Question

How do you solve `1/x+1/(x+3)=2`?

Answer 1

`x=-1+-sqrt(5/2)`.

Explanation:

To add fractions, we need to find a common denominator. For example, to do `1/3+1/5`, we would multiply `1/3` by `5/5` to get `5/15` and `1/5` by `3/3` to get `3/15`, then add them to get `8/15`.

Similarly, `1/x=1/x*(x+3)/(x+3)=(x+3)/(x(x+3))`.

`1/(x+3)=1/(x+3)*x/x=x/(x(x+3))`.

So `1/x+1/(x+3)=(x+3)/(x(x+3))+x/(x(x+3))=(2x+3)/(x(x+3))`.

So now we have `(2x+3)/(x(x+3))=2`.

Multiply both sides by `x(x+3)` to get `2x+3=2x(x+3)`.

Expand the right side to get `2x+3=2x^2+6x`.

Rearrange terms to get `2x^2+4x-3=0`.

This is a quadratic equation with solutions `x=-1+-sqrt(5/2)`.