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Systems of Equations. 11-6. Course 3. Warm Up. Problem of the Day. Lesson Presentation. Systems of Equations. 11-6. 3 V. = A. 1. C – S. h. 3. t. Course 3. Warm Up Solve for the indicated variable. 1. P = R – C for R 2. V = Ah for A 3. R = for C.

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## 11-6

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Presentation Transcript

Systems of Equations

11-6

Course 3

Warm Up

Problem of the Day

Lesson Presentation

Systems of Equations

11-6

3V

= A

1

C – S

h

3

t

Course 3

Warm Up

Solve 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

Systems of Equations

11-6

Course 3

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

Systems of Equations

11-6

Course 3

Learn to solve systems of equations.

Systems of Equations

11-6

Course 3

Insert Lesson Title Here

Vocabulary

system of equations

solution of a system of equations

Systems of Equations

11-6

Course 3

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.

Systems of Equations

11-6

Caution!

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

Course 3

Systems of Equations

11-6

Course 3

Additional Example 1A: Solving Systems of Equations

Solve the system of equations.

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.

y= 4x – 6

y =x + 3

4x – 6 = x + 3

Systems of Equations

11-6

Course 3

Solve the equation to find x.

4x – 6 = x + 3

– x– x

Subtract x from both sides.

3x - 6 = 3

+ 6+ 6

3x9

Divide both sides by 3.

3 = 3

x = 3

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

y = x + 3 = 3 + 3 = 6

The solution is (3, 6).

Systems of Equations

11-6

Course 3

Additional Example 1B: Solving Systems of Equations

y = 2x + 9

y = -8 + 2x

2x + 9 = -8 + 2x

Transitive Property

Subtract 2x from both sides.

– 2x– 2x

9 ≠ -8

The system of equations has no solution.

Systems of Equations

11-6

Course 3

Check It Out: Example 1A

Solve the system of equations.

y = x – 5

y = 2x – 8

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

y=x – 5

y =2x – 8

x – 5 = 2x – 8

Systems of Equations

11-6

Course 3

Check It Out: Example 1A Continued

Solve the equation to find x.

x – 5 = 2x – 8

– x– x

Subtract x from both sides.

–5 = x – 8

+ 8+ 8

3 = x

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

y = x – 5 = 3 – 5 = –2

The solution is (3, –2).

Systems of Equations

11-6

Course 3

Check It Out: Example 1B

y = 3x + -7

y = 6 + 3x

3x + -7 = 6 + 3x

Transitive Property

Subtract 3x from both sides.

– 3x– 3x

-7 ≠ 6

The system of equations has no solution.

Systems of Equations

11-6

Course 3

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

Systems of Equations

11-6

Course 3

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

Solve the system of equations.

5x + y = 7 x – 3y = 11

Solve both equations for x.

5x + y = 7 x – 3y = 11

-y-y+ 3y+ 3y

5x = 7 - y x = 11 + 3y

5(11 + 3y)= 7 -y

55 + 15y = 7 – y

Subtract 15y from both sides.

- 15y- 15y

55 = 7 – 16y

Systems of Equations

11-6

Course 3

55 = 7 – 16y

Subtract 7 from both sides.

–7–7

48-16y

–16 = -16

Divide both sides by –16.

-3 = y

x = 11 + 3y

= 11 + 3(-3)Substitute –3 for y.

= 11 + –9 = 2

The solution is (2, –3).

Systems of Equations

11-6

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

Course 3

Systems of Equations

11-6

= –

–8

–2x

–2

10y

–2

–2

Course 3

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

Solve the system of equations.

–2x + 10y = –8 x – 5y = 4

Solve both equations for x.

–2x + 10y = –8 x – 5y = 4

–10y–10y+5y+5y

–2x = –8 – 10yx = 4 + 5y

x = 4 + 5y

Subtract 5y from both sides.

4 + 5y = 4 + 5y

- 5y- 5y

4 = 4

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

Systems of Equations

11-6

Course 3

Check It Out: Example 2A

Solve the system of equations.

x + y = 5 3x + y = –1

Solve both equations for y.

x + y = 5 3x + y = –1

–x–x– 3x– 3x

y = 5 – x y = –1 – 3x

5 – x = –1 – 3x

+ x+ x

5 = –1 – 2x

Systems of Equations

11-6

Course 3

Check It Out: Example 2A Continued

5 = –1 – 2x

+ 1+ 1

6 = –2x

–3 = x

Divide both sides by –2.

y = 5 – x

= 5 – (–3)Substitute –3 for x.

= 5 + 3 = 8

The solution is (–3, 8).

Systems of Equations

11-6

Course 3

Check It Out: Example 2B

Solve the system of equations.

x + y = –2 –3x + y = 2

Solve both equations for y.

x + y = –2 –3x + y = 2

– x– x+ 3x+ 3x

y = –2 – xy = 2 + 3x

–2 – x = 2 + 3x

Systems of Equations

11-6

Course 3

Check It Out: Example 2B Continued

–2 – x = 2 + 3x

+ x+ x

–2 = 2 + 4x

Subtract 2 from both sides.

–2–2

–4 = 4x

Divide both sides by 4.

–1 = x

y = 2 + 3x

Substitute –1 for x.

= 2 + 3(–1) = –1

The solution is (–1, –1).

Systems of Equations

11-6

1

2

( , 2)

Course 3

Insert Lesson Title Here

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.

no solution

(–2,3)

15 and 8