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Chapter 1 Linear Equations and Graphs Section 1 Linear Equations and Inequalities Learning Objectives for Section 1.1 Linear Equations and Inequalities The student will be able to solve linear equations. The student will be able to solve and graph linear inequalities. The student will be able to use inequality and interval notation. The student will be able to solve applications involving linear equations and inequalities. Barnett/Ziegler/Byleen Calculus 12e 2 Linear Equations, Standard Form In general, a first-degree, or linear, equation in one variable is any equation that can be written in the form ax b 0 where a is not equal to zero. This is called the standard form of the linear equation. Convert this linear equation to standard form: x 3 2( x 3) 5 3 Barnett/Ziegler/Byleen Calculus 12e 3 Example of Solving a Linear Equation Example: Solve x2 x 5 2 3 Barnett/Ziegler/Byleen Calculus 12e 4 Example of Solving a Linear Equation Example: Solve x2 x 5 2 3 Solution: Since the LCD of 2 and 3 x2 x is 6, we multiply both sides of the 6 65 3 equation by 6 to clear the fractions. 2 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. Barnett/Ziegler/Byleen Calculus 12e 3( x 2) 2 x 30 3 x 6 2 x 30 x 6 30 x 24 5 Solving a Formula for a Particular Variable Example: Solve M=Nt+Nr for N. Barnett/Ziegler/Byleen Calculus 12e 6 Solving a Formula for a Particular Variable Example: Solve M=Nt+Nr for N. Factor out N: Divide both sides by (t + r): Barnett/Ziegler/Byleen Calculus 12e M N (t r ) M N tr 7 Linear Inequalities If 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. Solve the linear inequality shown below: x 5 1 3x 2 2 Barnett/Ziegler/Byleen Calculus 12e 8 Solving Linear Inequalities Don’t forget! The direction of the inequality reverses if we multiply or divide both sides by a negative number. Barnett/Ziegler/Byleen Calculus 12e 9 Interval and Inequality Notation Interval notation is another way to write inequalities, as shown in the following table. This will be the notation that we use. 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 Barnett/Ziegler/Byleen Calculus 12e 𝐿𝑖𝑛𝑒 𝐺𝑟𝑎𝑝ℎ 10 Interval and Inequality Notation and Line Graphs Write each inequality using Interval Notation and graph: (A) x ≥ –2 [−2, ∞) (B) x < 3 (−∞, 3) (C) -1 ≤ x < 2 [−1, 2) [ −3 − 2 ) 2 3 [ −1 0 −1 4 ) 2 1 Write each interval using Inequality Notation and graph: (A) (2, 5] 2<x≤5 (B) (6, ) 𝑥>6 (C) (- , 0] 𝑥≤0 Barnett/Ziegler/Byleen Calculus 12e ( 2 4 3 ( 6 5 −1 ] 5 ] 0 7 1 11 Example for Solving a Linear Inequality Solve the inequality 3(x – 1) < 5(x + 2) – 5 And write your answer using interval notation. Barnett/Ziegler/Byleen Calculus 12e 12 Example for Solving a Linear Inequality Solution: 3(x –1) < 5(x + 2) – 5 3x – 3 < 5x + 10 – 5 3x – 3 < 5x + 5 –2x < 8 x > -4 (-4, ∞) Barnett/Ziegler/Byleen Calculus 12e 𝐼𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 13 Example for Solving a Double Inequality Solve the inequality -10 < -4x + 2 ≤ 6 Write your answer using interval notation. 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: −12 < −4𝑥 ≤ 4 3 > 𝑥 ≥ -1 −1 ≤ 𝑥 < 3 [−1, 3) Barnett/Ziegler/Byleen Calculus 12e 𝐼𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑁𝑜𝑡𝑎𝑡𝑖𝑜𝑛 14 Procedure for Solving Word Problems 1. 2. 3. 4. 5. Read the problem carefully and introduce a variable to represent an unknown quantity in the problem. Identify other quantities in the problem (known or unknown) and express unknown quantities in terms of the variable you introduced in the first step. Write a verbal statement using the conditions stated in the problem and then write an equivalent mathematical statement (equation or inequality.) Solve the equation or inequality and answer the questions posed in the problem. Check the solutions in the original problem. Barnett/Ziegler/Byleen Calculus 12e 15 Example: Ticket Sales The TO Civic Arts Center charges $20 for adult tickets and $15 for senior citizens tickets for a particular play. If $2225 in revenue is collected and a total of 115 tickets were sold, how many of each ticket were sold? Barnett/Ziegler/Byleen Calculus 12e 16 Ticket Sales (solution) Let x = the number of adult tickets sold Then 115 – x = the number of senior citizen tickets sold Since prices are $20 per adult ticket, $15 per senior citizen ticket, and the total revenue was $2225, then we can set up this equation: 20𝑥 + 15 115 − 𝑥 = 2225 20𝑥 + 1725 − 15𝑥 = 2225 1725 + 5𝑥 = 2225 5𝑥 = 500 𝑥 = 100 𝑆𝑜, 100 𝑎𝑑𝑢𝑙𝑡 𝑎𝑛𝑑 15 𝑠𝑒𝑛𝑖𝑜𝑟 𝑐𝑖𝑡𝑖𝑧𝑒𝑛 𝑡𝑖𝑐𝑘𝑒𝑡𝑠 𝑤𝑒𝑟𝑒 𝑠𝑜𝑙𝑑. Check: 20 100 + 15 15 = 2225 Barnett/Ziegler/Byleen Calculus 12e 17 Example: Break-Even Analysis A recording company produces compact disks (CDs). Onetime 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? Barnett/Ziegler/Byleen Calculus 12e 18 Break-Even Analysis (continued) Solution Step 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 = $8.70x Barnett/Ziegler/Byleen Calculus 12e 19 Break-Even Analysis (continued) Step 3. The company breaks even if R = C, that is if $8.70x = $24,000 + $6.20x Step 4. 8.7x = 24,000 + 6.2x 2.5x = 24,000 Subtract 6.2x from both sides Divide both sides by 2.5 x = 9,600 The company must make and sell 9,600 CDs to break even. Barnett/Ziegler/Byleen Calculus 12e 20 Break-Even Analysis (continued) Step 5. Check: Costs = $24,000 + $6.2 ∙ 9,600 = $83,520 Revenue = $8.7 ∙ 9,600 = $83,520 Barnett/Ziegler/Byleen Calculus 12e 21 Homework Barnett/Ziegler/Byleen Calculus 12e 22