2(d+5)+38(4d-2)+8=2d-1+7+10(2-d)+d

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Solution for 2(d+5)+38(4d-2)+8=2d-1+7+10(2-d)+d equation:



2(d+5)+38(4d-2)+8=2d-1+7+10(2-d)+d
We move all terms to the left:
2(d+5)+38(4d-2)+8-(2d-1+7+10(2-d)+d)=0
We add all the numbers together, and all the variables
2(d+5)+38(4d-2)-(2d-1+7+10(-1d+2)+d)+8=0
We multiply parentheses
2d+152d-(2d-1+7+10(-1d+2)+d)+10-76+8=0
We calculate terms in parentheses: -(2d-1+7+10(-1d+2)+d), so:
2d-1+7+10(-1d+2)+d
determiningTheFunctionDomain 2d+10(-1d+2)+d-1+7
We add all the numbers together, and all the variables
3d+10(-1d+2)+6
We multiply parentheses
3d-10d+20+6
We add all the numbers together, and all the variables
-7d+26
Back to the equation:
-(-7d+26)
We add all the numbers together, and all the variables
154d-(-7d+26)-58=0
We get rid of parentheses
154d+7d-26-58=0
We add all the numbers together, and all the variables
161d-84=0
We move all terms containing d to the left, all other terms to the right
161d=84
d=84/161
d=12/23

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