gale bach intermediate algebra math 155 fall 2013 santa rosa junior college mathematics

Gale BachIntermediate Algebra

Math 155Fall 2013

Santa Rosa Junior College Mathematics

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Why Math?Why Math?

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Example Solve: 2( 1) 5( 1) 13x x x

Check:

Solution

2 2 5 5 13x x x 2( 1) 5( 1) 13x x x

2 5 2 5 13x x x 3 7 13x x 4 7 13x

4 20x 5x

2( 5 1) 5( 5 1) 5 13 12 20 8

The solution is x = – 5.

Types of Equations

An identity is an equation that is true for all replacements that can be used on both sides of the equation.

A contradiction is an equation that is never true.

A conditional equation is sometimes true and sometimes false, depending on what the replacement is.

Example Solve: 8 3 2 3(2 1).x x x

The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity equation.

Solution

8 3 2 3(2 1)x x x

8 2 33 6x x x Using the Distributive Law

8 3 8 3x x Combining like terms

0 = 0

Example Solve: 3 2 3( 5).x x x

The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction.

Solution3 2 3( 5)x x x

3 15

2 53 3 1x x x Using the Distributive Law

3 15x x Combining like terms

Example Solve: 11 3( 21).x x x

There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation.

Solution

11 3( 21)x x x

9 63x 7x

3 6311x x x Using the Distributive Law

8 63x x Combining like terms

Solve

How about technology!

0.5x

1 0.00458 1.7787 (13.002 1.005)Y x x

.21402244x

.214x

Solving FormulasA formula is an equation that uses variables to represent a relationship between two or more quantities.

Example

r

The area of a circle of radius r is given by the formula:

2.A r

Example

Solve , for (volume formula)V lwh w

V l w h

Solution

V l w h

l h l h

Divide by l and h

Vw

l h

Simplify

We want this variable alone.

Example

Solve 1, for (equation of line).x y

ya b

Solution

1x y

a b

1y x

b a Subtract x/a

1x

y ba

Multiply by b

Can you think of another strategy for solving for y?

Example

2 1Solve , for .Y mx mx m

Solution

2 1Y mx mx

To Solve a Formula for a Specified Variable

1. Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses.

• To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law.

Example

2 1Solve , for .Y mx mx m

Solution

2 1Y mx mx

Subtracting1 2Y mx mx

To Solve a Formula for a Specified Variable

1. Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses.

• To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law.

2. When all the terms with the specified variable are on the same side, factor (if necessary) so that the variable is written only once.

Example

2 1Solve , for .Y mx mx m

Solution

2 1Y mx mx

Subtracting1 2Y mx mx

Factoring 1 2Y m x x

To Solve a Formula for a Specified Variable

1. Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses.

• To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law.

2. When all the terms with the specified variable are on the same side, factor (if necessary) so that the variable is written only once.

3. Solve for the variable in question by dividing both sides by the multiplier of that variable.

Example

2 1Solve , for .Y mx mx m

Solution

2 1Y mx mx

Subtracting1 2Y mx mx

Factoring 1 2Y m x x

Dividing 1 2

1 2 1 2

m x xY

x x x x

1 2

Ym

x x

Simplifying

The Three-Step Strategy

Three Steps for Problem Solving with Algebra

1. Introduction

2. Body

3. ConclusionThe 3 amigos use this format.

The First Step in Problem Solving

Introduction: 1. If the problem is written, read it carefully. Then read it again,

perhaps aloud. Verbalize the problems to yourself.

2. List the information given and restate the question being asked.

3. Select a variable or variables to represent any unknown(s) and clearly state what each variable represents. Be descriptive! For example, let t = the flight time, in hours; let p = Paul’s weight, in pounds; and so on. Let the variable represent the value that you know least about.

4. Find additional information. Look up formulas or definitions with which you are not familiar. Geometric formulas appear on the inside back cover of this text; important words appear in the index.

The First Step in Problem Solving (continued)

Introduction:

5. Create a table, using variables, in which both known and unknown information is listed. Look for patterns.

6. Make and label a drawing.

The Second Step in Problem Solving

Body:

1. Translate the problem to mathematical language. In some cases, translation can be done by writing an algebraic expression, but most problems in this course are best solved by translating to an equation.

2. Solve the equation.

The Third Step in Problem Solving

Conclusion:1. State your results in sentence form.

2. Make sure your answer makes sense.

Translating to Algebraic Expressions

per of less than more than

ratio twicedecreased byincreased by

quotient of

times minus plus divided by

product ofdifference of sum of

divide multiply subtract add

DivisionMultiplicationSubtractionAddition

Key Words

Example

Translate to an algebraic expression:

“Eight more than twice the product of 5 and a number.”

Solution 8 2 5 n

Eight more than the product of 5 and a number.twice

Example

The faculty discount at a bookstore is 15%. If a sweatshirt after the discount was $32.30. What was the original price of the sweatshirt?

Solution

1. Introduction. The original price minus the discount is equal to sale price.

Let S = the sweatshirt’s original price. (US $)

2. Body.

(0.15) $32.30S S

Solve the equation.

(0.15) 32.30S S

(0.85) 32.30S 38S

the sweatshirt price minus

the discount is

the sale price

$38 $5.70 $32.30

Check. Note that 15% of $38 would be (0.15)($38) = $5.70. When this is subtracted from $38 we have

Thus $38 checks with the original problem.

3. Conclusion. The original price of the sweatshirt was $38.

Example

One angle of a triangle has the same measure as a second angle. The third angle measures 10o more than three times the measure of the first angle. Find the measures of the angles.

Solution

1. Introduction. We need to find three angles. Let x = first angle, in degrees.

First The first angle x

Second The same as the first x

Third 10o more than three times the first 3x + 10

1. Introduction (continued).

First Angle + Second Angle + Third Angle = 180o

First x x

Second The same as the first x

Third 10o more than three times the first 3x + 10

2. Body.

First Angle + Second Angle + Third Angle = 180o

(3 10) 180x x x

2. Solve the equation.(3 10) 180x x x

5 10 180x 5 170x

34x

34x first angle

34x second angle

3 10 3(34) 10 112x third angle

Check.

34 34 112 180

3. Conclusion. The three angles are 34o, 34o, and 112o.