linear equations and linear inequalities chapter 1 phase 1

Linear Equations and Linear Inequalities

Chapter 1Phase 1

EquationsStatement where two expressions are

equalExamples

x + 2 = 9x = 5x – 311x = 5x + 6x

Solutions to EquationsTo solve an equation means to find all numbers

that make the equation a true statementThese numbers are called the “solution”

Equation have “equal signs”What is on the left of the equal sign must have

the same value as what is on the right of the equal sign

Examplex = 4 for the equation x + 1 = 5 because if we

substitute 4 into the equation for x it is a true statement

x + 1 = 5 4 + 1 = 5 5 = 5

Properties of EquationsEquivalent equations (Equal) are a result of

Same quantity is added or subtracted on both sides of the equal sign

Same nonzero quantity is multiplied or divided on both sides of the equal sign

Linear EquationsFirst Degree Equations

Highest exponent on the x variable is a 1When there is not an exponent shown, the

default exponent is 1Example

x = x1

Solving Linear Equations – Starting SimpleSolve the following linear equation for x

x + 5 = 8 - 5 -5

========== x = 3

Check it by plugging in x = 3 into the original equation x + 5 = 8

3 + 5 = 8 8 = 8

Subtract 5 from both sides of the equation

Solving Linear Equations – Little HarderSolve the following linear equation for x

5x + 4 = 12 + 3x -3x - 3x ============ 2x + 4 = 12 -4 -4 ============ 2x = 8 /2 /2 ============ x = 4

Collect “like terms”

Subtract 3x on both sides of the equationCollect “like

terms”Subtract 4 on both sides of the equation

“Isolate x” by dividing both

sides by 2

Reminder: Forward slash represents division

Solving Linear Equations – Little HarderSolve the following linear equation for x

(CHECK IT!)5x + 4 = 12 + 3x

5(4) + 4 = 12 + 3(4) 20 + 4 = 12 + 12 24 = 24

We just found x = 4

PLUG IT IN!

Solving Linear Equations – Wow!

Solve and check the following linear equation5x – 2(x – 3) = 2(x – 2) + 15

Remove parentheses5x – 2x + 6 = 2x – 4 + 15

Remember a negative and a negative make a positiveCollect like terms on each side

3x + 6 = 2x + 11Subtract 6 from both sides

3x + 6 – 6 = 2x + 11 – 63x = 2x + 5

Subtract 2x from both sides3x – 2x = 2x + 5 – 2x x = 5

CHECK IT!Verify results by plugging in x = 5

5x – 2(x – 3) = 2(x – 2) + 155(5) – 2(5 – 3) = 2(5 – 2) + 15

Remember Order of Operations, PEMDAS25 – 2(2) = 2(3) + 1525 – 4 = 6 + 1521 = 21

Solving Linear Equations – Horrid FractionsSolve the following linear equation for x

Find common denominatorHow about 15? Multiply every fraction by 15!

3

1

3

2

5

x

3

115

3

215

515

x

Solving Linear Equations – Horrid Fractions

3

115

3

215

515

x

5103 x

10510103 x

153 x3

15

3

3x

5x

Solving Linear Equations – Horrid FractionsCHECK IT! Plug in x = 5

3

1

3

2

5

x

3

1

3

2

5

5

Common Denominator = 15

15

5

15

10

15

15

15

5

15

5

Solving for a Particular VariableSolve the following equation for y

6x – 2y = 4 - 6x -6x =========== - 2y = 4 – 6x /(-2) /(-2) =========== y = -2 + 3x

Subtract both sides by 6x

Divide both sides by

NEGATIVE 2

NOTE: You are dividing every

term on the right side by -2

Solving for a Particular VariableThe following equation can be used for

converting temperatures from Fahrenheit (F) to Celsius (C).

Solve the equation for C F = (9/5) C + 32 - 32 - 32 ============ F – 32 = (9/5) C * (5/9) *(5/9) ============= (5/9)(F-32) = C

Subtract both sides by 32

Multiply both sides by (5/9)

Linear InequalitiesInequality Symbols

< Less Than> Greater Than<= Less Than or Equal>= Great Than or Equal

Solving Linear Inequalities Properties

Inequality will stay the same ifAdd or Subtract a real valueMultiply or divide a positive value

Inequality will reverse ifMultiply or divide a negative value

NOTE: Multiplication by 0 is not allowed

Solving Linear InequalitiesSolve the following inequality

Subtract 3 on both sides of the inequality

Divide negative value of -2 to both sides (sign reverses)

432 x

34332 x

72 x

2

7

2

2

x

2

7x

Double InequalitiesVariable is between two valuesExample:

a < x <bWhich means

a < xand

x < bx is between a and b

Solving Double InequalitiesSolve: -9 < 2x + 1 < 7Looking for all numbers x such that 2x + 1 is

between -9 and 7Isolate x in the middle-9 < 2x + 1 <7

Subtract 1-9 – 1 < 2x + 1 – 1 < 7 – 1 -10 < 2x < 6

Divide 2-10/2 < 2x/2 < 6/2-5 <x < 3 or (-5,3)

Solving Word ProblemsUnderstand what the question is asking you

to findThat becomes the variable

Use only needed information to solve problem

See if answer makes sense!

