# Why am I losing solutions when solving Inequalities like $\frac{2y-3}{y}>0$ by multiplying both sides by $y$?

So, I need to solve the following Inequality-

$$\frac{2y-3}{y}>0$$

I proceeded in the following manner-

$$2y-3>0\qquad \text{[Multiplying both sides by }y.]$$

$$y>\frac{3}{2}\qquad \text{[Adding 3 to both sides and and dividing by 2.]}$$

So my solution set is ($\frac{3}{2},\infty$)

But clearly, the inequality also holds true whenever y has any negative value. Hence, my solution set should in fact be: $(-\infty,0)\cup(\frac{3}{2},\infty)$.

My question is- What is the proper way of solving such inequalities in order to get all values of the variable? I am looking for a simple explanation(pre-college level) if possible.

• If you multiply both sides of an inequality by a negative number, the direction of the sign switches! So... when you multiplied both sides of $\frac{2y-3}{y}>0$ by $y$... do you switch the direction to make it $2y-3\color{red}{<}0$? Or do you keep it the same as $2y-3\color{red}{>}0$? The answer is... you do both separately while keeping track of what assumption you made at the time, so $\frac{2y-3}{y}>0\implies \begin{cases} 2y-3<0&\text{if}~y<0\\2y-3=0&\text{if}~y=0\\2y-3>0&\text{if}~y>0\end{cases}$ (the second case obviously presents a contradiction) Jun 3, 2018 at 19:06
• ""$2y−3>0$[Multiplying both sides by y.]" That will not be true if $y < 0$. If $y< 0$ then multiplying both sides by $y$ yields $2y - 3 < 0$. Jun 3, 2018 at 19:08
• Insead of multiplying both sides by $y$ what happens if you multiply both sides by $-3y$. Jun 3, 2018 at 20:19

[Multiplying both sides by y.]

Here. You multiplied both sides of an inequality by some number, but you forgot to check whether that number was positive (in which case the sign remains ">") or negative (in which case the sign becomes "<").

If you want to make sure you have all the solutions, make sure you don't accidentally multiply or divide by something that might be non-positive, or each time you do it, split it into two cases ("Case 1: y > 0" and "Case 2: y < 0").

• Note also that the whole inequality is nonsense if $y=0$, which means you don't need to account for that case. But in general, you should consider it as a third case (especially when dealing with quadratics and the like). Jun 3, 2018 at 20:39

Consider the following general inequality:

$$\frac{f(y)}{g(y)} > 0.$$

Suppose that:

1. $f(y) > 0$ for all $y \in F \subseteq \mathbb{R}$;
2. $g(y) > 0$ for all $y \in G \subseteq \mathbb{R}$.

Starting from the sets $F$ and $G$, you must build up a third set as follows:

$$H = \{y \in \mathbb{R} : [f(y)>0 ~\text{and}~ g(y)>0]~\text{or}~[f(y)<0 ~\text{and}~ g(y)<0]\}.$$

This means that $y \in H$ if one of the following holds true:

1. $y \in F$ and $y \in G$;
2. $y \not\in F$ and $y \not\in G$.

For the particular case:

$$\frac{2y-3}{y}> 0.$$

$$f(y) = 2y-3>0 \Rightarrow F = \left\{y > \frac{3}{2}\right\}.$$ $$g(y) = y>0 \Rightarrow G = \left\{y > 0\right\}.$$

Then:

1. If $y>\frac{3}{2}$, then $y \in F$ and $y \in G$. Therefore, $y \in H$.
2. If $y < 0$, then $y \not\in F$ and $y \not\in G$. Therefore, $y \in H$.

Finally, the set H is:

$$H = \left\{y > \frac{3}{2} \vee y < 0\right\}.$$

If $y<0$ (clearly valid), when you multiply both sides by $y$, the inequality flips over; however, you stick to only one case of possible results.

One trick to get rid of this, is to multiply both sides by $y^2$, which is guaranteed to be positive: $$y(2y-3)>0\implies y<0\quad\text{or}\quad y>3/2$$

The correct way to do it is.

Case 1: $y > 0$. then $2y - 3 > 0$ and $y > 0$.

Case 2: $y = 0$. This is impossible.

Case 3: $y < 0$ and $2y - 3 < 0$ and $y < 0$.

In case 1: $2y -3 > 0$ so

$2y > 3$

$y > \frac 32$ and $y > 0$ so $y> \frac 32 > 0$

OR

Case 3: $2y - 3< 0$ so

$2y < 3$

$y < \frac 32$ and $y < 0$. So $y < 0 < \frac 32$.

So either $y > \frac 32$ OR $y < 0$.