Question 11 • 25-26 TX-Algebra II A - Period 1 - SY26-27 - RM2204
Set the function equal to 15: 4|x − 5| + 3 = 15. Subtract 3 and divide by 4 to get |x − 5| = 3, so x − 5 is either 3 or −3. Therefore x = 8 or x = 2, matching choice A, “x = 2, x = 8.”
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