module 1 ~ topic 1 solving equations table of contents slides 6-14: solving linear equations ...

Module 1 ~ Topic 1Solving Equations

Table of Contents

Slides 6-14: Solving Linear Equations Slides 15-29 : Practice Questions

Audio/Video and Interactive Sites

Slides 2: Algebra Cheat SheetSlides 3: Graphing Calculator Use Guide Slide 5: VideoSlide 7: Gizmos

Algebra Cheat Sheet

You may want to download and print this sheet for reference throughout the

course.

Special InstructionsSpecial InstructionsThis module includes graphing calculator work.

Refer to the website, TI-83/84 calculator instructions, for resources and instructions on how to use your calculator to obtain the results we do throughout the lessons. I suggest that you bookmark this site if you haven’t done so already.

Take careful notes and following along with every example in each lesson. I encourage you to ask me questions and think deeply as you’re studying these concepts!

Topic #1: Solving Linear Topic #1: Solving Linear EquationsEquationsMany real-life phenomena can be described

by linear functions. It is important to learn how to solve equations involving such functions to provide us with information about these phenomena.

Definition: A linear equation in one variable is an equation that can be written in the form ax + b = c, where a, b and c are real numbers, and a ≠ 0.

(The letter x is often used as the variable, but it is not required to be the variable.)

Video Break!!!!!Video Break!!!!!

Click on this link to watch videos on solving equations.

Recall General RulesRecall General RulesOrder of Operations: PEMDAS

Multiplication/Division is done in order, left to rightAddition/Subtraction is done in order, left to right

Solving Equations

First, you must know what you are solving for so you can isolate it.To do that:Take care of any exponents/FOIL or distribution/simplificationGet common denominators if necessaryCombine like terms on each side of the equal signAddition/Subtraction across = is done next to isolate the variableMultiplication/Division across =is done last and the variable should now be isolated

Gizmos Gizmos

Gizmo: Modeling 2-Step Equations

Gizmo: Modeling 1-Step Equations B

Gizmo: Modeling 1-Step Equations A

Gizmo: Solving 2-Step Equations

How to Solve EquationsHow to Solve Equations

As with any journey, you need to know where you want to

end up before you start off, or you will never get there!

1) Figure out what you are solving for. In this case we want to end up with x = #

1684 xEx: SolveEx: Solve

1684 x

?x

2) Move everything that is NOT x to the other side. When solving equations, do alladdition/subtraction first. Then all multiplication/division last.

8 8

244 x_____________

4 4

6xMake sure you are where you wanted to end up,x = some number.

Example 1Example 1Solving an equation in one variable algebraically means to find the value of the

variable that makes the mathematical statement true using appropriate algebraic

operations.

Example 1:Example 1: Solve the equation 5x - 4 = 0 .

5x – 4 = 0 + 4 +4

5x = 4

5x = 4 5 5 x =

5

4

This means makes

5x – 4 = 0 true.

5

4

One of the great things about solving equations is that we can always check our

solutions! We do this by plugging the value(s) we get for our variable into the

original equation and verify that we have a true statement.

Example 1:Example 1: Solve the equation 5x - 4 = 0 .

5x – 4 = 0 + 4 +4

5x = 4

5x = 4 5 5 x = 5

4

Check:

04445

4545

x

Original Equation

Now, plug in the value of x that you just found

We simplify and get 4-4, which is 0.This is what our original equation stated. This is true.

Example 2Example 2

Example 1:Example 1: Solve the equation 5(h – 2) = -4(3-h).

Solution: We have a couple of choices as to how to solve this equation. You may notice that it is not in the form introduced earlier, but it still a linear equation in one variable, since it can be written in the form described above. Our goal is to isolate the variable h, so that it appears on one side of the equation, and its value appears on the other. We will need to distribute on both sides of the equation before we can do that.

5(h - 2) = -4(3 – h)

5h – 10 = -12 + 4h +10 +10

5h = -2 + 4h

5h = -2 + 4h-4h - 4h

h = -2

Check:

5(h - 2) = -4(3 – h)

5((-2) – 2) = -4(3 – (-2))

5( - 4) = -4 ( 3 + 2)

-20 = - 4 (5)

-20 = -20

Since both sides of the equation yield the same result, we know that our answer is correct!

Example 1Example 1

Example 1: Solve the equationExample 1: Solve the equation10

1

9

1

2

1

3

5 mm

Remember: Solving Equations

First, you must know what you are solving for so you can isolate it.To do that:Take care of any exponents/FOIL or distribution/simplificationGet common denominators if necessaryCombine like terms on each side of the equal signAddition/Subtraction across = is done next to isolate the variableMultiplication/Division across =is done last and the variable should now be isolated

Example 1: SolutionExample 1: Solution

10

1

9

1

2

1

3

5 mm

10

190

9

190

2

190

3

590 mm

10

1

9

190

2

1

3

590 mm

91045150 mm-45 -45

5410150 mm-10m -10m

54140 m

140 140

70

27

140

54m

Check:Check: 5 1 1 1

3 2 9 10

5 27 1 9 1 9 7 2 1

3 70 2 14 2 14 14 14 7

1 27 1 3 1 3 7 10 1

9 70 10 70 10 70 70 70 7

?m m

We find that both sides of the equation give us the same result when we plug our answer in, which means that we obtained the correct result!

