Lesson 9.7 Solve Systems with Quadratic EquationsEssential Question: How do you solve systems that include a quadratic equation?

CC.9.12.A.REI.4bSolve quadratic equations by inspection, by taking square roots, completing thesquare, the quadraticformula and factoring, as appropriate to theinitial form of theequation. Recognizewhen the quadratic formula gives complexsolutions.

Warm-up:

1. Solve the linear system using substitution. y = 3 – 2x y = x + 9

2. Solve the linear system using elimination. x – y = 10 4x + y = -15

4-15-13

2 8

10

x y

x y

Systems of Linear EquationsWe had 3 methods to solve them.

Method 1 - Graphing

Solve for y.

2 8y x

10y x (6, – 4)

Systems of Linear Equations

Method 2 - Substitution

2 8

10

x y

x y

2 8y x

10x y 10x 2 8x

2 8 10x x 3 18x

6x 6

2 8y x 2( ) 8y 12 8y 4y

(6, – 4)

Systems of Linear Equations

Method 3 - Elimination

2 8

10

x y

x y

2 8

10

x y

x y

3 18x

6x 6

10x y ( ) 10y 4y 4y

(6, – 4) All 3 methods

giving us the

same answer

(6,–4).

Solving Systems of Linear and Quadratic Equations

Substitution Method

• Step 1: Solve one of the equations for one of its variables.

• Step 2: Substitute the expression from step 1 into the other equation and solve for the other variable.

• Step 3: Substitute the value from step 2 into one of the original equations and solve.

Example 1 Use Substitution Method

• Solve the system:263

232

xxy

xy

Systems with One Linear Equation and One Quadratic Equation

No solution One Solution Two Solutions

Solve the System AlgebraicallyUse Substitution

1

12

xy

xy

1 xy( ) 1x

Answer: (0,1) (1,2)

2 1x

02 xx

1 0x x 0 1 0x or x 0 1x or x

0x 20 1

1

y

y

(0, 1)

1x 21 1

2

y

y

(1, 2)

Solve the System AlgebraicallyUse Substitution

2 2y x 2( ) 2 2x x 22 x

Answer: (2,2)

2 2 2y x x 2x

(2, 2)

22

222

xy

xxy

20 4 4x x 0 2 2x x

0 2x 2x

2 2 2

4 2

2

y

y

y

Solve the System AlgebraicallyUse Substitution

6y x

22 6 26x x 2 2 12 36 26x x x

6

2622

yx

yx

22 12 10 0x x 2 6 5 0x x

5 1 0x x

5 1x or x

5x 5 6

1

1

y

y

y

1x 1 6

5

5

y

y

y

(5, –1)

(1, – 5)

Answer: (5, –1) (1, – 5)

Solve the System AlgebraicallyUse Substitution

3

4y x

22 3

254

x x

2 2925

16x x

2 216 9 400x x 225 400x 2 16x

4x

4x 4 3 4

4 12

3

y

y

y

4x 4 3 4

4 12

3

y

y

y

(4, 3)

(– 4, – 3)

Answer: (4, 3) (–4, – 3)

xy

yx

34

2522

Now let’s look at the Graphs of these Systems!

Classify each equation as linear/quadratic.

What does the graph of each look like?Line

Parabola

Linear

Quadratic

What is the solution to the system?

Point of Intersection (-2, 0)

Point of Intersection (1, -3 )

1

12

xy

xy

Answer: (0,1) (1,2)

22

222

xy

xxy

Answer: (2,2)

Now use your calculator to check it graphically.

We have solved the following algebraically

Classwork/Homework

•9.7 Exercises •Pages 625 •2- 20 even•Quiz on Quadratic formula on Tuesday!