3(2q-2)+2q=6-2(-2+q)

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Solution for 3(2q-2)+2q=6-2(-2+q) equation: 



3(2q-2)+2q=6-2(-2+q)

We move all terms to the left:

3(2q-2)+2q-(6-2(-2+q))=0

We add all the numbers together, and all the variables

3(2q-2)+2q-(6-2(q-2))=0

We add all the numbers together, and all the variables

2q+3(2q-2)-(6-2(q-2))=0

We multiply parentheses

2q+6q-(6-2(q-2))-6=0



We calculate terms in parentheses: -(6-2(q-2)), so:

6-2(q-2)

determiningTheFunctionDomain
-2(q-2)+6

We multiply parentheses

-2q+4+6

We add all the numbers together, and all the variables

-2q+10

Back to the equation:

-(-2q+10)



We add all the numbers together, and all the variables

8q-(-2q+10)-6=0

We get rid of parentheses

8q+2q-10-6=0

We add all the numbers together, and all the variables

10q-16=0

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

10q=16

q=16/10

q=1+3/5

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