3(40a-5)+15=2(33-2a)

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Solution for 3(40a-5)+15=2(33-2a) equation: 



3(40a-5)+15=2(33-2a)

We move all terms to the left:

3(40a-5)+15-(2(33-2a))=0

We add all the numbers together, and all the variables

3(40a-5)-(2(-2a+33))+15=0

We multiply parentheses

120a-(2(-2a+33))-15+15=0



We calculate terms in parentheses: -(2(-2a+33)), so:

2(-2a+33)

We multiply parentheses

-4a+66

Back to the equation:

-(-4a+66)



We add all the numbers together, and all the variables

120a-(-4a+66)=0

We get rid of parentheses

120a+4a-66=0

We add all the numbers together, and all the variables

124a-66=0

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

124a=66

a=66/124

a=33/62

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