lichting wrote:
Can anyone give me more detail explaination please
I still dont understand how to solve this
For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
STEP-BY-STEP DETAILED SOLUTION:There are three transition points at -7, 5 and 8 (transition points are points for which expression in modulus is equal to 0). So, four ranges to check:
1. \(x < -7\)For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is negative, so |x + 7| = -(x + 7).
Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > -(x + 7), which in turn is x < 20. Since, we consider x < -7 range, then finally for it we get
x < -7.
2. \(-7 \leq x \leq 5\)For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is positive, so |5 - x| = 5 - x.
x + 7 is positive, so |x + 7| = x + 7.
Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > x + 7, which in turn is x < 2. Since, we consider \(-7 \leq x \leq 5\) range, then finally for it we get
\(-7 \leq x < 2\).
3. \(5 < x < 8\)For this range:
x - 8 is negative, so |x - 8| = -(x - 8).
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.
Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) -(5 - x) > x + 7, which in turn is x < -4. Since, we consider \(5 < x < 8\) range and x < -4 is out of it, then
for this range the inequality does not have a solution (no x from this range satisfied the inequality).
3. \(x \geq 8\)For this range:
x - 8 is positive, so |x - 8| = x - 8.
5 - x is negative, so |5 - x| = -(5 - x).
x + 7 is positive, so |x + 7| = x + 7.
Thus, |x – 8| + |5 – x| > |x + 7| becomes x - 8 -(5 - x) > x + 7, which in turn is
x > 20 (already entirely in the range we consider)
So, we got that |x – 8| + |5 – x| > |x + 7| is true for x < 2 (combined range from cases 1 and 2) and x > 20. Thus, infinitely many integer values of x satisfy |x – 8| + |5 – x| > |x + 7|.
Answer: E.
Hope it's clear.