11-6 Systems of Equations

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpSolve for the indicated variable.

1. P = R – C for R

2. V = Ah for A

3. R = for C

R = P + C

Rt + S = C

Course 3

11-6 Systems of Equations

13C – S

t

= A3Vh

Problem of the Day

At an audio store, stereos have 2 speakers and home-theater systems have 5 speakers. There are 30 sound systems with a total of 99 speakers. How many systems are stereo systems and how many are home-theater systems?

17 stereo systems, 13 home-theater systems

Course 3

11-6 Systems of Equations

Learn to solve systems of equations.

Course 3

11-6 Systems of Equations

Vocabulary

system of equationssolution of a system of equations

Insert Lesson Title Here

Course 3

11-6 Systems of Equations

A system of equations is a set of two or more equations that contain two or more variables. A solution of a system of equations is a set of values that are solutions of all of the equations. If the system has two variables, the solutions can be written as ordered pairs.

Course 3

11-6 Systems of Equations

Course 3

11-6 Systems of Equations

When solving systems of equations, remember to find values for all of the variables.

Caution!

Additional Example 1A: Solving Systems of Equations

Solve the system of equations.

y = 4x – 6

y = x + 3

y = 4x – 6 y = x + 3

The expressions x + 3 and 4x – 6 both equal y. So by the Transitive Property they equal each other.

4x – 6 = x + 3

Course 3

11-6 Systems of Equations

Additional Example 1A Continued

To find y, substitute 3 for x in one of the original equations.y = x + 3 = 3 + 3 = 6

The solution is (3, 6).Course 3

11-6 Systems of Equations

Solve the equation to find x.

4x – 6 = x + 3 – x – x Subtract x from both sides.

3x 6 = 3

3x 9 6 6 Add 6 to both sides.

3 = 3 x = 3

Divide both sides by 3.

The system of equations has no solution.

Course 3

11-6 Systems of Equations

2x + 9 = 8 + 2x – 2x – 2x

Transitive Property

Subtract 2x from both sides. 9 ≠ 8

Additional Example 1B: Solving Systems of Equations

y = 2x + 9y = 8 + 2x

Check It Out: Example 1A

Solve the system of equations.

y = x – 5y = 2x – 8

y = x – 5 y = 2x – 8

x – 5 = 2x – 8

Course 3

11-6 Systems of Equations

The expressions x – 5 and 2x – 8 both equal y. So by the Transitive Property they equal each other.

Check It Out: Example 1A Continued

To find y, substitute 3 for x in one of the original equations.

y = x – 5 = 3 – 5 = –2

The solution is (3, –2).

Course 3

11-6 Systems of Equations

Solve the equation to find x.

x – 5 = 2x – 8– x – x Subtract x from both sides.

–5 = x – 8

3 = x

+ 8 + 8 Add 8 to both sides.

The system of equations has no solution.

Course 3

11-6 Systems of Equations

3x + 7 = 6 + 3x – 3x – 3x

Transitive Property

Subtract 3x from both sides. 7 ≠ 6

Check It Out: Example 1B

y = 3x + 7y = 6 + 3x

To solve a general system of two equations with two variables, you can solve both equations for x or both for y.

Course 3

11-6 Systems of Equations

Additional Example 2A: Solving Systems of Equations by Solving for a Variable

Solve the system of equations.

5x + y = 7 x – 3y = 115x + y = 7 x – 3y = 11

y y 3y 3y

Solve both equations for x.

5x = 7 y x = 11 + 3y

15y 15y55 = 7 – 16y

Subtract 15y from both sides.

Course 3

11-6 Systems of Equations

5(11 + 3y)= 7 y

55 + 15y = 7 – y

Additional Example 2A Continued

–7 –7

48 16y

Subtract 7 from both sides.

Divide both sides by –16.

–16 = 16

x = 11 + 3y = 11 + 3(3) Substitute –3 for y. = 11 + –9 = 2The solution is (2, –3).

Course 3

11-6 Systems of Equations

55 = 7 – 16y

3 = y

Course 3

11-6 Systems of Equations

You can solve for either variable. It is usually easiest to solve for a variable that has a coefficient of 1.

Helpful Hint

Additional Example 2B: Solving Systems of Equations by Solving for a Variable

Solve the system of equations.

–2x + 10y = –8 x – 5y = 4–2x + 10y = –8 x – 5y = 4 –10y –10y +5y +5y

Solve both equations for x.

–2x = –8 – 10y x = 4 + 5y

= ––8–2

10y–2

–2x–2

x = 4 + 5y4 + 5y = 4 + 5y

Course 3

11-6 Systems of Equations

5y 5ySubtract 5y from both sides.4 = 4

Since 4 = 4 is always true, the system of equations has an infinite number of solutions.

Check It Out: Example 2A

Solve the system of equations.

x + y = 5 3x + y = –1x + y = 5 3x + y = –1

–x –x – 3x – 3x

Solve both equations for y.

y = 5 – x y = –1 – 3x

5 – x = –1 – 3x+ x + x

5 = –1 – 2x

Add x to both sides.

Course 3

11-6 Systems of Equations

Check It Out: Example 2A Continued

5 = –1 – 2x

+ 1 + 1

6 = –2x

Add 1 to both sides.

Divide both sides by –2.

–3 = x

y = 5 – x = 5 – (–3) Substitute –3 for x. = 5 + 3 = 8The solution is (–3, 8).

Course 3

11-6 Systems of Equations

Check It Out: Example 2B

Solve the system of equations.

x + y = –2 –3x + y = 2x + y = –2 –3x + y = 2

– x – x + 3x + 3x

Solve both equations for y.

y = –2 – x y = 2 + 3x

–2 – x = 2 + 3x

Course 3

11-6 Systems of Equations

+ x + x Add x to both sides.

–2 = 2 + 4x–2 –2

–4 = 4x

–2 – x = 2 + 3x

Subtract 2 from both sides.

Divide both sides by 4.–1 = x

y = 2 + 3x= 2 + 3(–1) = –1 Substitute –1

for x.The solution is (–1, –1).

Course 3

11-6 Systems of Equations

Check It Out: Example 2B Continued

Lesson Quiz

Solve each system of equations.1. y = 5x + 10

y = –7 + 5x

2. y = 2x + 1 y = 4x

3. 6x – y = –15 2x + 3y = 5

4. Two numbers have a sum of 23 and a difference of 7. Find the two numbers.

Insert Lesson Title Here

(–2,3)

15 and 8

( , 2)12

Course 3

11-6 Systems of Equations

no solution