3(d-1)+2=3(d+2)-5

Solution for 3(d-1)+2=3(d+2)-5 equation:

3(d-1)+2=3(d+2)-5

We move all terms to the left:

3(d-1)+2-(3(d+2)-5)=0

We multiply parentheses

3d-(3(d+2)-5)-3+2=0


We calculate terms in parentheses: -(3(d+2)-5), so:

3(d+2)-5

We multiply parentheses

3d+6-5

We add all the numbers together, and all the variables

3d+1

Back to the equation:

-(3d+1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3d-3d-1-1=0

We add all the numbers together, and all the variables

-2!=0

There is no solution for this equation