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Find all c such that |c + 5| - 3c = 10 + 2|c - 4| - 6|c|. Enter all the solutions, separated by commas.

 Jun 5, 2024

Best Answer 

 #1
avatar+759 
+1

First, let's find every possible interval with this equation. 

We have \(c<-5,\:\:-5\le \:c<0,\:\:0\le \:c<4,\:\:c\ge \:4\)

 

Now, we solve the equation for each inequality. 

 

Case 1 - \(c<-5\)

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

This root is greater than -5, so it is invalid. 

 

Case 2 - \(-5 \leq c < 0\)

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This works!

 

Case 3 - \(0 \leq c < 4\)

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This works!

 

Case 4 - \(c>4\)

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

This is invalid. 

 

Now, we have \(c = -\frac{13}6, \frac{13}{6}\)

 

Thanks! :)

 Jun 5, 2024
 #1
avatar+759 
+1
Best Answer

First, let's find every possible interval with this equation. 

We have \(c<-5,\:\:-5\le \:c<0,\:\:0\le \:c<4,\:\:c\ge \:4\)

 

Now, we solve the equation for each inequality. 

 

Case 1 - \(c<-5\)

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

This root is greater than -5, so it is invalid. 

 

Case 2 - \(-5 \leq c < 0\)

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This works!

 

Case 3 - \(0 \leq c < 4\)

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This works!

 

Case 4 - \(c>4\)

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

This is invalid. 

 

Now, we have \(c = -\frac{13}6, \frac{13}{6}\)

 

Thanks! :)

NotThatSmart Jun 5, 2024
 #2
avatar+129401 
0

Very nice.....I would have to rely on a graph to solve this.....your approach is WAY more elegant !!!

 

cool cool cool

CPhill  Jun 5, 2024
edited by CPhill  Jun 5, 2024

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