How do you solve #3/(x-3)=x/(x-3)-3/2#?

2 Answers
Apr 27, 2018

Answer:

There are no solutions.

Explanation:

put the RHS over a common denominator

#x/[x-3]-3/2#=#[2x-3[x-3]]/[2[x-3]]#=#[2x-3x+9]/[2x-6]#=#[9-x]/[2x-6]#

that gives us

#3/[x-3]=[9-x]/[2x-6]#

cross multiply to make it a linear problem (no fractions)

#3(2x-6)=(9-x)(x-3)#

#6x-18=9x-27-x^2+3x#

#6x-18=12x-27-x^2#

collect all terms onto one side

#x^2+6x-12x-18+27=0#

#x^2-6x+9=0#

Factorise

#(x-3)(x-3)=0#

so #x=3#

If #x=3# we would have a denominator of zero so there are no solutions.

The graph of #y=x^2-6x+9# is above the x axis.

Thanks @georgec for the update.

Apr 27, 2018

Answer:

There is no value of #x# which satisfies this equation.

Explanation:

Given:

#3/(x-3) = x/(x-3)-3/2#

Adding #3/2 - 3/(x-3)# to both sides, we get:

#3/2 = (x-3)/(x-3) = 1" "(x != 3)#

Since this is false, there is no value of #x# that satisfies the given equation.

Here are the graphs of the left hand side and right hand side of the given equation plotted together:

graph{(y-3/(x-3))(y - (x/(x-3)-3/2)) = 0 [-10, 10, -5, 5]}

The two hyperbolas do not intersect, but have a common vertical asymptote at #x=3#