Page 130

hendricks_beginning_algebra_1e_ch1_3

()* 100(0.05x ) = ( 1 0 0 ) ( 0 . 0 5 ) x =5x Two decimal places Multiplying an expression with two decimal places by 102 or 100 clears the decimal from the expression. 1000(2.34a + 12.567) ('')''* = 1000(2.34a) + 1000(12.567) = 2340a + 12,567 Two and three decimal places Multiplying an expression with three decimal places by 103 or 1000 clears the decimal from the expression. Note that when we multiply the preceding expressions by an appropriate power of 10, the decimals are cleared from the expression. We will apply this same concept to remove decimals from an equation; that is, we will multiply each side of the equation by a power of 10 (10, 100, 1000, and so on) that will eliminate the decimal from the number with the most decimal places. To solve equations with decimals, we apply the general strategy that was stated at the beginning of the section. �� ������������������������ ������������������ Solve each equation by clearing the decimals. Check each answer. 2a. x + 0.15x = 36.80 2b. 0.04x + 0.05(500 - x) = 23 Solutions 2a. x + 0.15x = 36.80 100(x + 0.15x) = 100(36.80) Multiply each side by 100. 100(x) + 100(0.15x) = 3680 Distribute and simplify. 100x + 15x = 3680 Simplify. 115x = 3680 Combine like terms. 115x 115 = 3680 115 Divide each side by 115. x = 32 Simplify. Check: x + 0.15x = 36.80 Original equation 32 + 0.15(32) = 36.80 Replace x with 32. 32 + 4.8 = 36.80 Simplify. 36.8 = 36.80 Add. Since x = 32 makes the equation true, the solution set is {32}. ������������ We could also solve the equation without removing decimals by combining like terms on the left. 1x + 0.15x = 36.80 1.15x = 36.80 1.15x 1.15 = 36.80 1.15 x = 32 Recall that x = 1x. Combine like terms. Divide each side by 1.15. Simplify. 128 Chapter 2 Linear Equations and Inequalities in One Variable


hendricks_beginning_algebra_1e_ch1_3
To see the actual publication please follow the link above