# What is the solution to 2/(3-x)=1/3-1/x?

Mar 16, 2017

$x = \pm 3 i$

#### Explanation:

You are correct. there are no real solutions. They are complex.

There are different approaches to solving this equation.

I have decided on the following approach.

Collect fractions in x together on the left side of the equation.

$\text{add "1/x" to both sides}$

$\frac{2}{3 - x} + \frac{1}{x} = \frac{1}{3} \cancel{- \frac{1}{x}} \cancel{+ \frac{1}{x}}$

$\Rightarrow \frac{2}{3 - x} + \frac{1}{x} = \frac{1}{3}$

add the fractions on the left by creating a $\textcolor{b l u e}{\text{common denominator}}$

$\left(\frac{2}{3 - x} \times \frac{x}{x}\right) + \left(\frac{1}{x} \times \frac{3 - x}{3 - x}\right) = \frac{1}{3}$

$\Rightarrow \frac{2 x}{x \left(3 - x\right)} + \frac{3 - x}{x \left(3 - x\right)} = \frac{1}{3}$

Now there is a common denominator, we can add the numerators, leaving the denominator as it is.

$\Rightarrow \frac{2 x + 3 - x}{x \left(3 - x\right)} = \frac{1}{3}$

$\Rightarrow \frac{x + 3}{x \left(3 - x\right)} = \frac{1}{3}$

At this stage we can $\textcolor{b l u e}{\text{cross-multiply}}$

$\Rightarrow 3 \left(x + 3\right) = 3 x - {x}^{2}$

$\Rightarrow \cancel{3 x} + 9 = \cancel{3 x} - {x}^{2}$

$\Rightarrow {x}^{2} = - 9$

There is no real number when squared produces a negative.

$\Rightarrow \text{ there are no real solutions}$

I don't know if you have covered $\textcolor{b l u e}{\text{complex numbers}}$

for completeness I will provide the solution.

${x}^{2} = - 9 \to \textcolor{red}{\left(1\right)}$

Take the $\textcolor{b l u e}{\text{square root of both sides}}$

$\sqrt{{x}^{2}} = \pm \sqrt{- 9}$

• sqrt(-9)=sqrt(9xx-1)=sqrt9xxsqrt(-1)

$\sqrt{- 1} \text{ is defined as an imaginary number and given the symbol i}$

$\Rightarrow \sqrt{9} \times \sqrt{- 1} = 3 i$

$\text{Finally returning to } \textcolor{red}{\left(1\right)}$

${x}^{2} = - 9 \Rightarrow x = \pm 3 i$

Mar 16, 2017

$x = \pm 3 i$

#### Explanation:

Given:$\text{ } \frac{2}{3 - x} = \frac{1}{3} - \frac{1}{x}$

Multiply both sides by $\left(3 - x\right)$

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

$2 = 1 - \frac{x}{3} - \frac{3}{x} + 1$

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

The only way this can work is if we set$- \frac{x}{3} - \frac{3}{x} = 0$

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

$- \left({x}^{2}\right) = {3}^{2}$

Multiply both sides by $\left(- 1\right)$

${x}^{2} = - 9$

$x = \sqrt{9 \times \left(- 1\right)}$

$x = \pm 3 i$