How do you solve #6/(x-1) = 9/(x+1)#?

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Answer 1 · 2016-10-05T07:52:42

#x=5#

This is a rational equation therefore we have to lead it in a canonical form like:

#(N(x))/(D(x))=0#

In order to do this we have to move all the terms on LHS:

#6/(x-1)=9/(x+1)=>6/(x-1)-9/(x+1)=0#

Now we could find LCD (less common denominator) that is:

#LCD=(x-1)(x+1)#

Since a denominator cannot never be zero we have to find the #x# values that make it zero and exclude them from the solutions of the equation.

# (x-1)(x+1)=0#

This is a product, then it's zero when the factors are zero.

#(x+1)=0=>x=-1#
#(x-1)=0=>x=1#

#:. x!=+-1#

Now we can simplyfy the fractions:

#(6(x+1)-9(x-1))/((x+1)(x-1))=0#

and we can simplify the LCD

#(6(x+1)-9(x-1))/color(green)cancel((x+1)(x-1))=0#

Move all the #x# therms on the LHS and the other one on RHS

#6x-9x=-6-9#

#-3x=-15#

#cancel(-)3x=cancel(-)15#

#3x=15#

#x=cancel(15)^5/cancel(3)=5#

#x=5#

graph{6/(x-1)-9/(x+1) [4.438, 6.335, -0.688, 0.26]}