1/(m+1)+(1)/(m-5)=(6)(m-5)

Solution for 1/(m+1)+(1)/(m-5)=(6)(m-5) equation:

1/(m+1)+(1)/(m-5)=(6)(m-5)

We move all terms to the left:

1/(m+1)+(1)/(m-5)-((6)(m-5))=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-5)!=0

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

m!=5

m∈R


We calculate fractions


-(6(m-5))+(1*(m-5))/((m+1)*(m-5))+(1*(m+1))/((m+1)*(m-5))=0


We calculate terms in parentheses: -(6(m-5)), so:

6(m-5)

We multiply parentheses

6m-30

Back to the equation:

-(6m-30)



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

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

We multiply all the terms by the denominator


1*(m-5))

Back to the equation:

+(1*(m-5)))



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

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

We multiply all the terms by the denominator


1*(m+1))

Back to the equation:

+(1*(m+1)))


We get rid of parentheses

-6m+(1*(m-5)))+(1*(m+1)))+30=0