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

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Solution for 7(d-2)=5(d-3)+(2d+1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

7d-(5(d-3)+(2d+1))-14=0



We calculate terms in parentheses: -(5(d-3)+(2d+1)), so:

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

We multiply parentheses

5d+(2d+1)-15

We get rid of parentheses

5d+2d+1-15

We add all the numbers together, and all the variables

7d-14

Back to the equation:

-(7d-14)



We get rid of parentheses

7d-7d+14-14=0

We add all the numbers together, and all the variables

=0

d=0/1

d=0

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