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
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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.

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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.

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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.

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

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