solving systems of equations algebraically chapter 3.2

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Solving Systems of Equations Algebraically Chapter 3.2

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Page 1: Solving Systems of Equations Algebraically Chapter 3.2

Solving Systems of Equations Algebraically

Chapter 3.2

Page 2: Solving Systems of Equations Algebraically Chapter 3.2

Alternatives to Graphing

Sometimes, graphing systems is not the best way to go– Lines don’t intersect at a discernable point– Don’t have enough room to graph– Etc.

Alternate methods of solving systems– Substitution– Elimination

Page 3: Solving Systems of Equations Algebraically Chapter 3.2

Substitution

Substitution: – One equation is solved for one variable in terms

of the other. This expression can be substituted for the variable in the other equation.

Page 4: Solving Systems of Equations Algebraically Chapter 3.2

Example 1

Solve either equation for one variable

Substitute this expression into the other equation

– Now, there should only be one variable

After solving for one variable, plug it into either equation to solve for other variable

x + 4y = 26

x - 5y = -10

x = 5y – 10

5y – 10 + 4y = 26

9y – 10 = 26

9y = 36

y = 4

x + 4y = 26x + 4(4) = 26x + 16 = 26x = 10

Solution: (10, 4)

Page 5: Solving Systems of Equations Algebraically Chapter 3.2

Example 2

Solve either equation for one variable

Substitute this expression into the other equation

– Now, there should only be one variable

After solving for one variable, plug it into either equation to solve for other variable

2x + y = 43x + 2y = 1

y = -2x + 4

3x + 2(-2x + 4) = 13x - 4x + 8 = 1-x + 8 = 1-x = -7x = 7

2x + y = 42(7) + y = 414 + y = 4y = -10

Solution: (7, -10)

Page 6: Solving Systems of Equations Algebraically Chapter 3.2

Elimination

Elimination:– Eliminate one of the variables by adding or

subtracting the two equations together

Page 7: Solving Systems of Equations Algebraically Chapter 3.2

Example 3

Instead of using Substitution, you can subtract one equation from the other

– Subtract the 2nd from the 1st

After solving for one variable, plug it into either equation to find the other

x + 2y = 10

x + y = 6

0 + y = 4

y = 4

x + y = 6

x + 4 = 6

x = 2

Solution:

(2, 4)

Page 8: Solving Systems of Equations Algebraically Chapter 3.2

Example 4

Instead of using Substitution, you can add one equation to the other

After solving for one variable, plug it into either equation to find the other

2x + y = 5

3x - y = 20

5x + 0 = 25

x = 5

2x + y = 5

2(5) + y = 5

10 + y = 5

y = -5

Solution:

(5, -5)

Page 9: Solving Systems of Equations Algebraically Chapter 3.2

Elimination with Multiplication

When one variable cannot be easily eliminated using simple addition or subtraction, multiply one or both equations by constants so that a variable CAN be eliminated.

Page 10: Solving Systems of Equations Algebraically Chapter 3.2

Example 5

Simple addition or subtraction isn’t going to help here.

Decide which variable to eliminate

– Let’s do x

Now subtract 2nd from 1st

Use one variable to solve for the other

2x + 3y = 12

5x - 2y = 11

10x + 15y = 60

10x - 4y = 22

0 + 19y = 38

y = 2

x5 x2

2x + 3y = 12

2x + 3(2) = 12

2x + 6 = 12

2x = 6

x = 3

Solution:

(3, 2)

Page 11: Solving Systems of Equations Algebraically Chapter 3.2

Example 6

Simple addition or subtraction isn’t going to help here.

Decide which variable to eliminate

– Let’s do h

Now add together

Use one variable to solve for the other

2g + h = 6

3g - 2h = 16

4g + 2h = 12

3g - 2h = 16

7g + 0 = 28

g = 4

x2

2g + h = 6

2(4) + h = 6

8 + h = 6

h = -2

Solution:

(4, -2)

Page 12: Solving Systems of Equations Algebraically Chapter 3.2

Inconsistent and Dependent Systems

If you add or subtract two equations in a system and the result is an equation that is never true, then the system is inconsistent and it has no solution.

– Examples 1 = 2 -1 = 1

If the result is an equation that is always true, then the system is dependent and has infinitely many solutions.

– Examples 1 = 1 9 = 9

Page 13: Solving Systems of Equations Algebraically Chapter 3.2

Example 7 (substitution)

Solve either equation for one variable

Substitute this expression into the other equation– Now, there

should only be one variable

x + 3y = 8

1/3 x + y = 9

x = -3y + 8

1/3(-3y + 8) + y = 9

-y + 8/3 + y = 9

8/3 = 9

Not True No Solutions!!

Page 14: Solving Systems of Equations Algebraically Chapter 3.2

Example 8 (substitution)

Solve either equation for one variable

Substitute this expression into the other equation– Now, there

should only be one variable

2a - 4b = 6

-a + 2b = -3

a = 2b + 3

2(2b + 3) - 4b = 6

4b + 6 - 4b = 6

6 = 6

Always True

Infinitely Many Solutions!!

Page 15: Solving Systems of Equations Algebraically Chapter 3.2

Example 9 (elimination)

Simple addition or subtraction isn’t going to help here.

Decide which variable to eliminate

– Let’s do x

Now add together

4x - 2y = 5

-2x + y = 1

-4x + 2y = 2

4x - 2y = 5

0 + 0 = 7

Not True

x2

No Solutions