How do you solve #1/(m^2-m)+1/m=5/(m^2-m)# and check for extraneous solutions?

Question

4461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-24T12:39:11

#m=0# is an extraneous solution and #m=5# is a solution.

# 1/(m^2-m) +1/m= 5/ (m^2-m)#. Multiplying by #m(m^2-m)#

on both sides we get, # m +(m^2-m)= 5 m# or

# cancel m +m^2- cancel m- 5 m =0 # or

#m(m-5)=0 :. m=0 , m=5#

On check , # m != 0# , since function is undefined at #m=0#

So #m=0# is an extraneous solution and #m=5# is a solution. [Ans]

Answer 2 · 2018-05-03T21:49:19

In a problem like this we can assume the denominators are not zero, so we never really run into extraneous solutions #m=0# or #m=1# and get right to

#m = 5#

Sometimes extraneous solutions are unavoidable, as often happens when we have to square both sides of an equation.

That's not the case in a problem like this. We can at the outset assume the denominators are not zero, i.e assume

# m^2 - m ne 0#

#m(m-1) ne 0 #

# m ne 0# and #m ne 1#.

The #1/m# tells us #m ne 0# which we already know.

Usually we must always factor, never cancel. But when we know factors are non-zero, we can cancel them.

#1/{m^2 - m} + 1/m = 5/{m^2-m}#

#{1 + 1(m-1)}/{m(m-1)} = 5/{m(m-1)}#

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

#m = 5#

Check:

# 1/{5^2-5} + 1/5 = 1/20+4/20 = 5/20 = 1/4#

#5/{5^2-5}=5/20 quad sqrt#