120 Chapter 2 Linear Equations and Inequalities in One Variable 2b. ` y - 1 5 ` = ` y 2 ` y - 1 5 = y 2 or 10a y - 1 5 b = 10a y 2 b 2(y - 1) = 5y 2y - 2 = 5y 2y - 2 - 2y = 5y - 2y -2 = 3y -2 = 3 3y 3 - 2 3 = y So, the solution set is e- 2 3 , 2 7 y - 1 5 =- y 2 10a y - 1 5 b = 10ay 2 b 2(y - 1) = -5y 2y - 2=-5y 2y - 2 - 2y=-5y - 2y -2=-7y -2 = -7 -7y -7 2 7 = y f . We can check by replacing y with - 2 3 and 2 7 . Student Check 2 Solve each equation. a. u5z - 7u = uz + 1u b. ` x + 3 2 ` = ` 2x 5 ` Applications Since the absolute value of a number represents that number’s distance from zero on a number line, distance between objects is a key application of absolute value equations. In measurements, this distance is often referred to as absolute error. In science-related fields, accuracy in measurement is vital. However, there will almost always be error in measurement due to the precision of the measuring instruments and human error. Definition: Absolute error is the absolute value (sometimes called magnitude) of the difference between the measured value and the exact value. We can use the following formula to represent the absolute error. Eabs = ux - au, where a is the approximated value and x is the exact measure. Objective 3 Examples Solve each problem. 3a. A memory card is measured with a ruler with centimeter gradations. The width is reported to be 2.5 cm. The absolute error is 0.1 cm. Find the possible values for the exact width of the memory card. Solution 3a. Let w represent the exact width of the memory card. The approximated value is 2.5 cm and the absolute error is 0.1 cm. To find the exact width, we must solve the following equation. uw - 2.5u = 0.1 w - 2.5 = 0.1 or w - 2.5 = -0.1 Apply property 1. w - 2.5 + 2.5 = 0.1 + 2.5 w - 2.5 + 2.5=-0.1 + 2.5 Add 2.5 to each side. w = 2.6 w = 2.4 Simplify. So, the exact width of the memory card is at most 2.6 cm and at least 2.4 cm. Objective 3 ▶ Solve applications of absolute value equations.
hendricks_intermediate_algebra_1e_ch1_3
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