Let's solve the equation for c by considering the absolute value signs. There are three cases to analyze depending on the sign of c + 5.

Case 1: c + 5 ≥ 0

In this case, the absolute value simplifies to c + 5. The equation becomes:

(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-2c + 5 = 10 - 4c - 8 2c = -3 c = -\dfrac{3}{2} (Not a solution because c + 5 < 0)

Case 2: -5 ≤ c + 5 < 0

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4}

Case 3: c + 5 < -5

Here, the absolute value becomes -(c + 5). The equation becomes:

-(c + 5) - 3c = 10 + 2(c - 4) - 6c

Simplifying:

-c - 5 - 3c = 10 - 4c - 8 -4c - 5 = 2 -4c = 7 c = -\dfrac{7}{4} (Solution)

We saw that only c = -7/4 satisfies the equation when -5 ≤ c + 5 < 0.

Summary:

Therefore, the only solution to the equation is c = -\dfrac{7}{4}.

Case 1: \(c \leq -5\)

Then 

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

But then c is not less than or equal to -5. This root is rejected.

 

Case 2: \(-5 < c \leq 0\)

Then

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

This root is in the range -5 < c <= 0. So we keep it.

 

Case 3: \(0 < c \leq 4\)

Then

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

This root is in range 0 < c <= 4. So we keep it.

 

Case 4: \(c > 4\)

Then

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

But then c is not greater than 4. This root is rejected.

 

Hence, the solutions are: \(c = \dfrac{13}6\), \(c = -\dfrac{13}6\).