7(a-5)+11=3(a-4)

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Solution for 7(a-5)+11=3(a-4) equation: 



7(a-5)+11=3(a-4)

We move all terms to the left:

7(a-5)+11-(3(a-4))=0

We multiply parentheses

7a-(3(a-4))-35+11=0



We calculate terms in parentheses: -(3(a-4)), so:

3(a-4)

We multiply parentheses

3a-12

Back to the equation:

-(3a-12)



We add all the numbers together, and all the variables

7a-(3a-12)-24=0

We get rid of parentheses

7a-3a+12-24=0

We add all the numbers together, and all the variables

4a-12=0

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

4a=12

a=12/4

a=3

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