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Part 1 A Library of Elementary Functions Chapter 1 – Linear Equations and Graphs Chapter 2 – Functions and Graphs Section 1 Simple Interest Chapter 1 Linear Equations and Graphs Section 1 - Linear Eq. & Ineq. Section 2 – Graphs & Lines Section 3 – Linear Regression 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 linear inequalities. The student will be able to solve applications involving linear equations and inequalities. Barnett/Ziegler/Byleen College Mathematics 12e 3 Table of Content Linear Equations Linear Inequalities Applications Barnett/Ziegler/Byleen College Mathematics 12e 4 Terms first-degree, or linear equation, inequality standard form solution, solution set equivalent equations sense o direction [inequality] double inequality Barnett/Ziegler/Byleen College Mathematics 12e interval notation inequality notation endpoint [interval] closed/open interval break-even analysis • Costs [fixed/variable] • revenues • loss/profit 5 Linear Equations and Inequalities 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. For example, the equation x 3 2( x 3) 5 3 is a linear equation because it can be converted to standard form by clearing of fractions and simplifying. Barnett/Ziegler/Byleen College Mathematics 12e 6 Linear Equations and 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. For example x 5 1 3x 2 2 is a linear inequality. Barnett/Ziegler/Byleen College Mathematics 12e 7 Linear Equations - Solution A solution of an equation (or inequality) involving a simple variable is a number that when substituted for the variable makes the equation (or inequality) true. The set of solutions is called the solution set. Barnett/Ziegler/Byleen College Mathematics 12e 8 Equivalent Equations Two 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. Barnett/Ziegler/Byleen College Mathematics 12e 9 Example (1) of Solving a Linear Equation Example: Solve 8x – 3(x – 4) = 3(x -4) + 6 Solution: 8x – 3(x – 4) = 3(x -4) + 6 Use la Propiedad Distributiva. 8x – 3x + 12 = 3x -12 + 6 Combine términos iguales. 5x + 12 = 3x - 6 Reste 3x de ambos lados. 2x + 12 = -6 Reste 12 de ambos lados. 2x = -18 Divida ambos lados por 2. x = -9 Barnett/Ziegler/Byleen College Mathematics 12e 10 Example (2) 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 of 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 College Mathematics 12e 3( x 2) 2 x 30 3 x 6 2 x 30 x 6 30 x 24 11 Solving a Formula for a Particular Variable Example: Solve M =Nt +Nr for N. Barnett/Ziegler/Byleen College Mathematics 12e 12 Solving a Formula for a Particular Variable (1) Example: Solve M=Nt+Nr for N. Factor out N: Divide both sides by (t + r): M N (t r ) M N tr Barnett/Ziegler/Byleen College Mathematics 12e 13 Solving a Formula for a Particular Variable (2) Example: Solve A = P + Prt for (a) r in terms of A, P and t (b) P in terms de A, r and t Barnett/Ziegler/Byleen College Mathematics 12e 14 Solving a Formula for a Particular Variable (2) (a) r in terms of A, P y t A = P + Prt P + Prt = A Reverse the equation Subtract P from both sides Divide both members by Pt Barnett/Ziegler/Byleen College Mathematics 12e Prt = A – P A–P r = --------Pt 15 Solving a Formula for a Particular Variable (2) (b) P in terms of A, r y t A = P + Prt P + Prt = A Reverse the equation Factor out P Divide both sides by (1 + rt). Barnett/Ziegler/Byleen College Mathematics 12e P(1 +rt) = A A P = --------1 + rt 16 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. For example x 5 1 3x 2 2 is a linear inequality. Barnett/Ziegler/Byleen College Mathematics 12e 17 Linear Inequalities The inequality symbols have a very clear geometric interpretation on the real number line. If a < b, then a is to the left of b on the number line; if c > d, then c is to the right of d in the number line as indicated in the graph. a d b 0 c a < b, c > d Barnett/Ziegler/Byleen College Mathematics 12e 18 Solving Linear Inequalities The procedures used to solve linear inequalities in one variable are almost the same as those used to solve linear equations in one variable, but with one important exception, as follows: An equivalent inequality will result, and the same sense or direction will remains the same if each side of the original inequality 1. has same real number added to or subtracted from it. 2. is multiplied or divided by the same positive number. Barnett/Ziegler/Byleen College Mathematics 12e 19 Solving Linear Inequalities But, in the other hand … An equivalent inequality will result, and the same sense or direction will reverse if each side of the original inequality 3. is multiplied or divided by the same negative number. Multiplication by 0 and division by 0 are not permitted. Barnett/Ziegler/Byleen College Mathematics 12e 20 Solving Linear Inequalities We 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 write 6 < 27 to have a true statement. The sense of the inequality reverses. Barnett/Ziegler/Byleen College Mathematics 12e 21 Interval and Inequality Notation If a < b, the double inequality a < x < b means that a < x and x < 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. Barnett/Ziegler/Byleen College Mathematics 12e 22 Interval and Inequality Notation Interval Inequality [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 Barnett/Ziegler/Byleen College Mathematics 12e Line Graph x a b x a b a b x x a b x a x a x b x b 23 Interval 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. Barnett/Ziegler/Byleen College Mathematics 12e 24 Interval 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 [ ) -5 2 x (B) x ≥ –2 is equivalent to [–2, ∞) [ x -2 Barnett/Ziegler/Byleen College Mathematics 12e 25 Example for Solving a Linear Inequality Solve the inequality 3(x – 1) < 5(x + 2) – 5 Barnett/Ziegler/Byleen College Mathematics 12e 26 Example for Solving a Linear Inequality Solve the inequality 3(x – 1) < 5(x + 2) – 5 Solution: 3(x –1) < 5(x + 2) – 5 3x – 3 < 5x + 10 – 5 3x – 3 < 5x + 5 Distribute the 3 and the 5 Combine like terms. –2x < 8 Subtract 5x from both sides, and add 3 to both sides x > -4 Notice that the sense of the inequality reverses when we divide both sides by -2. Barnett/Ziegler/Byleen College Mathematics 12e 27 Example for Solving a Double Inequality Solve and graph: -3 < 2x + 3 ≤ 9 We are looking for all numbers x such that 2x + 3 is between -3 and 9, including 9 but not -3. We proceed as before except that we try to isolate x in the middle: Solution: -3 < 2x + 3 ≤ 9 -3 - 3 < 2x + 3 - 3 ≤ 9 - 3 -6 < 2x ≤ 6 -6/2 < 2x/2 ≤ 6/2 -3 < x ≤ 3 ó (-3, 3] ( ] -3 3 x 28 Procedure for Solving Word Problems 1. 2. 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. 3. 4. 5. Barnett/Ziegler/Byleen College Mathematics 12e 29 Example: Break-Even Analysis A 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? Barnett/Ziegler/Byleen College Mathematics 12e 30 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 College Mathematics 12e 31 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 College Mathematics 12e 32 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 College Mathematics 12e 33 Chapter 1 Linear Equations and Graphs Section 1 Linear Equations and Inequalities END Last Update: Juanuary 24/2013