5(7-30-5c)-7=(35(7c-6)+28)-50

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Solution for 5(7-30-5c)-7=(35(7c-6)+28)-50 equation: 



5(7-30-5c)-7=(35(7c-6)+28)-50

We move all terms to the left:

5(7-30-5c)-7-((35(7c-6)+28)-50)=0

We add all the numbers together, and all the variables

5(-5c-23)-((35(7c-6)+28)-50)-7=0

We multiply parentheses

-25c-((35(7c-6)+28)-50)-115-7=0



We calculate terms in parentheses: -((35(7c-6)+28)-50), so:

(35(7c-6)+28)-50



We calculate terms in parentheses: +(35(7c-6)+28), so:

35(7c-6)+28

We multiply parentheses

245c-210+28

We add all the numbers together, and all the variables

245c-182

Back to the equation:

+(245c-182)



We get rid of parentheses

245c-182-50

We add all the numbers together, and all the variables

245c-232

Back to the equation:

-(245c-232)



We add all the numbers together, and all the variables

-25c-(245c-232)-122=0

We get rid of parentheses

-25c-245c+232-122=0

We add all the numbers together, and all the variables

-270c+110=0

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

-270c=-110

c=-110/-270

c=11/27

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