How can I solve the inequality (-3/x) < 3?

[Qube]

Homework Statement

(-3/x) < 3

Homework Equations

Dividing/multiplying an inequality causes the inequality sign to change.

The Attempt at a Solution

I keep getting the wrong solution. I tried two methods. I cannot get the textbook solution (x < -1)

Method one:

-3 < 3x
-1 < x

Method two:

Divide both sides by negative one, resulting in:

3/x > -3
3 > -3x
-1 < x

 

[Mark44]

Homework Statement

(-3/x) < 3

Homework Equations

Dividing/multiplying an inequality causes the inequality sign to change.

The Attempt at a Solution

I keep getting the wrong solution. I tried two methods. I cannot get the textbook solution (x < -1)

Method one:

-3 < 3x

Here you have multiplied by x, which could be positive or negative (or zero). Since you don't know the sign of x, you need two cases - one where x is assumed to be positive and the other where x is assumed to be negative.

-1 < x


Method two:

Divide both sides by negative one, resulting in:

3/x > -3
3 > -3x

Again, you've multiplied by x, whose sign you don't know. Same comments as above apply here.

-1 < x

 

[Qube]

All right! Now I have -1 < x if x is positive and -1 > x if x is negative. How do I determine which is correct? Is the time for trial-and-error?

 

[Mark44]

The book's answer is wrong or at least incomplete.

If x > 0, you get x > - 1. This means that x > 0 AND x > -1. Together, these mean that x > 0.
If x < 0, you get x < -1. This means that x < 0 AND x < -1. Together, these mean that x < -1.

If you look at the graph of y = -3/x, you'll see that it has two parts. For the curve on the left, -3/x < 3 when x < -1. For the curve on the right, -3/x < 3 for any x > 0.

 

[Qube]

So I guess I'll have to do a bit of trial and error to determine the exact intervals? As in, x > -1 is technically correct, but the best answer is that x is actually greater than 0 or x > 0?

How do you recommend solving these problems? I'm having massive trouble.

 

[verty]

Just remember the rule, NEVER multiply or divide by x or an expression containing x, it could be negative. Read Mark44's reply again, the answer is: x < -1 OR x > 0.

Example: $\frac{x+2}{x-3} > 0$. If $x-3 > 0, x+2 > 0$, and if $x-3 < 0, x+2 < 0$. Can you give the answer for this example?

 

[rock.freak667]

Not to detract from what the others are saying, if you multiply by x, you are multiplying by an unknown value which can be +ve or -ve and will thus change the sign of the inequality (i.e. < might become >). So what you can do to prevent this is multiply instead by x² as this will always be positive.

Then you can solve it as you would using normal algebraic means. You will just need to be careful when doing this as you might pick up an extra solution. But simply analyzing your solution set for the problem would lead you in the correct path.

 

[Ray Vickson]

All right! Now I have -1 < x if x is positive and -1 > x if x is negative. How do I determine which is correct? Is the time for trial-and-error?

Turn your inequality into one of the form
\frac{1}{x} &lt; a
or
\frac{1}{x} &gt; a
I will let you figure out which inequality applies, and what is the value of $a$.

Anyway, once you have a simple inequality in $1/x$, you can envision the graph $y = 1/x$, and then figure out what the x-region must be for the corresponding inequality in y.

Note: this suggested method is safe; it never has you multiplying or dividing by an x of unknown sign, at least, not until the very end.

 

[vanhees71]

I think it's time to give the solution to clarify the issue:

You just have to do a case differentiation. First you can simplyfy the inequality by divding through -3:
-\frac{3}{x} &lt; 3 \Leftrightarrow \frac{1}{x}&gt;-1.
Now for x&gt;0 the inequality is always fulfilled and for x&lt;0 you get by multiplying with (-x)&gt;0
-1&gt;x \; \Leftrightarrow \; x&lt;-1.
Thus the inequality is fulfilled for either x&gt;0 or x&lt;-1.

 

[statdad]

You can also try
<br /> \begin{align*}<br /> \frac{-3}{x} &amp; &lt; 3 \tag{Given} \\<br /> \frac{-3}{x} \cdot x^2 &amp; &lt; 3x^2 \\<br /> -3x &amp; &lt; 3^2 \\<br /> 0 &amp; &lt; 3x^2 + 3x \tag{Now factor to obtain}\\<br /> 0 &amp; &lt; 3x\left(x + 1\right)<br /> \end{align*}<br />

If you set up a sign table (or graph 3x(x+1) you'll find that the inequality is solved
for numbers x &lt; -1 or x &gt; 0 as stated in other places. In short, it is not that one or the other of these two inequalities gives the solution set, is that the solution set consists of any number that satisfies either the first or the second of these two.

 

[Infrared]

I think this problem is being made much harder than it really is.

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

So x&gt;0 or x&lt;-1

 

[Ray Vickson]

I think this problem is being made much harder than it really is.

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

So x&gt;0 or x&lt;-1

You are doing exactly what I suggested the OP do, but he/she never reported back.

 

[statdad]

I think this problem is being made much harder than it really is.

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

So x&gt;0 or x&lt;-1

I would point out that your "So ..." requires the same amount of effort to verify as others have demonstrated, and it is likely that if a person is unsure of how to start the problem the notion that jumping from the final inequality to solution set is hardly obvious.

I do agree that this has been flogged enough.