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:

cbcaandcbca

thenbaif

,

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.

cbcathencbaif

0,

cb

ca

thencbaif

0,

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:

154535273

xx

x

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:

756491058

)715(753

xx

xx

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:

17125

1225

34245

44

3215

41

xx

xx

xx

xx

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:

1680530505

3055010305510

3.10005.10051.1003.05.51.

xxx

xxxx

xxxx

3.2 Solving Linear Equations Part II- Fractions/Decimals

• Eliminating fractions makes the calculation simpler:

Multiply by 94:

Cancel:

Distribute:Subtract x:Subtract 10:

1049410

941029452

19494945

4794

19415

471

xx

xxxx

xx

xx

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)

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

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.

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

43,42,41411233

12633126)2()1(

xxx

xxx

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$.6820$

3.4 Percent Increase/Decrease and Investment Problems

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

%35

35.06021%

216081

increase

increase

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

26.08121%

216081

decrease

decrease

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.

122.0.112%20

8.12

xx

xxxx

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

12001200)100(1280

1008012

x

xx

3.4 Percent Increase/Decrease and Investment Problems

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

PRTI

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$5.112.09500$

IPRTI

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

PPPRTI

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

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

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 complement3. 90 – x = 10 + x4. Subtract 10:

Add x: Divide by 2:

40280

80

xxxx

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

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

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

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

• Solve for x:

Solution:

79

381 x

609540

279567)3(9781

xxxx

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?

hourstt4.8

50420

50420

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:

)%()%()%( zcybxazyx

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 :

mixtureofgallonsxx

xxxxx

xx

%1066010

140200101402020010)10(14)10(2010

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

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

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

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

3.7 Solving Linear Inequalities in One Variable

• Example:36

121

32 xxx

1834 xxx18 xx

182 x9x

-9

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}

3.8 Solving Compound Inequalities

• Example:422 and 1013 xx

22 and 93 xx1 and 3 xx

31 x

1 3

3.8 Solving Compound Inequalities

• Example:422or 1013 xx

22or 93 xx1or 3 xx

1 3