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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

4(d+2)+02d-(3(d+2))=0

We add all the numbers together, and all the variables

02d+4(d+2)-(3(d+2))=0

We multiply parentheses

02d+4d-(3(d+2))+8=0



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

3(d+2)

We multiply parentheses

3d+6

Back to the equation:

-(3d+6)



We add all the numbers together, and all the variables

6d-(3d+6)+8=0

We get rid of parentheses

6d-3d-6+8=0

We add all the numbers together, and all the variables

3d+2=0

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

3d=-2

d=-2/3

d=-2/3

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