1. 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

2. *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)

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

4. 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)

5. 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)}

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

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

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

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

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

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

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

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