Solving Word Problems with Equations“Equilibrium point” or “Break-Even point”

Where supply = demand; sales = costs, etc.Set up the equationsSet the equations equal to each otherSolve for the desired variable

Solving Word Problems with EquationsA company produces fasteners that cost

two cents per fastener to make plus five dollars for fixed costs. The company sells the fasteners for twenty-seven cents each. Where is the break-even point?

Set up equationsLet x = the number of fasteners producedCost = 0.02x + 5Revenue = 0.27x

Set Cost = Revenue and solve for x

Solving Word Problems with EquationsSet Cost = Revenue and solve for x

Cost = 0.02x + 5Revenue = 0.27x

0.02x + 5 = 0.27x -0.02x -0.02x ============= 5 = 0.25x /0.25 /0.25 ============== 20 = x

Move x’s to one side by

subtracting .02x on both sides of

equationDivide both sides

by 0.25x to isolate x

Company must sell 20 fasteners to

break even

In your Math class you have completed four tests with grades of 68, 82, 87 and 89. If all tests are weighted equally, what grade range should you earn on the last test to ensure a B (80-89 range) in the class?How do you find an average?

895

8987826880

x

Solving Word Problems with Inequalities

Solving Word Problems with InequalitiesSolve the inequality for x

Multiply 5 to all sides of the inequality

895

8987826880

x

5*895*5

898782685*80

x

44589878268400 x

Add the current grades and subtract value from all sides

Need to make at least a 74 on the last testDid we really need the right side of the

inequality and does the value make sense?

44589878268400 x

326445326326326400 x

11974 x

Solving Word Problems with Inequalities

Interval Notation

Interval Notation Inequality Notation[a,b] a ≤ x ≤ b[a,b) a ≤ x < b(a,b] a < x ≤ b(a,b) a < x < b

(-∞,a] x ≤ a(-∞,a) x < a[b,∞) x ≥ b(b,∞) x > b

Rectangular / Cartesian Coordinate System

Page 29

Originx-axis

abscissa

y-axisordinate

Plotting PointsWhere would we graph the following points?(-5,1), (0,3), (4,3), (3,0), (3,-3), (0,-4), (-5,-2)

(-5,1)

(0,3)(4,3)

(3,0)

(0,-4)

(3,-3)(-5,-2)

Plotting (x,y) coordinates

Graphing Linear Equations vs. Linear Inequalities

Linear EquationsGraph of a lineOrdered pairs (x,y) solutions

to the equation Linear Inequalities

Graph of an “area” (line could be included)

Ordered pairs (x,y) solutions to the inequality

Linear Equation in Two VariablesStandard Form

Ax + By = CExample

3x – 2y = 6One solution x = 4 and y = 3 or the ordered

pair (4,3)Find other solutions by picking a value for one

variable and solving for the otherLet x = 23(2) – 2y = 6 6 – 2y = 6 -2y = 0 y = 0Thus, another solution is (2,0)

Graph Linear Equation in Two VariablesAx + By = C is a lineIf A ≠ 0 and B ≠ 0, then the equation can

be rewritten as

If A = 0 and B ≠ 0 graphs a horizontal line

If A ≠ 0 and B = 0 graphs a vertical line

B

Cx

B

Ay

B

Cy

B

Ay

Slope of a LineNumerical measurement of the steepness

of a lineLine passes through two distinct points

P1(x1,y1) and P(x2,y2)

(run) change horizontal

(rise) change vertical

12

12

xx

yym

Slope of a LineFind the slope of a line that passes through

the points (-2, 3) and (4, -5)

3

4

6

8

)2(4

35

12

12

xx

yym

Other Forms of Linear EquationsSlope-Intercept Form

y = mx + bm is the slope (steepness of the line)b is the y intercept (where the line crosses the y

axis)Example

y = 3x + 5Where 3 is the slope and 5 is the y-intercept

Other Forms of Linear EquationsPoint-Slope Form

y – y1 = m(x – x1)Example:

Line has slope of 3 and point (2,4)y – 4 = 3(x – 2)Can now solve for y to put in slope-intercept form

y – 4 = 3x – 6y – 4 + 4 = 3x – 6 + 4y = 3x – 2

04/19/23Page 38

InterceptsWhere the line crosses the axis

x-intercept is where the line crosses the x-axis or where y=0Example: (5,0)

y-intercept is where the line crosses the y-axis or where x=0Example: (0,3)

© Cindy Roberts 2007

04/19/23Page 39

y-intercept (0,5)

x-intercept (5,0)

Distance FormulaGiven two points: P1(x1,y1) and P(x2,y2)

Find the distance between the two points

212

212 )()( yyxxd

Distance FormulaFind the distance between (3,2) and (7,4)

