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

##### 2 Answers

#### 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#

#### Explanation:

Given:

Multiply both sides by

The only way this can work is if we set

Multiply both sides by