2(3d-3)+1=3(d+2)-2

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Solution for 2(3d-3)+1=3(d+2)-2 equation: 



2(3d-3)+1=3(d+2)-2

We move all terms to the left:

2(3d-3)+1-(3(d+2)-2)=0

We multiply parentheses

6d-(3(d+2)-2)-6+1=0



We calculate terms in parentheses: -(3(d+2)-2), so:

3(d+2)-2

We multiply parentheses

3d+6-2

We add all the numbers together, and all the variables

3d+4

Back to the equation:

-(3d+4)



We add all the numbers together, and all the variables

6d-(3d+4)-5=0

We get rid of parentheses

6d-3d-4-5=0

We add all the numbers together, and all the variables

3d-9=0

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

3d=9

d=9/3

d=3

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