-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

-2d-(-2d+3)+15=0

We get rid of parentheses

-2d+2d-3+15=0

We add all the numbers together, and all the variables

12!=0

There is no solution for this equation

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