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

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Solution for 4(d-3)-(d+1)=2(d-5) equation: 



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

We move all terms to the left:

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

We multiply parentheses

4d-(d+1)-(2(d-5))-12=0

We get rid of parentheses

4d-d-(2(d-5))-1-12=0



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

2(d-5)

We multiply parentheses

2d-10

Back to the equation:

-(2d-10)



We add all the numbers together, and all the variables

3d-(2d-10)-13=0

We get rid of parentheses

3d-2d+10-13=0

We add all the numbers together, and all the variables

d-3=0

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

d=3

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