# Solving inequality with fraction in one side confusion

In the book I'm using an example is given as follow:

$$\frac{2x - 5}{x-2}< 1$$

then it proceeds to say that we could multiple both sides by $$x-2$$ to get rid of the denominator in the left hand side (I understand that). But then it goes on to say that this method would require to consider the following cases,

$$x-2 > 0$$ and $$x-2<0$$ separately.

How did the author of the book reached to the conclusion that we will need to evaluate such cases? and why did the orientation of the inequalites change?

I tried doing some algebraic manipulation myself but I could reach a conclusion. I tried:

multiplying both sides by $$x-2$$

$$2x-5

subtract x from both sides

$$x-5 < -2$$

$$x-3 < 0$$

• Multiplying by a negative number flips the inequality Commented Feb 29, 2020 at 14:07
• A different possibly better approach could be to look at $\frac{2x-5}{x-2}-1<0$ so $\frac{x-3}{x-2}<0$ now it's clear that quotient of two numbers is negative if they are of opposite signs. Commented Feb 29, 2020 at 14:10
• @Riquelme because its implied by the inequality that the denominator must be a negative value? Commented Feb 29, 2020 at 14:16
• You can get $2x-5\lt x-2$ from $\frac{2x-5}{x-2}\lt 1$ only when $x-2$ is positive. You are implicitly assuming that $x-2$ is positive. Commented Feb 29, 2020 at 14:55
• @mathlove thanks now I think I understand. Commented Feb 29, 2020 at 15:01

How did the author of the book reached to the conclusion that we will need to evaluate such cases?

Because we don't know the sign of $$x-2.$$ All we know is that it may either be positive ($$>0$$) or negative ($$<0$$); the case $$x-2=0$$ not arising since then the fraction is not a real number.

why did the orientation of the inequalites change?

I don't know which of the inequalities you're talking about, but when you're considering the case when $$x-2<0,$$ you have to change the original inequality from $$\text{LHS}<\text{RHS}$$ to $$\text{LHS}>\text{RHS}$$ because whenever you multiply both sides of an inequality by a negative number, the order is reversed.

A way to proceed by not evaluating cases is simply to multiply both sides by $$(x-2)^2$$ without bothering about signs, since a square can never be negative.

$$\frac{2x-5}{x-2}<1$$ it is alwys better that you have 0 on the RHS. Then $$\frac{2x-5}{x-2}-1<0 \implies \frac{x-3}{x-2} <0$$ $$A/B <0 \implies AB<0 \implies (1):$$ $$A<0$$ and $$B>0$$ or (2): $$A>0$$ and $$B<0$$ So we get (1): $$x-3<0$$ and $$x-2>0 \implies 2 or $$(ii) x>3$$ and $$x<2 \implies$$ no solution (as the oprverlap is null).

So the the solution is $$2

Because $$\frac{a}{b}<1\iff\begin{cases}a0\end{cases}\quad \text{or}\quad \begin{cases}a>b\\ b<0\end{cases}.$$