(331/280)+(7/4)x=(25/7)x+1

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Solution for (331/280)+(7/4)x=(25/7)x+1 equation: 



(331/280)+(7/4)x=(25/7)x+1

We move all terms to the left:

(331/280)+(7/4)x-((25/7)x+1)=0



Domain of the equation: 4)x!=0

x!=0/1

x!=0

x∈R





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

x!=0/1

x!=0

x∈R



We add all the numbers together, and all the variables

(+7/4)x-((+25/7)x+1)+(+331/280)=0

We multiply parentheses

7x^2-((+25/7)x+1)+(+331/280)=0

We get rid of parentheses

7x^2-((+25/7)x+1)+331/280=0

We calculate fractions


7x^2+()/7x+1)*280)+2317x/7x+1)*280)=0

We calculate fractions


7x^2+(()*7x)/49x^2+1*280)+2317x*7x)/49x^2=0

We multiply all the terms by the denominator


7x^2*49x^2+(()*7x)+1*280)+2317x*7x)=0



We calculate terms in parentheses: +(()*7x), so:

()*7x



Wy multiply elements

343x^4+1960x^2+(()*7x)=0



We calculate terms in parentheses: +(()*7x), so:

()*7x



We do not support expression: x^4

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