Question

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

Answer 1

`x=+-3i`

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.

`"add "1/x" to both sides"`

`2/(3-x)+1/x=1/3cancel(-1/x)cancel(+1/x)`

`rArr2/(3-x)+1/x=1/3`

add the fractions on the left by creating a `color(blue)"common denominator"`

`(2/(3-x)xxx/x)+(1/x xx(3-x)/(3-x))=1/3`

`rArr(2x)/(x(3-x))+(3-x)/(x(3-x))=1/3`

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

`rArr(2x+3-x)/(x(3-x))=1/3`

`rArr(x+3)/(x(3-x))=1/3`

At this stage we can `color(blue)"cross-multiply"`

`rArr3(x+3)=3x-x^2`

`rArrcancel(3x)+9=cancel(3x)-x^2`

`rArrx^2=-9`

There is no real number when squared produces a negative.

`rArr" there are no real solutions"`

I don't know if you have covered `color(blue)"complex numbers"`

for completeness I will provide the solution.

`x^2=-9tocolor(red)((1))`

Take the `color(blue)"square root of both sides"`

`sqrt(x^2)=+-sqrt(-9)`

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

`sqrt(-1)" is defined as an imaginary number and given the symbol i"`

`rArrsqrt9xxsqrt(-1)=3i`

`"Finally returning to " color(red)((1))`

`x^2=-9rArrx=+-3i`

Answer 2

`x=+-3 i`

Explanation:

Given:`" "2/(3-x)=1/3-1/x`

Multiply both sides by `(3-x)`

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

`2=1-x/3-3/x+1`

`2=2 -x/3-3/x`

The only way this can work is if we set`-x/3-3/x=0`

`-x/3=+3/x`

`-(x^2)=3^2`

Multiply both sides by `(-1)`

`x^2=-9`

`x=sqrt(9xx(-1))`

`x=+-3 i`