Upload
cecil-oscar-craig
View
229
Download
1
Embed Size (px)
Citation preview
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