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April 25th, 2014, 07:56 PM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  Question on solving an inequality
In the inequality: $\displaystyle \frac{(x + 3)(x  1)}{x} \geq 0$ Why does the following manipulation lead to the wrong solution set? $\displaystyle \frac{(x + 3)(x  1)}{x} \geq 0$ $\displaystyle (x + 3)(x  1) \geq 0$ $\displaystyle x \leq 3, x \geq 1$ Apparently, the solution set is $\displaystyle 3 \leq x \leq 0, x\geq 1$. I don't understand how to get the correct solution. 
April 25th, 2014, 08:47 PM  #2  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Quote:
Solution: $\displaystyle x\in [3,0)\cup [+1,+\infty)$ Last edited by Dacu; April 25th, 2014 at 08:59 PM.  
April 25th, 2014, 08:48 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
The problem comes when you multiply by $x$. If $x<0$, then mutiplying by it will reverse the inequality. \begin{align*} \frac{(x + 3)(x  1)}{x} &\geq 0 \\ (x + 3)(x  1) &\leq 0 &\forall x \lt 0 \\ \end{align*} And $(u + 3)(u  1) \leq 0 \Longrightarrow 3 \leq u \leq 1$, but since we are limited to $x \lt 0$, we have $3 \leq x \lt 0$. 
April 26th, 2014, 04:43 AM  #4 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
When solving inequalities, always be careful about multiplication and division. Adding and subtracting terms is always okay, but when you multiply or divide you have to make sure you’re only doing it with positive quantities so as to preserve the direction of the inequality. To find the solution set for $\displaystyle \frac{(x+3)(x1)}x \geqslant 0$ one way is to divide the real line into into four intervals, $x < 3$, $3 < x < 0$, $0 < x < 1$, $x > 1$, and see whether the factors are positive or negative (for the moment we’ll ignore the end points of the intervals): $$\begin{array}{ccccc} & x+3 & x1 & x & \frac{(x+3)(x1)}x \\ \hline \\ x < 3 & \mathrm{ve} & \mathrm{ve} & \mathrm{ve} & \mathrm{ve} \\ \hline \\ 3 < x < 0 & \mathrm{+ve} & \mathrm{ve} & \mathrm{ve} & \mathrm{+ve} \\ \hline \\ 0 < x < 1 & \mathrm{+ve} & \mathrm{+ve} & \mathrm{ve} & \mathrm{ve} \\ \hline \\ x > 1 & \mathrm{+ve} & \mathrm{+ve} & \mathrm{+ve} & \mathrm{+ve} \end{array}$$ Finally we note that $\displaystyle \frac{(x+3)(x1)}x = 0$ when $x=3$ or $x=1$ so we can add these points. This gives the solution set as $[3,\,0]\cup[1,\,\infty)$. Last edited by Olinguito; April 26th, 2014 at 05:27 AM. 

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