Chapter 1 Linear Equations and GraphsLinear Equations, Standard FormEquivalent EquationsExample of Solving a Linear EquationSlide 5Solving a Formula for a Particular VariableSlide 7Linear InequalitiesSolving Linear InequalitiesExample for Solving a Linear InequalitySlide 11Interval and Inequality NotationInterval and Inequality Notation and Line GraphsSlide 14Procedure for Solving Word ProblemsExample: Break-Even AnalysisBreak-Even Analysis (continued)Slide 18Slide 19Chapter 1Linear Equations and GraphsSection 1Linear Equations and Inequalities2Linear Equations, Standard Form0 bax53)3(23 xxwhere a is not equal to zero. This is called the standard form of the linear equation.For example, the equationis a linear equation because it can be converted to standard form by clearing of fractions and simplifying.In general, a first-degree, or linear, equation in one variable is any equation that can be written in the form3Equivalent EquationsTwo equations are equivalent if one can be transformed into the other by performing a series of operations which are one of two types:1. The same quantity is added to or subtracted from each side of a given equation.2. Each side of a given equation is multiplied by or divided by the same nonzero quantity.To solve a linear equation, we perform these operations on the equation to obtain simpler equivalent forms, until we obtain an equation with an obvious solution.4Example of Solving a Linear EquationExample: Solve5322 xx5Example of Solving a Linear Equation2430630263302)2(3563226xxxxxxxxExample: SolveSolution: Since the LCD of 2 and 3 is 6, we multiply both sides of the equation by 6 to clear of fractions.Cancel the 6 with the 2 to obtain a factor of 3, and cancel the 6 with the 3 to obtain a factor of 2.Distribute the 3.Combine like terms.5322 xx6Solving a Formula for a Particular VariableExample: Solve M =Nt +Nr for N.7Solving a Formula for a Particular VariableExample: Solve M=Nt+Nr for N.( )M N t rMNt r= +=+Factor out N:Divide both sides by (t + r):8Linear InequalitiesIf the equality symbol = in a linear equation is replaced by an inequality symbol (<, >, ≤, or ≥), the resulting expression is called a first-degree, or linear, inequality. For exampleis a linear inequality. ( )5 1 3 22xx� - +9Solving Linear InequalitiesWe can perform the same operations on inequalities that we perform on equations, except that the sense of the inequality reverses if we multiply or divide both sides by a negative number. For example, if we start with the true statement –2 > –9 and multiply both sides by 3, we obtain–6 > –27.The sense of the inequality remains the same. If we multiply both sides by -3 instead, we must write6 < 27to have a true statement. The sense of the inequality reverses.10Example for Solving a Linear InequalitySolve the inequality 3(x – 1) < 5(x + 2) – 511Example for Solving a Linear InequalitySolve the inequality 3(x – 1) < 5(x + 2) – 5Solution:3(x –1) < 5(x + 2) – 53x – 3 < 5x + 10 – 5 Distribute the 3 and the 53x – 3 < 5x + 5 Combine like terms.–2x < 8 Subtract 5x from both sides, and add 3 to both sidesx > -4 Notice that the sense of the inequality reverses when we divide both sides by -2.12Interval and Inequality NotationInterval Inequality Interval Inequality[a,b] a ≤ x ≤ b (–∞,a] x ≤ a[a,b) a ≤ x < b (–∞,a) x < a(a,b] a < x ≤ b [b,∞) x ≥ b(a,b) a < x < b (b,∞) x > bIf a < b, the double inequality a < x < b means that a < x andx < b. That is, x is between a and b. Interval notation is also used to describe sets defined by single or double inequalities, as shown in the following table.13Interval and Inequality Notation and Line Graphs(A) Write [–5, 2) as a double inequality and graph .(B) Write x ≥ –2 in interval notation and graph.14Interval and Inequality Notation and Line Graphs(A) Write [–5, 2) as a double inequality and graph .(B) Write x ≥ –2 in interval notation and graph.(A) [–5, 2) is equivalent to –5 ≤ x < 2[ ) x-52(B) x ≥ –2 is equivalent to [–2, ∞)[ x-215Procedure for Solving Word Problems1. Read the problem carefully and introduce a variable to represent an unknown quantity in the problem.2. Identify other quantities in the problem (known or unknown) and express unknown quantities in terms of the variable you introduced in the first step.3. Write a verbal statement using the conditions stated in the problem and then write an equivalent mathematical statement (equation or inequality.)4. Solve the equation or inequality and answer the questions posed in the problem.5. Check the solutions in the original problem.16Example: Break-Even AnalysisA recording company produces compact disk (CDs). One-time fixed costs for a particular CD are $24,000; this includes costs such as recording, album design, and promotion. Variable costs amount to $6.20 per CD and include the manufacturing, distribution, and royalty costs for each disk actually manufactured and sold to a retailer. The CD is sold to retail outlets at $8.70 each. How many CDs must be manufactured and sold for the company to break even?17Break-Even Analysis(continued)SolutionStep 1. Let x = the number of CDs manufactured and sold.Step 2. Fixed costs = $24,000 Variable costs = $6.20x C = cost of producing x CDs = fixed costs + variable costs = $24,000 + $6.20x R = revenue (return) on sales of x CDs = $8.70x18Break-Even Analysis(continued)Step 3. The company breaks even if R = C, that is if $8.70x = $24,000 + $6.20xStep 4. 8.7x = 24,000 + 6.2x Subtract 6.2x from both sides 2.5x = 24,000 Divide both sides by 2.5 x = 9,600 The company must make and sell 9,600 CDs to break even.19Break-Even Analysis(continued)Step 5. Check: Costs = $24,000 + $6.2 ∙ 9,600 = $83,520 Revenue = $8.7 ∙ 9,600 =