fraction equation inequality

• August 13th 2008, 07:11 PM
Craka
fraction equation inequality
I'm trying to solve the equality $\frac{{2x - 1}}{{x + 1}} < 3 \$

$
\begin{array}{l}
\frac{{2x - 1}}{{x + 1}} < 3 \\
2x - 1 < 3(x + 1) \\
2x - 1 < 3x + 3 \\
- 4 < x \\
\end{array}
$

I can't work out how to get the second solution of x > -1
• August 13th 2008, 07:19 PM
Quick
You messed up in your second step. (Crying)

When you multiply both sides of an inequality by a negative number (which happens when x is less than -1) you have to flip the sign, meaning that the second step would actually yield two inequalities:

$2x-1<3(x+1)\quad\text{when}\quad x>-1$

and

$2x-1>3(x+1)\quad\text{when}\quad x<-1$

Notice that $x \neq -1$ because then you would have division by zero, which isn't allowed.
• August 13th 2008, 07:21 PM
o_O
$\begin{array}{rcl} \frac{2x-1}{x+1} - 3 & < & 0 \\ \frac{2x - 1 - 3(x+1)}{x+1} & < & 0 \\ \frac{2x - 1 - 3x - 3}{x+1} & < & 0 \\ \frac{-x - 4}{x+1} & < & 0 \\ \frac{x+4}{x+1}& > & 0\end{array}$

(Divided -1 from both sides in that last step there which switches the inequality).

So you have your critical points x = -4 and -1. Test below, in between, and above to see which regions are greater than 0 and you should have your answer.
• November 27th 2008, 05:15 PM
jerome_janet
ineoualities with fraction
(Happy)
Quote:

Originally Posted by Craka
I'm trying to solve the equality $\frac{{2x - 1}}{{x + 1}} < 3 \$

$
\begin{array}{l}
\frac{{2x - 1}}{{x + 1}} < 3 \\
2x - 1 < 3(x + 1) \\
2x - 1 < 3x + 3 \\
- 4 < x \\
\end{array}
$

I can't work out how to get the second solution of x > -1