solving linear systems algebraically

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Solving Linear Systems Algebraically Substitution - Section 3.2 06/13/22 16:41 3-2 - Solving Systems through Substitution 1 Solving Linear Systems Algebraically

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Solving Linear Systems Algebraically. Solving Linear Systems Algebraically. Substitution - Section 3.2. Steps in Substitution. Steps: SOLVE for one equation into one variable REPLACE one equation into other equation SUBSTITUTE the value into either equation CHECK the solution. - PowerPoint PPT Presentation

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Page 1: Solving Linear Systems Algebraically

Solving Linear Systems Algebraically

Substitution - Section 3.2

04/21/23 04:433-2 - Solving Systems through

Substitution1

Solving Linear Systems Algebraically

Page 2: Solving Linear Systems Algebraically

Steps in Substitution

• Steps:1.SOLVE for one equation into one variable2.REPLACE one equation into other equation3.SUBSTITUTE the value into either equation4.CHECK the solution

04/21/23 04:433-2 - Solving Systems through

Substitution2

Page 3: Solving Linear Systems Algebraically

Example 1• Solve using Substitution

04/21/23 04:433-2 - Solving Systems through

Substitution3

2 9

3 4 8

x y

x y

2 9

3 4 8

x y

x y

1. SOLVE for one equation into one variable

2 9y x

Page 4: Solving Linear Systems Algebraically

04/21/23 04:433-2 - Solving Systems through

Substitution

4

2 9

3 4 8y

x y

x

23 4( 89)xx

2 9y x

3 8 36 8x x 11 36 8x

11 44x

4x

2. REPLACE one equation into other equation

Page 5: Solving Linear Systems Algebraically

04/21/23 04:433-2 - Solving Systems through

Substitution5

2 9

3 4 8

x y

x y

3. SUBSTITUTE the value into either equation

(43 ) 4 8y

4x

3 4 8x y

12 4 8y 4 4y

1y

(4,1)

Page 6: Solving Linear Systems Algebraically

Example 1• Solve using Substitution

04/21/23 04:433-2 - Solving Systems through

Substitution6

2 9

3 4 8

x y

x y

4. CHECK the solution2 9

3 4 8

x y

x y

2(4) (1) 9 3(4) 4(1) 8

12 49 89

8 8

(4,1)

Page 7: Solving Linear Systems Algebraically

04/21/23 04:433-2 - Solving Systems through

Substitution7

3 2 8 2x x

2 8y x

5 8 2x5 10x

2x

Example 2• Solve using Substitution

2 8y x

2( 2) 8y

4y

( 2,4)

2 8

3 2

x y

x y

Page 8: Solving Linear Systems Algebraically

Example 2• Solve using Substitution

04/21/23 04:433-2 - Solving Systems through

Substitution8

4. CHECK the solution

2( 2) (4) 4 4 8

3( 2) (4) 6 4 2

( 2,4)

2 8

3 2

x y

x y

Page 9: Solving Linear Systems Algebraically

04/21/23 04:433-2 - Solving Systems through

Substitution9

3 63 6yy

3x y

3 9 6 6y y 3 9 6y

3 3y

1y

Your Turn• Solve using Substitution

4 4 12

3 6 6

x y

x y

3x y

1 3x

4y

(4,1)

Page 10: Solving Linear Systems Algebraically

04/21/23 04:433-2 - Solving Systems through

Substitution10

25 6( 2) 9xx

2 2y x

5 12 12 9x x 7 12 9x

7 21x

3x

Example 3• Solve using Substitution

5 6 9

2 2

x y

x y

2 2y x

2 3 2y

4y

(3, 4)

Page 11: Solving Linear Systems Algebraically

Example 4

A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans, which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?

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Substitution11

Let x represent the amount of the Sumatra beans in the blend.Let y represent the amount of the Kona beans in the blend.

Page 12: Solving Linear Systems Algebraically

Example 4A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans,

which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?

04/21/23 04:433-2 - Solving Systems through

Substitution12

Write one equation based on the amount of each bean:Amount of Sumatra

beans plus amount of Kona

beans equals

x y

50.

50+ =

Write another equation based on cost of the beans:Cost of Sumatra

beans pluscost of Kona

beans equals

5x 13y

cost of beans.

10(50)+ =

Page 13: Solving Linear Systems Algebraically

Example 4A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans,

which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?

04/21/23 04:433-2 - Solving Systems through

Substitution13

Solve the system.x + y = 50

5x + 13y = 500

x + y = 50

y = 50 – x

First equation

5x + 13(50 – x) = 500

5x + 650 – 13x = 500–8x = –150 x = 18.75

Solve the first equation for y.

Substitute (50 – x) for y.Distribute.Simplify.

Page 14: Solving Linear Systems Algebraically

Example 4A coffee blend contains Sumatra beans which cost $5/lb, and Kona beans,

which cost $13/lb. If the blend costs $10/lb, how much of each type of coffee is in 50 lb of the blend?

04/21/23 04:43 3-2 - Solving Systems through Substitution 14

Substitute the value of x into one equation.

Substitute x into one of the original equations to solve for y.

18.75 + y = 50

y = 31.25 Solve for y.

The mixture will contain 18.75 lb of the Sumatra beans and 31.25 lb of the Kona beans.

Solve the system.x + y = 50

5x + 13y = 500

y = 31.25

(18.75, 31.25)

Page 15: Solving Linear Systems Algebraically

Assignment

Page 194 (2-5, 15-18)

04/21/23 04:433-2 - Solving Systems through

Substitution15