In this problem, ABS(x) provides absolute value of x, without regard to its sign. For example ABS(3) = ABS(-3) = 3
The graphs of y = – ABS(x – a) + b and y = ABS(x – c) + d intersect at points (2,5) and (8, 3).
Find a + c
a. 8
b. 10
c. 7
d. 5

• Both curves intersect at (2, 5)
  So 5 = – |2 – a| + b ⇒ 5 + |2 – a| – b = 0 - - - (1)
  5 = – |2 – c| + d ⇒ 5 + |2 – c| – d = 0 - - - (2)
  Also,
  3 = – |8 – a| + b ⇒ 3 + |8 – a| – b = 0 - - - (4)
  3 = – |8 – c| + d ⇒ 3 + |8 – c| – d = 0- - - (4)
  Equating 1 and 3, 5 + |2 – a| = 3 + |8 – a| ⇒ 2 + |2 – a| = |8– a|
  So a = 4
  Similarly if we equate 2 and 4, we get c = 4
  a + c = 8
• 9 years agoHelpfull: Yes(8) No(18)
• 10 hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
• 9 years agoHelpfull: Yes(1) No(5)