Inequalities of rational expressions

[delsoo]

referring to the photo attached, can anyone expalin and give suitable numerical example please? i couldn't understand the notes.

 

[micromass]

We know that

\frac{a}{b}=\frac{c}{d}

implies $ad = cb$. Indeed, to prove it, we multiply both sides of the equation with $bd$.

The question is now whether \frac{a}{b}<\frac{c}{d}

implies $ad < cb$.

The answer is no. Indeed, we know that

\frac{-1}{-1} &lt; \frac{2}{1}

but it is not true that $(-1)\cdot 1 < 2\cdot (-1)$. In fact, it is the reverse inequality that holds.

So the rule

\frac{a}{b}&lt;\frac{c}{d}~\Rightarrow~ad&lt;bc

is only true if $b,d>0$.

So if you have something to solve like

\frac{x+1}{x+2}&lt; \frac{x}{x-3}

then you cannot solve this by cross-multiplying, that is you cannot say that

(x+1)(x-3)&lt;x(x+2)

because $x+2$ and $x-3$ might be negative. Now, if you know that it's positive, then you can do it, but you don't know this in general.

The way to solve it is to bring both to the same side, so the above is equivalent to

\frac{x}{x-3} - \frac{x+1}{x+2} &gt; 0

and thus

\frac{x(x+2) - (x+1)(x-3)}{(x-3)(x+2)}&gt;0

So we get that a fraction is positive. This can only happen in two situations:

1) Both numerator and denominator are positive. So the above implies
x(x+2)&gt;(x+1)(x-3)~\text{and}~(x-3)(x+2)&gt;0

Solving this yields possible values of $x$ which solve the inequality, but they are not the only solutions:

2) Both numerator and denominator are negative. So we get

x(x+2)&lt;(x+1)(x-3)~\text{and}~(x-3)(x+2)&lt;0

Solving this also yields solutions.