# Inequalities with variable in denominator

• May 10th 2008, 02:00 AM
eniuqvw
Inequalities with variable in denominator
I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.
• May 10th 2008, 02:09 AM
Moo
Quote:

Originally Posted by eniuqvw
I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.

Hello,

Ok, there is one thing you should know. When multiplying by a negative number, the direction of the inequality changes.

For example : 2<3
When multiplying by -1, you get -2>-3

So here, intuitively, you should separate the situations :
1) firstly, if x is positive
2) secondly, if x is negative

1) If x is positive :

1/x<4

Multiplying by x, we get : 1<4x, since x>0.
--> x>1/4

Solution to this is (1/4, pos infinity)

2) If x is negative :

Here, it's immediate because 1/x is negative, hence always inferior to 4.
But let's apply the method.

1/x<4

Multiplying by x, we get 1>4x, because x is negative !

--> x<1/4

Because x is negative, you gotta find all the negative numbers which verify x<1/4.
This is true for all x negative.

Thus the solution to this is (neg infinity, 0)
• May 10th 2008, 02:10 AM
mr fantastic
Quote:

Originally Posted by eniuqvw
I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.

Because my algebra is so poor I always solve inequalities by drawing graphs and then solving an equality.

In your case I'd draw the graphs of y = 1/x and y = 1/4. Then I'd solve 1/x = 4. Then I'd use the graphs to get the solution to the inequality.
• May 10th 2008, 09:14 AM
Krizalid
Quote:

Originally Posted by eniuqvw

I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity).

First Method: make $0$ the RHS,

\begin{aligned}
\frac{1}{x}&<4 \\
\frac{1}{x}-4&<0 \\
\frac{1-4x}{x}&<0.
\end{aligned}

To get a negative quantity we must study two cases:
• If $1-4x<0$ & $x>0.$ Intersect these conditions and we have $S_1=\left] \frac{1}{4},\infty \right[.$

• If $1-4x>0$ & $x<0.$ Intersect these conditions and we have $S_2=\bigg]-\infty,0 \bigg[.$

Final solution we'll be given by $S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty \right[.$

Second Method: we obviously discard $x=0,$ otherwise the initial inequality makes no sense. So we're gonna multiply both sides by $x^2,$ since this quantity is positive, the inequality stays as it's written. Following up on this we have:

\begin{aligned}
\frac{1}{x}&<4 \\
x&<4x^{2} \\
x(4x-1)&>0.
\end{aligned}

From here study two cases again:
• If $4x-1>0$ & $x>0\implies S_{1}=\left] \frac{1}{4},\infty \right[.$

• If $4x-1<0$ & $x<0\implies S_{2}=\bigg] -\infty ,0 \bigg[.$

Hence $S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty \right[,$ which is we got before.
• Jul 24th 2008, 02:55 AM
redmafiya
A much easier approach is to multiply both sides by x^2 instead of x.

As x^2 is always greater than 0 (x = 0 is not allowed in this case), there is no need to worry about the direction of the inequality sign.

For 1/x < 4
Multiplying both sides by x^2, we have
x < 4x^2
4x^2 - x > 0
x(4x - 1) > 0

The two roots of x(x/4 - 1) are x = 0 and x = 1/4.

So the solution to the inequality is
x < 0 or x > 1/4
• Jul 25th 2008, 01:07 PM
Krizalid
Then you're not gonna have trouble to check the second method above.