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

Apr 27, 2018

There are no solutions.

#### Explanation:

put the RHS over a common denominator

$\frac{x}{x - 3} - \frac{3}{2}$=$\frac{2 x - 3 \left[x - 3\right]}{2 \left[x - 3\right]}$=$\frac{2 x - 3 x + 9}{2 x - 6}$=$\frac{9 - x}{2 x - 6}$

that gives us

$\frac{3}{x - 3} = \frac{9 - x}{2 x - 6}$

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

$3 \left(2 x - 6\right) = \left(9 - x\right) \left(x - 3\right)$

$6 x - 18 = 9 x - 27 - {x}^{2} + 3 x$

$6 x - 18 = 12 x - 27 - {x}^{2}$

collect all terms onto one side

${x}^{2} + 6 x - 12 x - 18 + 27 = 0$

${x}^{2} - 6 x + 9 = 0$

Factorise

$\left(x - 3\right) \left(x - 3\right) = 0$

so $x = 3$

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

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

Thanks @georgec for the update.

Apr 27, 2018

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

#### Explanation:

Given:

$\frac{3}{x - 3} = \frac{x}{x - 3} - \frac{3}{2}$

Adding $\frac{3}{2} - \frac{3}{x - 3}$ to both sides, we get:

$\frac{3}{2} = \frac{x - 3}{x - 3} = 1 \text{ } \left(x \ne 3\right)$

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 $x = 3$