-30c+3c-7(2-3c)=4(c-5)+3c+27

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Solution for -30c+3c-7(2-3c)=4(c-5)+3c+27 equation:



-30c+3c-7(2-3c)=4(c-5)+3c+27
We move all terms to the left:
-30c+3c-7(2-3c)-(4(c-5)+3c+27)=0
We add all the numbers together, and all the variables
-30c+3c-7(-3c+2)-(4(c-5)+3c+27)=0
We add all the numbers together, and all the variables
-27c-7(-3c+2)-(4(c-5)+3c+27)=0
We multiply parentheses
-27c+21c-(4(c-5)+3c+27)-14=0
We calculate terms in parentheses: -(4(c-5)+3c+27), so:
4(c-5)+3c+27
We add all the numbers together, and all the variables
3c+4(c-5)+27
We multiply parentheses
3c+4c-20+27
We add all the numbers together, and all the variables
7c+7
Back to the equation:
-(7c+7)
We add all the numbers together, and all the variables
-6c-(7c+7)-14=0
We get rid of parentheses
-6c-7c-7-14=0
We add all the numbers together, and all the variables
-13c-21=0
We move all terms containing c to the left, all other terms to the right
-13c=21
c=21/-13
c=-1+8/13

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