Solving Systems of Linear-Quadratic Equations

Front

First step in solving: y = x² + 3x - 5 y = x + 3

Back

x² + 3x - 5 = x + 3

Front

Solutions to the system:
y = x² + 3x - 5
y = x + 3

Back

(-4, -1) and (2, 5)

Front

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

Back

(1, 3) and (4, 6)

Front

Solution to the given linear-quadratic function: (0, 0), (3, 0), (2, -3), (-3, 2)

Back

(2, -3)

Front

Solve the system of equations: y = 5 and y = 2x^2 - 16x + 29

Back

(6, 5) and (2, 5)

Front

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

Back

2 solutions

Front

Back

(-1,-5) and (5,-5)

Front

Solutions to the graphed system: (0,-2) and (3,1), (-2,0) and (2,0), (0,-2) and (3,1), (2,2) and (1,0)

Back

(0,-2) and (3,1)

Front

What is the solution? Options: (1, -1), (-1, 1), (0, -2), (0, 1)

Back

(1, -1)

Front

Given the system of equations:
y = x² + 3x - 5
y = x + 3
When x = -4, what is the value of y?

Back

-1