Question

How do I do 28-(3c+4)=2(c+6)+c?

Answer 1

Hello, call me Emac.

Indeed, that is a very hard question! Here is how to answer it. Please hear this if it helps so I know. This helps, read it and if you don’t understand reply. I do have the answer, but try figuring it out first!

Simplifying
28 + -1(3c + 4) = 2(c + 6) + c

Reorder the terms:
28 + -1(4 + 3c) = 2(c + 6) + c
28 + (4 * -1 + 3c * -1) = 2(c + 6) + c
28 + (-4 + -3c) = 2(c + 6) + c

Combine like terms: 28 + -4 = 24
24 + -3c = 2(c + 6) + c

Reorder the terms:
24 + -3c = 2(6 + c) + c
24 + -3c = (6 * 2 + c * 2) + c
24 + -3c = (12 + 2c) + c

Combine like terms: 2c + c = 3c
24 + -3c = 12 + 3c

Solving
24 + -3c = 12 + 3c

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '-3c' to each side of the equation.
24 + -3c + -3c = 12 + 3c + -3c

Combine like terms: -3c + -3c = -6c
24 + -6c = 12 + 3c + -3c

Combine like terms: 3c + -3c = 0
24 + -6c = 12 + 0
24 + -6c = 12

Add '-24' to each side of the equation.
24 + -24 + -6c = 12 + -24

Combine like terms: 24 + -24 = 0
0 + -6c = 12 + -24
-6c = 12 + -24

Combine like terms: 12 + -24 = -12
-6c = -12

Divide each side by '-6'.
c =?

Answer 2

c=2

28-(3c+4)=2(c+6)+c

1. get #'s out of parentheses

28-3c-4=2c+12+c

2. combine like terms

24-3c=3c+12

3. put the C's on one side and #'s on the other

subtract -12 from 24 and at +3c to the 3c

12=6c

4. then divide the 6 off to leave C all by itself

12/6=c

5. that gives you your answer

2=c