Consider the following inequality

\(a-bc+dc>0\)

One can arrange it in the following way

Subtract both side by \(a\)

\(-bc+dc>-a\)

Multiply both side by \(-1\)

\(bc-dc<a\)

Factorize \(c\) on LHS and divide both side by \((b-d)\) and we end up with

\(\boxed{c<\frac{a}{b-d}}\)

Now consider the original inequality, \(a-bc+dc>0\) again. We can also rearrange it in the following way:

Again, subtract both side by \(a\)

\(-bc+dc>-a\)

Now instead of multiplying both side by \(-1\), we factorize \(c\) on left hand side and divide both side by \((-b+d)\) and we can obtain the following result

\(c>\frac{-a}{-b+d}\)

= \(\boxed{c>\frac{a}{b-d}}\)

It seems that the first and second result contradicts each other, so what is going on? Was there any flaw in the rearrangement?

## Comments

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TopNewestIn the first result, you divided both sides by \(b-d\), while in the second, you divided by \(d-b\). One of the differences must be negative, so you need to reverse the inequality sign when dividing. – Alex Li · 1 year, 9 months ago

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– Tan Wei Xin · 1 year, 9 months ago

Nice observation. Thanks!Log in to reply