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Equations and Inequalities Copyright © Cengage Learning. All rights reserved. 2 Section 2.1 Solving Basic Linear Equations in One Variable Copyright © Cengage Learning. All rights reserved. Objectives 1. Determine whether a statement is an 1 expression or an equation. 22. Determine whether a number is a solution of an equation. 33. Solve a linear equation in one variable by applying the addition or subtraction property of equality. 4. Solve a linear equation in one variable by 4 applying the multiplication or division property of equality. 3 Objectives 5 Application 1. 6 Application 2 7 5. Application 3 4 1. Determine whether a statement is an expression or an equation 5 Expression or Equation Equation: a statement indicating that two quantities are equal. • • • x + 5 = 21 2x – 5 = 11 3x2 – 4x + 5 = 0 Expression: Combination of variables and constants which evaluates to a single value • • • 6x – 1 3x2 – x – 2 –8(x + 1) 6 Your Turn Determine whether the following are expressions or equations. a. 9x2 – 5x = 4 b. 3x + 2 c. 6(2x – 1) + 5 Solution: a.9x2 – 5x = 4 is an equation b.3x + 2 is an expression c.6(2x – 1) + 5 is an expression 7 2. Determine whether a number is a solution of an equation 8 Solving an Equation Given: x + 5 = 21 The letter x is called the variable (or the unknown). Solution of the equation: A value for a variable which makes the equation true 16 is a solution, because (16) + 5 = 21 3 is not a solution, because (3) + 5 ≠ 21 9 Your Turn Determine whether 6 is a solution of 3x – 5 = 2x. Solution: To see whether 6 is a solution, we can substitute 6 for x and simplify. 3x – 5 = 2x 36–5≟ 26 18 – 5 ≟ 12 13 = 12 Substitute 6 for x. Do the multiplication. False. Since 13 = 12 is a false statement, 6 is not a solution. 10 3. Solve a linear equation in one variable by applying the addition or subtraction property of equality 11 Solving an Equation (Using add/subt prop) Addition Property of Equality Suppose that a, b, and c are real numbers. Then, If a = b, then a + c = b + c. Subtraction Property of Equality Suppose that a, b, and c are real numbers. Then, If a = b, then a – c = b – c. 12 Solving an Equation Comment The subtraction property of equality is a special case of the addition property. Instead of subtracting a number from both sides of an equation, we could add the opposite of the number to both sides. When we use the properties described above, the resulting equation will have the same solution set as the original one. We say that the equations are equivalent. Equivalent Equations Two equations are called equivalent equations when they have the same solution set. 13 Example Solve: x – 5 = 2 Solution: x–5=2 x–5+5=2+5 Add 5 to both sides of the equation. x+0=7 x=7 Apply the additive inverse property. Apply the additive identity property. 14 Example cont’d We check by substituting 7 for x in the original equation and simplifying. x–5=2 7–5≟2 2=2 Substitute 7 for x. True. Since the previous statement is true, 7 is a solution. The solution set of this equation is {7}. 15 4. Solve a linear equation in one variable by applying the multiplication or division property of equality 16 Solving an Equation (Using mult/div property) Multiplication Property of Equality Suppose that a, b, and c are real numbers. If a = b, then ca = cb. Division Property of Equality Suppose that a, b, and c are real numbers and c 0. If a = b, then = . 17 Example Solve: Solution: Multiply both sides by 3. 3 = x and 3 12 = 36. 18 5. Application 1 19 Example – Buying Furniture A sofa is on sale for $650. If it has been marked down $325, find its regular price. Solution: We can let r represent the regular price and substitute 650 for the sale price and 325 for the markdown in the following formula. 20 Example 8 – Solution cont’d We can use the addition property of equality to solve the equation. 650 = r – 325 650 + 325 = r – 325 + 325 975 = r Add 325 to both sides. 650 + 325 = 975 and –325 + 325 = 0. The regular price is $975. 21 6. Application 2 22 Application 2 A percent is the numerator of a fraction with a denominator of 100. For example, percent (written as ) is the fraction or the decimal 0.0625. In problems involving percent, the word of usually means multiplication. For example, 8,500. of 8,500 is the product of 0.0625 and of 8,500 = 0.0625 8,500 = 531.25 23 Percent In the statement of 8,500 = 531.25, the percent is called a rate, 8,500 is called the base, and their product, 531.25, is called the amount. Every percent problem is based on the equation rate base = amount. Percent Formula If r is the rate, b is the base, and a is the amount, then rb = a 24 Percent Percent problems involve questions such as the following. • What is 30% of 1,000? We must find the amount. • 45% of what number is 405? We must find the base. • What percent of 400 is 60? We must find the rate. When we substitute the values of the rate, base, and amount into the percent formula, we will obtain an equation that we can solve. 25 Example 10 What is 30% of 1,000? Solution: In this example, the rate r is 30% and the base is 1,000. We must find the amount. We can substitute these values into the percent formula and solve for a. rb = a 26 Example 10 – Solution 30% 1,000 = a Substitute 30% for r and 1,000 for b. 0.30 1,000 = a Change 30% to the decimal 0.30. 300 = a cont’d Multiply. Thus, 30% of 1,000 is 300. 27 7. Solve an application involving percents 28 Example – Investing At a stockholders meeting, members representing 4.5 million shares voted in favor of a proposal for a mandatory retirement age for the members of the board of directors. If this represented 75% of the number of shares outstanding, how many shares were outstanding? Solution: Let b represent the number of outstanding shares. Then 75% of b is 4.5 million. We can substitute 75% for r and 4.5 million for a in the percent formula and solve for b. rb = a 29 Example 13 – Solution 75% b = 4,500,000 0.75b = 4,500,000 = b = 6,000,000 cont’d 4.5 million = 4,500,000 Change 75% to a decimal. To undo the multiplication of 0.75, divide both sides by 0.75. Divide. There were 6 million shares outstanding. 30