(2m)/(m-1)+(m-5)/m=1

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



(2m)/(m-1)+(m-5)/m=1

We move all terms to the left:

(2m)/(m-1)+(m-5)/m-(1)=0



Domain of the equation: (m-1)!=0

We move all terms containing m to the left, all other terms to the right

m!=1

m∈R





Domain of the equation: m!=0

m∈R



We calculate fractions


2m^2/(m^2-1m)+((m-5)*(m-1))/(m^2-1m)-1=0



We calculate terms in parentheses: +((m-5)*(m-1))/(m^2-1m), so:

(m-5)*(m-1))/(m^2-1m

We add all the numbers together, and all the variables

-1m+(m-5)*(m-1))/(m^2

We multiply all the terms by the denominator


-1m*(m^2+(m-5)*(m-1))

Back to the equation:

+(-1m*(m^2+(m-5)*(m-1)))



We multiply all the terms by the denominator


2m^2+((-1m*(m^2+(m-5)*(m-1))))*(m^2-1m)-1*(m^2-1m)=0



We calculate terms in parentheses: +((-1m*(m^2+(m-5)*(m-1))))*(m^2-1m), so:

(-1m*(m^2+(m-5)*(m-1))))*(m^2-1m

We add all the numbers together, and all the variables

-1m+(-1m*(m^2+(m-5)*(m-1))))*(m^2

Back to the equation:

+(-1m+(-1m*(m^2+(m-5)*(m-1))))*(m^2)

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