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

ineoualities with fraction

(Happy) Quote:

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