1/5(3x-4)+1/6(3x+1)=7/15

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Solution for 1/5(3x-4)+1/6(3x+1)=7/15 equation: 



1/5(3x-4)+1/6(3x+1)=7/15

We move all terms to the left:

1/5(3x-4)+1/6(3x+1)-(7/15)=0



Domain of the equation: 5(3x-4)!=0

x∈R





Domain of the equation: 6(3x+1)!=0

x∈R



We add all the numbers together, and all the variables

1/5(3x-4)+1/6(3x+1)-(+7/15)=0

We get rid of parentheses

1/5(3x-4)+1/6(3x+1)-7/15=0

We calculate fractions


(90x3/(5(3x-4)*6(3x+1)*15)+(75x3/(5(3x-4)*6(3x+1)*15)+(-210x^23/(5(3x-4)*6(3x+1)*15)=0



We calculate terms in parentheses: +(90x3/(5(3x-4)*6(3x+1)*15)+(75x3/(5(3x-4)*6(3x+1)*15)+(-210x^23/(5(3x-4)*6(3x+1)*15), so:

90x3/(5(3x-4)*6(3x+1)*15)+(75x3/(5(3x-4)*6(3x+1)*15)+(-210x^23/(5(3x-4)*6(3x+1)*15

We do not support expression: x^23

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