3.1 Solving Linear Equations Part I

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3.1 Solving Linear Equations Part I A linear equation in one variable can be written in the form: Ax + B = 0 Linear equations are solved by getting “x” by itself on one side of the equation Addition (Subtraction) Property of Equality: c b c a and c b c a then b a if ,

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3.1 Solving Linear Equations Part I. A linear equation in one variable can be written in the form: Ax + B = 0 Linear equations are solved by getting “x” by itself on one side of the equation Addition (Subtraction) Property of Equality:. 3.1 Solving Linear Equations Part I. - PowerPoint PPT Presentation

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Page 1: 3.1 Solving Linear Equations Part I

3.1 Solving Linear Equations Part I

• A linear equation in one variable can be written in the form: Ax + B = 0

• Linear equations are solved by getting “x” by itself on one side of the equation

• Addition (Subtraction) Property of Equality:

cbca

andcbca

thenbaif

,

Page 2: 3.1 Solving Linear Equations Part I

3.1 Solving Linear Equations Part I

• Multiplication Property of Equality:

• Since division is the same as multiplying by the reciprocal, you can also divide each side by a number.

• General rule: Whatever you do to one side of the equation, you have to do the same thing to the other side.

cbca

thencbaif

0,

c

b

c

a

thencbaif

0,

Page 3: 3.1 Solving Linear Equations Part I

3.1 Solving Linear Equations Part I

• Example: Solve by getting x by itself on one side of the equation.

Subtract 7 from both sides:

Divide both sides by 3:

15

453

5273

x

x

x

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3.2 Solving Linear Equations Part II- Fractions/Decimals

• As with expressions, you need to combine like terms and use the distributive property in equations.Example:

7

56491058

)715(753

x

x

xx

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3.2 Solving Linear Equations Part II- Fractions/Decimals

• Fractions - Multiply each term on both sides by the Least Common Denominator (in this case the LCD = 4):

Multiply by 4:

Reduce Fractions:

Subtract x:

Subtract 5:

17

125

1225

342

45

4

4

32

15

4

1

x

x

xx

xx

xx

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3.2 Solving Linear Equations Part II- Fractions/Decimals

• Decimals - Multiply each term on both sides by the smallest power of 10 that gets rid of all the decimals

Multiply by 100:Cancel:Distribute:Subtract 5x:Subtract 50:Divide by 5:

16805

30505

3055010

305510

3.10005.10051.100

3.05.51.

xx

x

xx

xx

xx

xx

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3.2 Solving Linear Equations Part II- Fractions/Decimals

• Eliminating fractions makes the calculation simpler:

Multiply by 94:

Cancel:

Distribute:Subtract x:Subtract 10:

104

9410

94102

9452

19494

945

47

94

194

15

47

1

x

x

xx

xx

xx

xx

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3.2 Solving Linear Equations Part II

• 1 – Multiply on both sides to get rid of fractions/decimals

• 2 – Use the distributive property• 3 – Combine like terms• 4 – Put variables on one side, numbers on the

other by adding/subtracting on both sides• 5 – Get “x” by itself on one side by multiplying or

dividing on both sides• 6 – Check your answers (if you have time)

Page 9: 3.1 Solving Linear Equations Part I

3.2 Solving Linear Equations Part II

• Example:Clear fractions:

Combine like terms:

Get variables on one side:

Solve for x:

361

21

32 xxx

1834 xxx

187 xx

186 x3x

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3.3 Applications of Linear Equations to General Problems

• 1 – Decide what you are asked to find• 2 – Write down any other pertinent information

(use other variables, draw figures or diagrams )• 3 – Translate the problem into an equation.• 4 – Solve the equation.• 5 – Answer the question posed.• 6 – Check the solution.

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3.3 Applications of Linear Equations to General Problems

• Example: The sum of 3 consecutive integers is 126. What are the integers?x = first integer, x + 1 = second integer, x + 2 = third integer

43,42,41

41

1233

12633

126)2()1(

x

x

x

xxx

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3.3 Applications of Linear Equations to General Problems

• Example: Renting a car for one day costs $20 plus $.25 per mile. How much would it cost to rent the car for one day if 68 miles are driven?$20 = fixed cost, $.25 68 = variable cost

37$

17$20$

25$.6820$

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3.4 Percent Increase/Decrease and Investment Problems

• A number increases from 60 to 81. Find the percent increase in the number.

%35

35.060

21%

216081

increase

increase

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3.4 Percent Increase/Decrease and Investment Problems

• A number decreases from 81 to 60. Find the percent increase in the number.

