y + 16 + 2y² - 11y - 44 - 3y + 31 = 0
3y² - 6y + 3 = 0
y² - 2y + 1 = 0
(y-1)² = 0
y = 1, x = 5

Solution: (5, 1)

Problem 8:
4y² + 34x + y - 52 = 0
2x + y - 4 = 0 → y = 4 - 2x

4(4-2x)² + 34x + (4-2x) - 52 = 0
4(16 - 16x + 4x²) + 34x + 4 - 2x - 52 = 0
64 - 64x + 16x² + 34x + 4 - 2x - 52 = 0
16x² - 32x + 16 = 0
x² - 2x + 1 = 0
(x-1)² = 0
x = 1, y = 2

Solution: (1, 2)

Problem 10:
-2x² + y² + 24y + 76 = 0
x + 3y + 2 = 0 → x = -3y - 2

-2(-3y-2)² + y² + 24y + 76 = 0
-2(9y² + 12y + 4) + y² + 24y + 76 = 0
-18y² - 24y - 8 + y² + 24y + 76 = 0
-17y² + 68 = 0
y² = 4
y = ±2

If y = 2: x = -8
If y = -2: x = 4

Solutions: (-8, 2) and (4, -2)

Problem 11:
5x² + 20x + 9y - 7 = 0
5x² + 10y² + 20x - y - 67 = 0

Subtract first from second:
(5x² + 10y² + 20x - y - 67) - (5x² + 20x + 9y - 7) = 0
10y² - 10y - 60 = 0
y² - y - 6 = 0
(y-3)(y+2) = 0
y = 3 or y = -2

If y = 3: 5x² + 20x + 27 - 7 = 0 → 5x² + 20x + 20 = 0 → x² + 4x + 4 = 0 → x = -2
If y = -2: 5x² + 20x - 18 - 7 = 0 → 5x² + 20x - 25 = 0 → x² + 4x - 5 = 0 → x = 1 or x = -5

Solutions: (-2, 3), (1, -2), (-5, -2)

Problem 12:
-x² - 3x + y = 0 → y = x² + 3x
-12x² - 3x + y = 0

Substitute:
-12x² - 3x + (x² + 3x) = 0
-11x² = 0
x = 0, y = 0

Solution: (0, 0)

Problem 13:
3x² - 12x - 2y - 2 = 0
3x² + 3y² - 12x + 22y + 19 = 0

Subtract first from second:
(3x² + 3y² - 12x + 22y + 19) - (3x² - 12x - 2y - 2) = 0
3y² + 24y + 21 = 0
y² + 8y + 7 = 0
(y+7)(y+1) = 0
y = -7 or y = -1

If y = -7: 3x² - 12x + 14 - 2 = 0 → 3x² - 12x + 12 = 0 → x² - 4x + 4 = 0 → x = 2
If y = -1: 3x² - 12x + 2 - 2 = 0 → 3x² - 12x = 0 → x(x-4) = 0 → x = 0 or x = 4

Solutions: (2, -7), (0, -1), (4, -1)

Problem 14:
7y² + 25x + 42y - 137 = 0
24x² + 7y² - 191x + 42y + 55 = 0

Subtract first from second:
24x² - 216x + 192 = 0
x² - 9x + 8 = 0
(x-8)(x-1) = 0
x = 8 or x = 1

If x = 8: 7y² + 200 + 42y - 137 = 0 → 7y² + 42y + 63 = 0 → y² + 6y + 9 = 0 → y = -3
If x = 1: 7y² + 25 + 42y - 137 = 0 → 7y² + 42y - 112 = 0 → y² + 6y - 16 = 0 → y = 2 or y = -8

Solutions: (8, -3), (1, 2), (1, -8)

Final Answer:
5) (9, 2), (-3, -2)
6) (-6, 2), (3, -1)
7) (5, 1)
8) (1, 2)
9) (-1, -1)
10) (-8, 2), (4, -2)
11) (-2, 3), (1, -2), (-5, -2)
12) (0, 0)
13) (2, -7), (0, -1), (4, -1)
14) (8, -3), (1, 2), (1, -8)