If any of the variables were zero, then the fractions in the problem would be undefined. This the solutions are in the form (a, a, -a), where a is a real number not equal to zero.

@Aditya Parson
–
We cannot clear out any factor or expression unless we know about the values it takes. Since we are given the expression, we will have to keep in mind that it is the base expression and solve accordingly.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest\(\large \frac{1}{m+n+k}-\frac{1}{k}=\frac{1}{m}+\frac{1}{n}\)

\(\large \frac{-(m+n)}{k(m+n+k)}=\frac{m+n}{mn}\)

\(-mn=k(m+n+k)\)

\(k^2+k(m+n)+mn=0\)

Solving the quadratic gives us:

\(k=-n\) or \(k=-m\)

An analogous argument gives us that \(m=-n\) is also a solution.

My doubt is that whether we can have \(m,n,k=0\) as solutions or not?

Although the equation actually satisfies, I am not sure.

Log in to reply

If any of the variables were zero, then the fractions in the problem would be undefined. This the solutions are in the form (a, a, -a), where a is a real number not equal to zero.

Log in to reply

But, what if we simplify the expression such that the variables are only in the numerator?

Log in to reply

Log in to reply

Log in to reply