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# Absolute value

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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

#1
+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
+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! :)

NotThatSmart Jun 5, 2024
#2
+129401
0

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

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