MATH 125 Section 2.2 - {2020} | MATH125 Section 2.2 - Graded A

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Solve Equations using the Division and Multiplication Properties of Equality Learning Objectives By the end of this section, you will be able to: Solve equations using the Division and Multiplicat... ion Properties of Equality Solve equations that require simplification Translate to an equation and solve Translate and solve applications Be Prepared! Before you get started, take this readiness quiz. 1. Simplify: −7⎛ ⎝−7 1 ⎞ ⎠. If you missed this problem, review Example 1.68. 2. Evaluate 9x + 2 when x = −3 . If you missed this problem, review Example 1.57. Solve Equations Using the Division and Multiplication Properties of Equality You may have noticed that all of the equations we have solved so far have been of the form x + a = b or x − a = b . We were able to isolate the variable by adding or subtracting the constant term on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant and so will require division to isolate the variable. Let’s look at our puzzle again with the envelopes and counters in Figure 2.5. Figure 2.5 The illustration shows a model of an equation with one variable multiplied by a constant. On the left side of the workspace are two instances of the unknown (envelope), while on the right side of the workspace are six counters. In the illustration there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope? How do we determine the number? We have to separate the counters on the right side into two groups of the same size to correspond with the two envelopes on the left side. The 6 counters divided into 2 equal groups gives 3 counters in each group (since 6 ÷ 2 = 3 ). What equation models the situation shown in Figure 2.6? There are two envelopes, and each contains x counters. Together, the two envelopes must contain a total of 6 counters. 212 Chapter 2 Solving Linear Equations and Inequalities This OpenStax book is available for free at http://cnx.org/content/col12116/1.2Figure 2.6 The illustration shows a model of the equation 2x = 6 . If we divide both sides of the equation by 2, as we did with the envelopes and counters, we get: We found that each envelope contains 3 counters. Does this check? We know 2 · 3 = 6 , so it works! Three counters in each of two envelopes does equal six! This example leads to the Division Property of Equality. The Division Property of Equality For any numbers a, b, and c, and c ≠ 0 , If a = b, then a c = bc When you divide both sides of an equation by any non-zero number, you still have equality. MANIPULATIVE MATHEMATICS Doing the Manipulative Mathematics activity “ Division Property of Equality” will help you develop a better understanding of how to solve equations by using the Division Property of Equality. The goal in solving an equation is to ‘undo’ the operation on the variable. In the next example, the variable is multiplied by 5, so we will divide both sides by 5 to ‘undo’ the multiplication. EXAMPLE 2.13 Solve: 5x = −27. Solution To isolate x , “undo” the multiplication by 5. Divide to ‘undo’ the multiplication. Simplify. Check: Chapter 2 Solving Linear Equations and Inequalities 213Substitute −27 5 for x. Since this is a true statement, x = − 27 5 is the solution to 5x = −27 . TRY IT : : 2.25 Solve: 3y = −41. TRY IT : : 2.26 Solve: 4z = −55. Consider the equation x 4 = 3 . We want to know what number divided by 4 gives 3. So to “undo” the division, we will need to multiply by 4. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal. The Multiplication Property of Equality For any numbers a, b, and c, If a = b, then ac = bc If you multiply both sides of an equation by the same number, you still have equality. EXAMPLE 2.14 Solve: y −7 = −14. Solution Here y is divided by −7 . We must multiply by −7 to isolate y . Multiply both sides by −7 . Multiply. Simplify. Check: y −7 = −14 Substitute y = 98 . Divide. TRY IT : : 2.27 Solve: a −7 = −42. TRY IT : : 2.28 Solve: b −6 = −24. 214 Chapter 2 Solving Linear Equations and Inequalities This OpenStax book is available for free at http://cnx.org/content/col12116/1.2EXAMPLE 2.15 Solve: −n = 9. Solution Remember −n is equivalent to −1n . Divide both sides by −1 . Divide. Notice that there are two other ways to solve −n = 9 . We can also solve this equation by multiplying both sides by −1 and also by taking the opposite of both sides. Check: Substitute n = −9 . Simplify. TRY IT : : 2.29 Solve: −k = 8. TRY IT : : 2.30 Solve: −g = 3. [Show More]

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