SOLUTION: Find all c such that |c + 5| - 3c = 10 - 5|c - 1| + 7|c + 8|.

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Question 1208820: Find all c such that |c + 5| - 3c = 10 - 5|c - 1| + 7|c + 8|.
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


abs%28c%2B5%29-3c=10-5%2Aabs%28c-1%29%2B7%2Aabs%28c%2B8%29

The "behavior" of the expression changes at each value of c for which one of absolute value expressions is equal to 0. Those values are -8, -5, and 1.

Use those three values of c to divide the number line into four intervals and look for values in each interval that satisfy the given equation.

To make the analysis a bit easier, I will change the form of the equation so I can look for a zero value for the entire expression:

abs%28c%2B5%29-3c-10%2B5%2Aabs%28c-1%29-7%2Aabs%28c%2B8%29=0

(1) On the interval (-infinity,-8)....
abs%28c%2B5%29=-c-5
abs%28c-1%29=-c%2B1
abs%28c%2B8%29=-c-8
The expression is

-c-5-3c-10%2B5%28-c%2B1%29-7%28-c-8%29=0
-c-5-3c-10-5c%2B5%2B7c%2B56=0
-2c%2B46=0
2c=46
c=23

This "solution" is not in the interval for this case, so it is not a solution.

(2) On the interval [-8,-5)....
abs%28c%2B5%29=-c-5
abs%28c-1%29=-c%2B1
abs%28c%2B8%29=c%2B8
The expression is

-c-5-3c-10%2B5%28-c%2B1%29-7%28c%2B8%29=0
-c-5-3c-10-5c%2B5-7c-56=0
-16c-66=0
16c=-66
c=-66%2F16

This "solution" is also not in the interval for this case, so it also is not a solution.

(3) On the interval [-5,1)....
abs%28c%2B5%29=c%2B5
abs%28c-1%29=-c%2B1
abs%28c%2B8%29=c%2B8
The expression is

c%2B5-3c-10%2B5%28-c%2B1%29-7%28c%2B8%29=0
c%2B5-3c-10-5c%2B5-7c-56=0
-14c-56=0
14c=-56
c=-56%2F14=-4

This "solution" IS in the interval for this case, so it IS a solution.

(4) On the interval [1,infinity)....
abs%28c%2B5%29=c%2B5
abs%28c-1%29=c-1
abs%28c%2B8%29=c%2B8
The expression is

c%2B5-3c-10%2B5%28c-1%29-7%28c%2B8%29=0
c%2B5-3c-10%2B5c-5-7c-56=0
-4c-66=0
4c=-66
c=-66%2F4

This "solution" is once again not in the interval for this case, so it is not a solution.

ANSWER: There is a single solution: c = -4