Linear Equations and Linear Inequalities Chapter 1 Phase 1 Linear Equations and Linear Inequalities

Equations Statement where two expressions are equal Examples x + 2 = 9 x = 5x – 3 11x = 5x + 6x

Solutions to Equations To solve an equation means to find all numbers that make the equation a true statement These 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 Example x = 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 Equations Equivalent 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 Equations First Degree Equations Example Highest exponent on the x variable is a 1 When there is not an exponent shown, the default exponent is 1 Example x = x1 Phase 1 DB Part A

Solving Linear Equations – Starting Simple Solve the following linear equation for x x + 5 = 8 - 5 -5 ========== x = 3 Check it by plugging in x = 3 into the original equation 3 + 5 = 8 8 = 8 Subtract 5 from both sides of the equation

Solving Linear Equations – Little Harder Solve 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 equation Collect “like terms” Subtract 4 on both sides of the equation MML Homework Question #1 “Isolate x” by dividing both sides by 2 Reminder: Forward slash represents division

Solving Linear Equations – Little Harder Solve 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! MML Homework Question #1

Solving Linear Equations – Wow! Solve and check the following linear equation 5x – 2(x – 3) = 2(x – 2) + 15 Remove parentheses 5x – 2x + 6 = 2x – 4 + 15 Remember a negative and a negative make a positive Collect like terms on each side 3x + 6 = 2x + 11 Subtract 6 from both sides 3x + 6 – 6 = 2x + 11 – 6 3x = 2x + 5 Subtract 2x from both sides 3x – 2x = 2x + 5 – 2x x = 5 MML Homework Chapter1 Question #5

CHECK IT! Verify results by plugging in x = 5 5x – 2(x – 3) = 2(x – 2) + 15 5(5) – 2(5 – 3) = 2(5 – 2) + 15 Remember Order of Operations, PEMDAS 25 – 2(2) = 2(3) + 15 25 – 4 = 6 + 15 21 = 21

Solving Linear Equations – Horrid Fractions Solve the following linear equation for x Find common denominator How about 15? Multiply every fraction by 15! MML Homework Question #4

Solving Linear Equations – Horrid Fractions

Solving Linear Equations – Horrid Fractions CHECK IT! Plug in x = 5 Common Denominator = 15

Solving for a Particular Variable Solve 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 MML Homework Question #6

Solving for a Particular Variable The 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) MML Homework Question #7

Linear Inequalities Inequality Symbols < Less Than > Greater Than <= Less Than or Equal >= Great Than or Equal Phase 1 DB Part A

Solving Linear Inequalities Properties Inequality will stay the same if Add or Subtract a real value Multiply or divide a positive value Inequality will reverse if Multiply or divide a negative value NOTE: Multiplication by 0 is not allowed

Solving Linear Inequalities Solve the following inequality Subtract 3 on both sides of the inequality Divide negative value of -2 to both sides (sign reverses) MML Homework Chapter1 Question #1

Double Inequalities Variable is between two values Example: a < x <b Which means a < x and x < b x is between a and b

Solving Double Inequalities Solve: -9 < 2x + 1 < 7 Looking for all numbers x such that 2x + 1 is between -9 and 7 Isolate 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) MML Homework Chapter1 Question #3

Solving Word Problems Understand what the question is asking you to find That 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 equations Set the equations equal to each other Solve for the desired variable

Solving Word Problems with Equations A 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 equations Let x = the number of fasteners produced Cost = 0.02x + 5 Revenue = 0.27x Set Cost = Revenue and solve for x

Solving Word Problems with Equations Set Cost = Revenue and solve for x Cost = 0.02x + 5 Revenue = 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 equation Divide both sides by 0.25x to isolate x Company must sell 20 fasteners to break even

Solving Word Problems with Inequalities 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? Phase 1 DB

Solving Word Problems with Inequalities Solve the inequality for x Multiply 5 to all sides of the inequality

Solving Word Problems with Inequalities Add the current grades and subtract value from all sides Need to make at least a 74 on the last test Did we really need the right side of the inequality and does the value make sense?