212

212 )()( yyxxd

22 )24()37( d22 )2()4( d

416 d47.420 d

Application of Linear EquationCost EquationThe management of a company that

manufacturers roller skates have fixed costs (costs at 0 output) of $300 per day and total costs of $4,300 per day at an output of 10 pairs of skates per day. Assume that cost is C is linearly related to output xFind the slope of the line joining the points

with outputs of 0 and 100; that is, the line passing through (0,300) and (100,4300)

Application of Linear EquationFind Slope from points (0,300) and

(100,4300)

40100

4000

0100

3004300

12

12

xx

yym

Application of Linear EquationThe management of a company that

manufacturers roller skates have fixed costs (costs at 0 output) of $300 per day and total costs of $4,300 per day at an output of 10 pairs of skates per day. Assume that cost is C is linearly related to output xFind an equation of the line relating output

to cost. Write the final answer in the form Cost = mx + b. Then graph the equation from 0 ≤ x ≤ 200

Application of Linear EquationCost Equation

Slope = 40Given the point (0,300) gives the y

intercept of 300Cost = 40x + 300

Graphing in Excel

x Cost = 40x + 3000 300

50 2300100 4300150 6300200 8300

x Cost = 40x + 3000 =40*A2+300

50 =40*A3+300100 =40*A4+300150 =40*A5+300200 =40*A6+300

Graph

Break—Even PointEquilibrium point

Supply = DemandCost = Revenue

Application – Supply and DemandThe following table lists the supply and

demand for barley in the United States during two recent years

Assume the relationship between supply and price is linear and between demand and price is linear

Find the supply and demand equationsFind the equilibrium pointYear Supply (mil bu) Demand (mil bu) Price ($/bu)

1990 7500 7900 2.28

1991 7900 7800 2.37

Application – Supply and DemandFind the supply equation of the form p =

mx + b, where p is the price per bushel in dollars and x is the corresponding supply in billions of bushelsUse two points to find the slope (7500, 2.28)

and (7900, 2.37)

Year Supply (mil bu) Demand (mil bu) Price ($/bu)

1990 7500 7900 2.28

1991 7900 7800 2.37

000225.0400

09.0

75007900

28.237.2

12

12

xx

yym

Application – Supply and DemandUse point-slope form to find the supply

equationp – p1 = m(x – x1)

p – 2.28 = 0.000225(x – 7500)p – 2.28 = 0.000225x – 1.6875p = 0.000225x + 0.5925

Application – Supply and DemandFind the demand equation of the form p =

mx + b, where p is the price per bushel in dollars and x is the corresponding demand in billions of bushelsUse two points to find the slope (7500, 2.28)

and (7900, 2.37)

Year Supply (mil bu) Demand (mil bu) Price ($/bu)

1990 7500 7900 2.28

1991 7900 7800 2.37

0009.0100

09.0

79007800

28.237.2

12

12

xx

yym

Application – Supply and DemandUse point-slope form to find the demand

equationp – p1 = m(x – x1)

p – 2.28 = -0.0009(x – 7900)p – 2.28 = -0.0009x + 7.11p = -0.0009x + 9.39

Application – Supply and DemandFind the equilibrium point set supply = demand

Supply: p = 0.000225x + 0.5925Demand: p = -0.0009x + 9.39

0.000225x + 0.5925 = -0.009x + 9.39Add 0.0009x on both sides

0.000225x + 0.5925 + 0.0009x = -0.009x + 9.39 + 0.0009x

0.001125x + 0.5925 = 9.39Subtract 0.5925 on both sides

0.001125x + 0.5925 – 0.5925 = 9.39 – 0.59250.001125x = 8.7975

Divide 0.001125 on both sides0.001125x / 0.001125 = 8.7975 / 0.001125x = 7820

Application – Supply and DemandPlug in x = 7820 into either equationp = 0.000225x + 0.5925p = 0.000225(7820) + 0.5925p = 2.352Equilibrium Point (7820, 2.352)

Mathematical ModelConstruct a mathematical model to

represent dataSolve the mathematical modelUse the model for predicting unknown

valuesModel using Excel’s Add Trendline FunctionLinear Regression

“Best Fit” Line that models data

Linear Regression ExampleThe following table contains recent average

and median purchase prices for a house in Texas.

Model the data in Excel to “find the regression equation”Let x = # of years since 2000Year Average Price (in thousands) Median Price (in thousands)

0 $146 $112 1 $150 $120 2 $156 $125 3 $160 $128 4 $164 $130 5 $174 $136

Graph Data in Excel (Scatterplot)

Find the Linear Regression EquationSelect GraphChart Tools / Layout / Trendline / More

Trendline OptionsTrend/Regression Type = LinearCheckbox Display Equation on chart

Linear Regression Equation

Linear Regression Equationy = 5.3143x + 145.05

Slope = 5.3143Every year the average price in thousands goes up

5.3143Y-intercept = 145.05

At year = 0 or year = 2000 the “predicted price” is $145.05 in thousands

PredictingPredict the average price in the year 2010

2010 would be 10 years after 2000 so x = 10y = 5.3143x + 145.05y = 5.3143(10) + 145.05y = 53.143 + 145.05y = 198.193 or $198,193 is the predicted

average price for the year 2010