5(2c-7)+3c=2(c-4)+6

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Solution for 5(2c-7)+3c=2(c-4)+6 equation: 



5(2c-7)+3c=2(c-4)+6

We move all terms to the left:

5(2c-7)+3c-(2(c-4)+6)=0

We add all the numbers together, and all the variables

3c+5(2c-7)-(2(c-4)+6)=0

We multiply parentheses

3c+10c-(2(c-4)+6)-35=0



We calculate terms in parentheses: -(2(c-4)+6), so:

2(c-4)+6

We multiply parentheses

2c-8+6

We add all the numbers together, and all the variables

2c-2

Back to the equation:

-(2c-2)



We add all the numbers together, and all the variables

13c-(2c-2)-35=0

We get rid of parentheses

13c-2c+2-35=0

We add all the numbers together, and all the variables

11c-33=0

We move all terms containing c to the left, all other terms to the right

11c=33

c=33/11

c=3

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