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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3d+5d-(3(-1d+1)+d)-20-5=0



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

3(-1d+1)+d

We add all the numbers together, and all the variables

d+3(-1d+1)

We multiply parentheses

d-3d+3

We add all the numbers together, and all the variables

-2d+3

Back to the equation:

-(-2d+3)



We add all the numbers together, and all the variables

8d-(-2d+3)-25=0

We get rid of parentheses

8d+2d-3-25=0

We add all the numbers together, and all the variables

10d-28=0

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

10d=28

d=28/10

d=2+4/5

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