132 Chapter 2 Linear Equations and Inequalities in One Variable �� ������������������������ ������������������ A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. 5a. Solve 3x 2 + 5 = 3x 2 1. To solve an equation with fractions, multiply both sides by the LCD of all fractions in the equation. This will create an equation that does not have any fractions. If the resulting equation has fractions, then the LCD was not correct or the equation was simplified incorrectly. 2. To solve an equation with decimals, multiply both sides by a power of 10 that will eliminate decimals in the number with the largest number of decimal places. 3x 2 3. Equations that produce a false statement have no solution. These are called contradictions and are denoted by the null set, ∅. 4. Equations that produce a true statement have infinitely many solutions and are called identities. The solution set is the set of all real numbers, . SUMMARY OF KEY CONCEPTS In previous sections, we learned how to check solutions of equations on the calculator by using the main screen, the store feature, and the table feature. We will now use the table feature to verify solutions when the proposed solution is large or a fraction. We can change the table settings in one of two ways as shown next. Example: Determine if y = 51 is a solution of y - 5 4 - 2y 9 = 1 6 . ���������������������������������������������������� x 2 . �������������������������������������� ���������������������������������������������������������������� The error was made in not applying the distributive property to each term on the left side of the equation. + 5 = x 2 2a3x 2 + 5b = 2a x 2 b 3x + 10 = x 2x=-10 x=- 10 2 x=-5 The solution set is {-5}. 5b. Solve 2x - 4(x - 1) = -2(x + 2). �������������������������������������� ���������������������������������������������������������������� The error was made in not recognizing that 0x = 0. The resulting statement should be 0 = -8, which is a contradiction. So, the solution set is the empty set, ∅. = 2 2a3x 2 5b = 2a x 2 b 3x + = x 2x=-5 x=- 2 2x - 4(x - 1) = -2(x + 2) 2x - 4x + 4 = -2x - 4 -2x + 4 = -2x - 4 + 2x - 4 + 2x - 4 0x = -8 The solution set is {-8}. ANSWERS TO STUDENT CHECKS Student Check 1 a. {4} b. e 2 3 f c. e 280 13 f d. {10} Student Check 2 a. {72} b. {100} Student Check 3 a. ∅ b. ∅ Student Check 4 a. b. + 5 x a3x + 5 = - 5 x x x
hendricks_beginning_algebra_1e_ch1_3
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