Why is this percent different than the last slide?

%26

26.081

21%

216081

decrease

decrease

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3.4 Percent Increase/Decrease and Investment Problems

• A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?

15$

128.

122.0.1

12%20

8.12

x

x

xx

xx

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3.4 Percent Increase/Decrease and Investment Problems

• Another Way: A flash drive is on sale for $12 after a 20% discount. What was the original price of the flash drive?Since $12 was on sale for 20% off, it is 100% - 20% = 80% of the original price set up as a proportion (see 3.6):

1580

1200

1200)100(1280100

8012

x

xx

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3.4 Percent Increase/Decrease and Investment Problems

• Simple Interest Formula:I = interestP = principalR = rate of interest per yearT = time in years

PRTI

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3.4 Percent Increase/Decrease and Investment Problems

• Example: Given an investment of $9500 invested at 12% interest for 1½ years, find the simple interest.

1710$

5.112.09500$

I

PRTI

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3.4 Percent Increase/Decrease and Investment Problems

• Example: If money invested at 10% interest for 2 years yields $84, find the principal.

420$2.

84$

2.210.084$

P

PP

PRTI

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3.5 Geometry Applications and Solving for a Specific Variable

• A = lw• • P = a + b + c• • •

• Area of rectangle• Area of a triangle• Perimeter of triangle• Sum of angles of a triangle• Area of a circle• Circumference of a circle

321180 mmm2rA rC 2

bhA 21

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3.5 Geometry Applications and Solving for a Specific Variable

• Complementary angles – add up to 90

• Supplementary angles – add up to 180

• Vertical angles – the angles opposite each other are congruent

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3.5 Geometry Applications and Solving for a Specific Variable

• Find the measure of an angle whose complement is 10 larger.

1. x = degree measure of the angle.

2. 90 – x = measure of its complement

3. 90 – x = 10 + x

4. Subtract 10: Add x: Divide by 2:

40

280

80

x

x

xx

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Ratio – quotient of two quantities with the same units

Note: percents are ratios where the second number is always 100:

ba

35.%35 10035

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Percents :

Example: If 70% of the marbles in a bag containing 40 marbles are red, how many of the marbles are red?:# of red marbles =

73.%73 10073

2840)70(.40%70 of

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Proportion – statement that two ratios are equal:

Solve using cross multiplication:

dc

ba

bcad

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Solve for x:

Solution:

79

381 x

60

9540

279567

)3(9781

x

x

x

x

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Example: d=rt (distance = rate time)How long will it take to drive 420 miles at 50 miles per hour?

hourst

t

4.8

50420

50420

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• General form of a mixture problem: x units of an a% solution are mixed with y units of a b% solution to get z units of a c% solution

Equations will always be:

)%()%()%( zcybxa

zyx

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3.6 Applications of Linear Equations to Proportions, d=rt, and Mixture Problems

• Example: How many gallons of a 10% indicator solution must be mixed with a 20% indicator solution to get 10 gallons of a 14% solution? Let x = # gallons of 10% solution,

then 10 - x = # gallons of 20% solution :

mixtureofgallonsx

x

xxx

xx

xx

%106

6010

140200101402020010

)10(14)10(2010

)10%(14)10%(20)%(10

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3.7 Solving Linear Inequalities in One Variable

• < means “is less than” means “is less than or equal to”

• > means “is greater than” means “is greater than or equal to”

note: the symbol always points to the smaller number

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3.7 Solving Linear Inequalities in One Variable

• A linear inequality in one variable can be written in the form:

ax < b (a0)

• Addition property of inequality:

if a < b then a + c < b + c

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3.7 Solving Linear Inequalities in One Variable

• Multiplication property of inequality:– If c > 0 then

a < b and ac < bc are equivalent

– If c < 0 thena < b and ac > bc are equivalent

note: the sign of the inequality is reversed when multiplying both sides by a negative number

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3.7 Solving Linear Inequalities in One Variable

• Example:36

121

32 xxx

1834 xxx

18 xx182 x

9x-9

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3.8 Solving Compound Inequalities

• For any 2 sets A and B, the intersection of A and B is defined as follows:

AB = {x x is an element of A and x is an element of B}

• For any 2 sets A and B, the union of A and B is defined as follows:

AB = {x x is an element of A or x is an element of B}

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3.8 Solving Compound Inequalities

• Example:422 and 1013 xx

22 and 93 xx

1 and 3 xx31 x

1 3

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3.8 Solving Compound Inequalities

• Example:422or 1013 xx

22or 93 xx

1or 3 xx

1 3