Practice ExamplesPractice Examples

Example 2:Example 2:

Example 3:Example 3:

Solve the equation 3p + 2 = 0 .

Solve the equation -7m – 1 = 0.

Solutions on next slide. Solve these on your own first.

Example 4:Example 4:Solve the equation 14z – 28 = 0.

Practice Examples AnswersPractice Examples AnswersExample 2:

Example 3:

Solve the equation 3p + 2 = 0 .

Solve the equation -7m – 1 = 0.

Example 4:Solve the equation 14z – 28 = 0.

3

2p

7

1m

2z

00

022

023

23

Check:

Check:

Check:

00

011

017

17

00

02828

028)2(14

More Practice ExamplesMore Practice ExamplesExample 5:

Example 6:

Solve the equation .


Solve these on your own first. Solutions on next slide.

19684 nn

542

1

3 x

x

More Practice Examples - More Practice Examples - AnswersAnswers

Example 5:

Example 6:



19684 nn

542

1

3 x

x

2

27

272

6 6

2764

8 8

19684

n

n

nn

nn

nn

2

27n

26

123

12326

120326

1202432

20462

1

36

2042

1

3

542

1

3

x

x

x

xx

xx

xx

xx

26

123x


Example 8:




xxx 45.0)200(10.005.0

364254 yyy

Example 7:

Example 8:



xxx 45.0)200(10.005.0

364254 yyy

x

x

x

xx

xx

xxx

45.36

45.36

55.005.20

45.01.005.20

45.01.02005.0

45.0)200(10.005.0

__

y

y

yy

yy

yyy

3

412

6122

6752

364254

x45.36

y 3

More Practice Examples - Answers


Example 10:

Solve the equation for m. .

Solve the equation for .


bmxy

hbbA 212

1 2b


Example 10:

Solve the equation for m. .

Solve the equation for .

bmxy

mx

by

mxby

bb

bmxy

mx

by

hbbA 212

1 2b

21

21

21

21

21

2

2

2

2

12)(2

2

1

bbh

A

bbh

A

hbbA

hbbA

hbbA

122

bh

Ab

More Practice ExamplesMore Practice ExamplesExample 11:Solve the equation for x. xxx 12

5

38

4

34

5

3

4

3

05

3

4

30

4 4

45

3

4

34

125

38

4

34

xx

xx

xxx

Obviously, , so the answer is No Solution5

3

4

3

No Solution

Example: The relationship between ºC and ºF can be represented by the equation

where F is the number of degrees Fahrenheit, and C is the number of degrees Celsius.

325

9 cF

a. Solve the above equation for C.b. Convert 98oF into degrees Celsiusa. Solve the above equation for C.b. Convert 98oF into degrees Celsius

Word Problem ExamplesWord Problem Examples

Solution a): We want to isolate C on one side of the equation. So, we apply the following operations on our original equation.

325

9 cF

-32 -32

cF5

932

cF

5

9

9

532

9

5

cF 329

5

or

)32(9

5 FC

Now we have an equation that allows us to compute degrees Celsius if we knew degrees Fahrenheit.

Solution b): We plug 98 in for F and solve for C.

325

9 cF

C

C

5

966

325

998

-32 -32

c

c

9330

5

95)66(5

9 9

067.333

110

9

330

9

330

c

or

c

So, 98ºF is approximately 36.67ºC.

Example: When you buy a new car, they say that the value of the car depreciates as soon as you drive it off the lot! Accountants use the following equation to measure depreciation of assets:

L

SCD

where …D is the depreciation of the asset per year, C is the initial cost of the asset, S is the salvage value, and L is the asset’s estimated life.

a. What is the salvage value of a machine that cost a company $40,000 initially, has an annual depreciation of $3000, and an estimated life of 10 years?

b. Solve the original equation for S, the salvage value, in general.

Solution a): We plug 30,000 in for C, 2000 for D, and 10 for L. We then solve for S.

L

SCD

10

000,403000

S

10

000,4010)3000(10

S

S 000,40000,30000,40000,40

S 000,10))(1()000,10)(1( S

S000,10So, the salvage value for the machine is $10,000.

Solution b): We want to isolate the variable S, treating all of the other letters in the equation as constants.

L

SCD

L

SCLDL )())((

SCLD -C -C

SCLD

))(1())(1( SCLD

SCLD

LDCS

or

This equation allows us to calculate the salvage value for any asset, given the initial cost, estimated life, and depreciation value.

module 1 ~ topic 1 solving equations table of contents slides 6-14: solving linear equations ...

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