3(a-5)-12a=40-2(a-3)

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Solution for 3(a-5)-12a=40-2(a-3) equation: 



3(a-5)-12a=40-2(a-3)

We move all terms to the left:

3(a-5)-12a-(40-2(a-3))=0

We add all the numbers together, and all the variables

-12a+3(a-5)-(40-2(a-3))=0

We multiply parentheses

-12a+3a-(40-2(a-3))-15=0



We calculate terms in parentheses: -(40-2(a-3)), so:

40-2(a-3)

determiningTheFunctionDomain
-2(a-3)+40

We multiply parentheses

-2a+6+40

We add all the numbers together, and all the variables

-2a+46

Back to the equation:

-(-2a+46)



We add all the numbers together, and all the variables

-9a-(-2a+46)-15=0

We get rid of parentheses

-9a+2a-46-15=0

We add all the numbers together, and all the variables

-7a-61=0

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

-7a=61

a=61/-7

a=-8+5/7

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