How to solve the inequality $\frac {5x+1}{4x-1}\geq1$ Please help me solve the following inequality.

\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}

I have tried the following method but it is wrong. Why?

\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}


Thank you for your attention

As  Carlos Eugenio Thompson Pinzón has commented, we  need $\displaystyle4x-1\ne0$
Method $1:$
If $\displaystyle4x-1\ne0, (4x-1)^2>0$
Multiplying either sides  of $\displaystyle\frac{5x+1}{4x-1}\ge1$ by  $(4x-1)^2$
we get $\displaystyle(5x+1)(4x-1)\ge(4x-1)^2$
$\displaystyle\implies (4x-1)\{5x+1-(4x-1)\}\ge0 $
$\displaystyle\implies \left(x-\frac14\right)\{x-(-2)\}\ge0$
We know if $(x-a)(x-b)\ge0$ where $a<b$  either $x\le a$ or $x\ge b$

Method $2:$
As we know $a\ge b\implies \begin{cases} ac\ge bc &\mbox{if } c> 0 \\
ac\le bc &\mbox{if } c< 0 \end{cases}$
If $\displaystyle 4x-1>0\iff x>\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\ge 4x-1\iff x\ge-2$
$\displaystyle\implies x>\frac14$ is one of the solutions
If $4x-1<0\iff x<\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\le 4x-1\iff x\le-2$
$\displaystyle\implies x\le -2$ is the other solution
