7(1d-2)-3d=5-2(d+3)+4d

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Solution for 7(1d-2)-3d=5-2(d+3)+4d equation:



7(1d-2)-3d=5-2(d+3)+4d
We move all terms to the left:
7(1d-2)-3d-(5-2(d+3)+4d)=0
We add all the numbers together, and all the variables
7(d-2)-3d-(5-2(d+3)+4d)=0
We add all the numbers together, and all the variables
-3d+7(d-2)-(5-2(d+3)+4d)=0
We multiply parentheses
-3d+7d-(5-2(d+3)+4d)-14=0
We calculate terms in parentheses: -(5-2(d+3)+4d), so:
5-2(d+3)+4d
determiningTheFunctionDomain -2(d+3)+4d+5
We add all the numbers together, and all the variables
4d-2(d+3)+5
We multiply parentheses
4d-2d-6+5
We add all the numbers together, and all the variables
2d-1
Back to the equation:
-(2d-1)
We add all the numbers together, and all the variables
4d-(2d-1)-14=0
We get rid of parentheses
4d-2d+1-14=0
We add all the numbers together, and all the variables
2d-13=0
We move all terms containing d to the left, all other terms to the right
2d=13
d=13/2
d=6+1/2

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