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 x-axis abscissa Origin y-axis ordinate

Where would we graph the following points? Plotting Points Where would we graph the following points? (-5,1), (0,3), (4,3), (3,0), (3,-3), (0,-4), (-5,-2) (0,3) (4,3) (-5,1) (3,0) (-5,-2) (3,-3) (0,-4) Plotting (x,y) coordinates

Graphing Linear Equations vs. Linear Inequalities Graph of a line Ordered 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 Phase 1 DB

Linear Equation in Two Variables Standard Form Ax + By = C Example 3x – 2y = 6 One 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 other Let x = 2 3(2) – 2y = 6  6 – 2y = 6  -2y = 0  y = 0 Thus, another solution is (2,0)

Graph Linear Equation in Two Variables Ax + By = C is a line If 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 MML Homework Chapter1 Question #14

Slope of a Line Numerical measurement of the steepness of a line Line passes through two distinct points P1(x1,y1) and P(x2,y2) MML Homework Chapter1 Question #16

Slope of a Line Find the slope of a line that passes through the points (-2, 3) and (4, -5) MML Homework Chapter1 Question #16

Other Forms of Linear Equations Slope-Intercept Form y = mx + b m is the slope (steepness of the line) b is the y intercept (where the line crosses the y axis) Example y = 3x + 5 Where 3 is the slope and 5 is the y-intercept MML Homework Question #10

Other Forms of Linear Equations Point-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 – 6 y – 4 + 4 = 3x – 6 + 4 y = 3x – 2 MML Homework Question #15

Intercepts Where the line crosses the axis x-intercept is where the line crosses the x-axis or where y=0 Example: (5,0) y-intercept is where the line crosses the y-axis or where x=0 Example: (0,3) 4/22/2017 © Cindy Roberts 2007

x-intercept (5,0) y-intercept (0,5) 4/22/2017 MML Homework Question #13 4/22/2017

Distance Formula Find the distance between the two points Given two points: P1(x1,y1) and P(x2,y2) Find the distance between the two points Phase 1 DB Part B #1

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

Application of Linear Equation Cost Equation The 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 x Find 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 Equation Find Slope from points (0,300) and (100,4300)

Application of Linear Equation The 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 x Find 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 Equation Cost Equation Slope = 40 Given the point (0,300) gives the y intercept of 300 Cost = 40x + 300

Graphing in Excel x Cost = 40x + 300 =40*A2+300 50 =40*A3+300 100 =40*A2+300 50 =40*A3+300 100 =40*A4+300 150 =40*A5+300 200 =40*A6+300 x Cost = 40x + 300 300 50 2300 100 4300 150 6300 200 8300

Graph

Break—Even Point Equilibrium point Supply = Demand Cost = Revenue

Application – Supply and Demand The 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 equations Find the equilibrium point Year Supply (mil bu) Demand (mil bu) Price ($/bu) 1990 7500 7900 2.28 1991 7800 2.37

Application – Supply and Demand Find 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 bushels Use 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 7800 2.37

Application – Supply and Demand Use point-slope form to find the supply equation p – p1 = m(x – x1) p – 2.28 = 0.000225(x – 7500) p – 2.28 = 0.000225x – 1.6875 p = 0.000225x + 0.5925

Application – Supply and Demand Find 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 bushels Use 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 7800 2.37

Application – Supply and Demand Use point-slope form to find the demand equation p – p1 = m(x – x1) p – 2.28 = -0.0009(x – 7900) p – 2.28 = -0.0009x + 7.11 p = -0.0009x + 9.39

Application – Supply and Demand Find the equilibrium point set supply = demand Supply: p = 0.000225x + 0.5925 Demand: p = -0.0009x + 9.39 0.000225x + 0.5925 = -0.009x + 9.39 Add 0.0009x on both sides 0.000225x + 0.5925 + 0.0009x = -0.009x + 9.39 + 0.0009x 0.001125x + 0.5925 = 9.39 Subtract 0.5925 on both sides 0.001125x + 0.5925 – 0.5925 = 9.39 – 0.5925 0.001125x = 8.7975 Divide 0.001125 on both sides 0.001125x / 0.001125 = 8.7975 / 0.001125 x = 7820

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

Mathematical Model Construct a mathematical model to represent data Solve the mathematical model Use the model for predicting unknown values Model using Excel’s Add Trendline Function Linear Regression “Best Fit” Line that models data

Linear Regression Example The 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 2000 Year Average Price (in thousands) Median Price (in thousands) $146 $112 1 $150 $120 2 $156 $125 3 $160 $128 4 $164 $130 5 $174 $136 Phase 1 IP MML Homework Chapter1 Question #19

Graph Data in Excel (Scatterplot)

Find the Linear Regression Equation Select Graph Chart Tools / Layout / Trendline / More Trendline Options Trend/Regression Type = Linear Checkbox Display Equation on chart

Linear Regression Equation

Linear Regression Equation y = 5.3143x + 145.05 Slope = 5.3143 Every year the average price in thousands goes up 5.3143 Y-intercept = 145.05 At year = 0 or year = 2000 the “predicted price” is $145.05 in thousands

Predicting Predict the average price in the year 2010 2010 would be 10 years after 2000 so x = 10 y = 5.3143x + 145.05 y = 5.3143(10) + 145.05 y = 53.143 + 145.05 y = 198.193 or $198,193 is the predicted average price for the year 2010