Use substitution to solve systems of equations .

1 / 24

# Use substitution to solve systems of equations . - PowerPoint PPT Presentation

Use substitution to solve systems of equations. Vocabulary. system of equations solution of a system of equations.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Use substitution to solve systems of equations .' - aelan

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Vocabulary

system of equations

solution of a system 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.

Caution!

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

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 are equal to each other.

y= 4x – 6

y =x + 3

4x – 6 = x + 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).

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.

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

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).

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.

Up to this point, we have used one or both equations in slope-intercept form: y = mx+ b

How do we solve equations written in general form: ax + by = c (notice that x and y are both on the same side of the equal sign.)

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

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

55 = 7 – 16y

Subtract 7 from both sides.

–7–7

Divide both sides by –16.

48– 16y

–16 = –16

–3 = y

x = 11 + 3y

= 11 + 3(–3)Substitute –3 for y.

= 11 + –9 = 2

The solution is (2, –3).

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

= –

–8

–2x

–2

10y

–2

–2

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.

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

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).

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

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).

Lesson Quizzes

Standard Lesson Quiz

Lesson Quiz for Student Response Systems

1

2

( , 2)

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

Lesson Quiz for Student Response Systems

1. Solve the given system of equations.

y = 11x + 20

y = –2 + 11x

A. (2, 2)

B. (1, 1)

C. (1, –1)

D. no solution

Lesson Quiz for Student Response Systems

2. Solve the given system of equations.

4x + y = 11

2x + 3y = –7

A. (4, –5)

B. (4, 5)

C. (2, –5)

D. (2, 5)

Lesson Quiz for Student Response Systems

3. Two numbers have a sum of 37 and a difference of 17. Identify the two numbers.

A. –27and–10

B. –27and10

C. 27 and 10

D. 27 and –10