Question

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

Answer 1

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.

Answer 2

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`