1/3(d-7)-7/2(5d-3)=7/12

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Solution for 1/3(d-7)-7/2(5d-3)=7/12 equation: 



1/3(d-7)-7/2(5d-3)=7/12

We move all terms to the left:

1/3(d-7)-7/2(5d-3)-(7/12)=0



Domain of the equation: 3(d-7)!=0

d∈R





Domain of the equation: 2(5d-3)!=0

d∈R



We add all the numbers together, and all the variables

1/3(d-7)-7/2(5d-3)-(+7/12)=0

We get rid of parentheses

1/3(d-7)-7/2(5d-3)-7/12=0

We calculate fractions


(24d5/(3(d-7)*2(5d-3)*12)+(-252dd/(3(d-7)*2(5d-3)*12)+(-42d^2d/(3(d-7)*2(5d-3)*12)=0



We calculate terms in parentheses: +(24d5/(3(d-7)*2(5d-3)*12)+(-252dd/(3(d-7)*2(5d-3)*12)+(-42d^2d/(3(d-7)*2(5d-3)*12), so:

24d5/(3(d-7)*2(5d-3)*12)+(-252dd/(3(d-7)*2(5d-3)*12)+(-42d^2d/(3(d-7)*2(5d-3)*12

We can not solve this equation

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