1. Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2

2. Copyright © Cengage Learning. All rights reserved. Section 2.1 Solving Basic Linear Equations in One Variable

3. 3 Objectives 1. Determine whether a statement is an expression or an equation. 2. Determine whether a number is a solution of an equation. 3. 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 applying the multiplication or division property of equality. 1 1 2 2 3 3 4 4

4. 4 Objectives Application 1. Application 2 5.Application 3 5 5 6 6 7 7

5. 5 Determine whether a statement is an expression or an equation 1.

6. 6 Expression or Equation Equation: a statement indicating that two quantities are equal. x + 5 = 21 2x – 5 = 11 3x 2 – 4x + 5 = 0 Expression: Combination of variables and constants which evaluates to a single value 6x – 1 3x 2 – x – 2 –8(x + 1)

7. 7 Your Turn Determine whether the following are expressions or equations. a. 9x 2 – 5x = 4 b. 3x + 2 c. 6(2x – 1) + 5 Solution: a.9x 2 – 5x = 4 is an equation b.3x + 2 is an expression c.6(2x – 1) + 5 is an expression

8. 8 Determine whether a number is a solution of an equation 2.

9. 9 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

10. 10 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 3  6 – 5 ≟ 2  6 18 – 5 ≟ 12 13 = 12 Since 13 = 12 is a false statement, 6 is not a solution. Substitute 6 for x. Do the multiplication. False.

11. 11 Solve a linear equation in one variable by applying the addition or subtraction property of equality 3.

12. 12 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.

13. 13 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.

14. 14 Example Solve: x – 5 = 2 Solution: x – 5 = 2 x – 5 + 5 = 2 + 5 x + 0 = 7 x = 7 Add 5 to both sides of the equation. Apply the additive identity property. Apply the additive inverse property.

15. 15 Example We check by substituting 7 for x in the original equation and simplifying. x – 5 = 2 7 – 5 ≟ 2 2 = 2 Since the previous statement is true, 7 is a solution. The solution set of this equation is {7}. Substitute 7 for x. True. cont’d

16. 16 Solve a linear equation in one variable by applying the multiplication or division property of equality 4.

17. 17 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 =.

18. 18 Example Solve: Solution: Multiply both sides by 3. 3  = x and 3  12 = 36.

19. 19 Application 1 5.

20. 20 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.

21. 21 Example 8 – Solution We can use the addition property of equality to solve the equation. 650 = r – 325 650 + 325 = r – 325 + 325 975 = r The regular price is $975. Add 325 to both sides. 650 + 325 = 975 and –325 + 325 = 0. cont’d

22. 22 Application 2 6.

23. 23 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, of 8,500 is the product of 0.0625 and 8,500. of 8,500 = 0.0625  8,500 = 531.25

24. 24 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

25. 25 Percent Percent problems involve questions such as the following. What is 30% of 1,000? 45% of what number is 405? What percent of 400 is 60? When we substitute the values of the rate, base, and amount into the percent formula, we will obtain an equation that we can solve. We must find the amount. We must find the base. We must find the rate.

26. 26 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

27. 27 Example 10 – Solution 30%  1,000 = a 0.30  1,000 = a 300 = a Thus, 30% of 1,000 is 300. Substitute 30% for r and 1,000 for b. Change 30% to the decimal 0.30. Multiply. cont’d

28. 28 Solve an application involving percents 7.

29. 29 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

30. 30 Example 13 – Solution 75%  b = 4,500,000 0.75b = 4,500,000 = b = 6,000,000 There were 6 million shares outstanding. 4.5 million = 4,500,000 Change 75% to a decimal. To undo the multiplication of 0.75, divide both sides by 0.75. cont’d Divide.