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

Question

extaneous solutions. I don't know how to solve it and find x and they all have to have the same denominator.

Answer 1

Subtracting 1/(m^2-m) from both sides:

 

 1/m = 4/(m^2-m)

 

cross multiplying:

     m^2-m = 4m

 

    m^2 - m - 4m = 0    <-- subtracts 4m from both sides

 

    m^2 - 5m = 0   <--- combines like terms

 

    m(m-5)=0   <-- factors

 

  m=0 or m-5 = 0

 

  m=0 or m=5

 

SO yes, M=0 is an extraneous solution because it causes division by zero in the second fraction,

rendering it invalid, unacceptable and absolutely useless. 

 

CHecking m=5, when substituted into the first fraction: 1/(5^2 - 5) = 1/(25-5) = 1/20

 

                          when substituted into the second fraction: 1/5

 

1/20 + 1/5 = 1/20 + 4/20 = 5/20 = 1/4

 

Now the right hand sides becomes:  5/(5^2-5) = 5/(25-5) = 5/20 = 1/4

 

The expression on the left side equals the expression on the right side when m=5.

So we have a good solution of m=5

 

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Now, I see you are asking about common denominator. m^2-m = m(m-1). So the first fraction on the left

and the fraction on the right have that common denominator. The second fraction on the left 1/m must be

changed so that it contains the common denominator. In order to do that, m-1 must be multiplied on the

bottom. But then you have to multiply the top by (m-1) as well. Note that multiplying that second fraction

by (m-1)/(m-1)=1 does not really change anything expect the denominator. Value does not change when

multiplying by 1. 

 

 So the equation becomes:

 

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

 

Multiplying both sides by m(m-1), the denominators cancel, provided that we

are not dividing by zero, which we must verify at the end***

 

The equation becomes:

 1 + (m-1) = 5

 

The ones cancel on the left, and the solution is m=5