Solving Quadratic Equations Algebraically

Which of the following represents the first step in solving:

y = x² + 3x - 5

y = x + 3

• x² - 3x + 5 = x - 3

• x² + 3x - 5 = x + 3

• x² + 3x - 5 + x + 3 = 0

• x² + 3x - 5 = 0

What are the solutions to the system:
y = x² + 3x - 5
y = x + 3

• (-4, 2) and (-1, 5)

• (-4, -1) and (2, 5)

• (2, -1 and (-4, 5)

• (-4, 1) and (2, -1)

Find the solution:
y = x² - 4x + 6
y = x + 2

• (3, 1) and (6, 4)

• (-1, 1) and (4, 2)

• (1, 3) and (4, 6)

• (1, 4) and (3, 6)

Solve the system of linear and quadratic equations using substitution:
y = 5
y = 2x² - 16x + 29

• (6, 5) and (2, 5)

• (2, 6)

• (6, 2)

• (5, 6) and (5, 2)

Determine the number of solutions to the given linear-quadratic system.

• 0 solutions

• 1 solution

• 2 solutions

• 3 solutions

Solve by factoring: x² + 3x - 18 = 0

• x = -3 and x = 6

• x = 3 and x = -6

• x = -9 and x = 2

• x = 9 and x = -2

Solve
x² + 2x - 24 = 0

• x = -6, x = 4

• x = 6, x = -4

• x = 12, x = -2

• x = 8, x = -3

Solve
x² + 9x + 20 = 0

• x = -2, x = -10

• x = -3, x = -4

• x = -4, x = -5

• x = -4, x = -6

Solve by factoring: x² - 9x + 20 = 0

• x = -4 and x = -5

• x = 20 and x = -9

• x = 4 and x = 5

• x = -4 and x = 5

Factor and solve:

z² − 14z + 45 = 0

• z = 1, 45

• z = −1, 45

• z = −5, 9

• z = 5, 9