Solving Quadratic-Linear Systems Section 4.9 Solving Quadratic-Linear Systems Objective: I will be able to Solve a quadratic-linear system by graphing and algebraically

*You can solve systems involving quadratic equations using methods similar to the ones used to solve systems of linear equations. 1. Graphically 2. Algebraically (Substitution/Elimination)

A system on one quadratic equation and one linear equation can have two solutions, one solution, or no solutions.

Solving a System of Equations We can use our calculator to solve two equations, a quadratic and linear equation. y = x² - 2x + 1 y = 2x - 3 Enter the equations in y = Enter graph to determine how many solutions Select 2nd Trace 5 First Curve? Enter ; Second Curve? Enter; Guess? Enter If there is more than one solution, move the cursor to the other intersection point and repeat this process to find the other solutions. Solution: (2,1)

So the solution set is: {(0,-2),(5,3)} Solve the following system of equations graphically:          y = x2 - 4x - 2          y = x - 2           Our graphs intersect at 2 points whose coordinates are: (0,-2)  and  (5,3) So the solution set is: {(0,-2),(5,3)}

Solve graphically:              y = x2 + 4x + 3              y = 2x + 6                                        

Solve graphically:              y = -x2 + 2x + 4              x + y = 4

Now let’s solve Systems of Linear and Quadratic Equations Algebraically!

Solve the System Algebraically Use Substitution (0, 1) (1, 2) Answer: (0,1) (1,2)

Solve the System Algebraically Use Substitution (2, 2) Answer: (2,2)

What is the solution of the system of equations? y = -x2 – x + 6 y = x + 3

What is the solution of the system of equations? y = -x2 – x + 12 y = x2 + 7x + 12

Exit Ticket Solve graphically: 1. y = -x2 + 2x + 4 y = -2x + 4 Solve Algebraically 2. y = -x2 – 3x + 10 3. y = x2 – 4x +5 y = x + 5 y = -x2 + 5