122 Chapter 2 Linear Equations and Inequalities in One Variable 4c. Solve: u3a + 2u =-7. Incorrect Solution Correct Solution and Explanation u3a + 2u =-7 Because the absolute value expression is equal to a negative number, this equation has no solution. The absolute value of any real number is always greater than or equal to zero. So, the solution set is the empty set, or . 3a + 2=-7 or 3a 2 = 4 3a=-9 3a = 2 a=-3 a = 2 3 The solution set is e-3, 2 3 . = 2 = + 3a 9 n se e f Student Check 1 a. 5±76 b. c. 5-7, 56 d. e- 14 3 , 10 3 f e. e 9 5 f Student Check 2 a. 51, 26 b. e-15, - 5 3 f Student Check 3 a. The length of the dog is at most 83 cm and at least 77 cm. b. The absolute error is 0.0183. ANSWERS TO STUDENT CHECKS SUMMARY OF KEY CONCEPTS 1. To solve an absolute value equation, we must write the equation so that it is in the form uXu = c, where c is a real number. If the constant c is positive, there are two solutions. If the constant c is zero, there is one solution. If the constant c is negative, there are no solutions. 2. To solve an equation in which the absolute value of one expression is equal to the absolute value of another expression, we must form two equations. One equation GRAPHING CALCULATOR SKILLS is formed by setting the two expressions equal to one another. The other equation is formed by setting one expression equal to the opposite of the other expression. 3. A key application of absolute value involves the error in measurement. The absolute error is calculated as the absolute value of the difference of the exact value and the approximated value. We can use the graphing calculator to verify solutions of absolute value equations. Example: Verify that the solutions of the equation u2x - 1u - 3 = 2 are -2 and 3. Solution: Method 1: Enter the solutions as the stored value for the variable x. Then compute the value of the left side of the equation. The result should be the right side of the equation. (–) 2 T X,T,u,n ENTER ATH 2 X,T,u,n 2 ) ENTER 2 3 3 T X,T,u,n ENTER ATH 2 X,T,u,n 2 ) 3 ENTER 2 Method 2: Enter the left side of the equation in the equation editor. Verify that the table shows that the corresponding y-value of x=-2 and x = 3 is y = 2. Since Y1 = 2 for both x=-2 and x = 3, we know they are solutions of the equation.
hendricks_intermediate_algebra_1e_ch1_3
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