How do you solve the equation and identify any extraneous solutions for #6/(m+5) = 1 - 1/(m-5)#?

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Answer 1 · 2015-06-18T23:40:50

I found:
#m_1=0#
#m_2=7#

Consider the common denominator #(m+5)(m-5)# and get:

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

#(6(m-5))/cancel((m+5)(m-5))=(1*(m+5)(m-5)-1*(m+5))/cancel((m+5)(m-5))#

#6m-30=m^2-25-m-5#

#m^2-7m=0#

#m(m-7)=0#

#m_1=0#
#m_2=7#