Thread: fraction equation inequality

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

$\displaystyle
\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

You messed up in your second step.

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:

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

and

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

Notice that $\displaystyle x \neq -1$ because then you would have division by zero, which isn't allowed.

$\displaystyle \begin{array}{rcl} \frac{2x-1}{x+1} - 3 & < & 0 \\ \frac{2x - 1 - 3(x+1)}{x+1} & < & 0 \\ \frac{2x - 1 - 3x - 3}{x+1} & < & 0 \\ \frac{-x - 4}{x+1} & < & 0 \\ \frac{x+4}{x+1}& > & 0\end{array}$

(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.

Originally Posted by Craka

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

$\displaystyle
\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