4(3c-11)=5(29-3c)

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Solution for 4(3c-11)=5(29-3c) equation: 



4(3c-11)=5(29-3c)

We move all terms to the left:

4(3c-11)-(5(29-3c))=0

We add all the numbers together, and all the variables

4(3c-11)-(5(-3c+29))=0

We multiply parentheses

12c-(5(-3c+29))-44=0



We calculate terms in parentheses: -(5(-3c+29)), so:

5(-3c+29)

We multiply parentheses

-15c+145

Back to the equation:

-(-15c+145)



We get rid of parentheses

12c+15c-145-44=0

We add all the numbers together, and all the variables

27c-189=0

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

27c=189

c=189/27

c=7

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