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