(1)/(6)(2r+8)+(1)/(4)(r-3)=(7)/(6)

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Solution for (1)/(6)(2r+8)+(1)/(4)(r-3)=(7)/(6) equation: 



(1)/(6)(2r+8)+(1)/(4)(r-3)=(7)/(6)

We move all terms to the left:

(1)/(6)(2r+8)+(1)/(4)(r-3)-((7)/(6))=0



Domain of the equation: 6(2r+8)!=0

r∈R





Domain of the equation: 4(r-3)!=0

r∈R



We add all the numbers together, and all the variables

1/6(2r+8)+1/4(r-3)-(+7/6)=0

We get rid of parentheses

1/6(2r+8)+1/4(r-3)-7/6=0

We calculate fractions


(4rr/(6(2r+8)*4(r-3)*6)+(36r2/(6(2r+8)*4(r-3)*6)+(-28rr/(6(2r+8)*4(r-3)*6)=0



We calculate terms in parentheses: +(4rr/(6(2r+8)*4(r-3)*6)+(36r2/(6(2r+8)*4(r-3)*6)+(-28rr/(6(2r+8)*4(r-3)*6), so:

4rr/(6(2r+8)*4(r-3)*6)+(36r2/(6(2r+8)*4(r-3)*6)+(-28rr/(6(2r+8)*4(r-3)*6

We can not solve this equation

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