(1/7(2x-5))=(2/7x+1)

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Solution for (1/7(2x-5))=(2/7x+1) equation: 



(1/7(2x-5))=(2/7x+1)

We move all terms to the left:

(1/7(2x-5))-((2/7x+1))=0



Domain of the equation: 7(2x-5))!=0

x∈R





Domain of the equation: 7x+1))!=0

x∈R



We calculate fractions


7x/(-245x^2+14x)+(-14x2/(-245x^2+14x)=0



We calculate terms in parentheses: +(-14x2/(-245x^2+14x), so:

-14x2/(-245x^2+14x

We multiply all the terms by the denominator


-14x2

We add all the numbers together, and all the variables

-14x^2

Back to the equation:

+(-14x^2)



We get rid of parentheses

7x/(-245x^2+14x)-14x^2=0

We multiply all the terms by the denominator


-14x^2*(-245x^2+14x)+7x=0

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