# How do you solve the inequality  6x - 5 < 6/x?

Mar 8, 2018

$6 x - 5 < \frac{6}{x}$

We need to get all the $x$s on one side

multiply both sides by $x$

$x \left(6 x - 5\right) < 6$

distribute

$6 {x}^{2} - 5 x < 6$

Looks like a quadratic. Let's bring the $6$ over and set the inequality to $0$

$6 {x}^{2} - 5 x - 6 < 0$

Now we need to factor $6 {x}^{2} - 5 x - 6$

$\textcolor{w h i t e}{. .} + 36$
$\textcolor{w h i t e}{. .} \times 5$
. . . . . . . . .
$\textcolor{w h i t e}{. .} 1 \times 36$
$\textcolor{w h i t e}{. .} 2 \times 18$
$\textcolor{w h i t e}{. .} 3 \times 12$
$\textcolor{w h i t e}{. .} \textcolor{\mathmr{and} a n \ge}{4} \times \textcolor{w h i t e}{0} \textcolor{\mathmr{and} a n \ge}{9} \textcolor{w h i t e}{0}$ $\implies \textcolor{\mathmr{and} a n \ge}{- 9} + \textcolor{\mathmr{and} a n \ge}{4}$

Since the leading coefficient of the expression isn't $1$, we need to factor by grouping

$\left(6 {x}^{2} + \textcolor{\mathmr{and} a n \ge}{4} x\right) + \left(\textcolor{\mathmr{and} a n \ge}{- 9} x - 6\right)$

$2 x \left(3 x + 2\right) + - 3 \left(3 x + 2\right)$

$\left(2 x - 3\right) \left(3 x + 2\right)$

So now we have

$\left(2 x - 3\right) \left(3 x + 2\right) < 0$

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

Solve for $x$ in $\left(2 x - 3\right)$:

$2 x - 3 < 0$

$2 x < 3$

$x < \frac{3}{2}$

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

Solve for $x$ in $\left(3 x + 2\right)$:

$3 x + 2 < 0$

$3 x < - 2$

$x < - \frac{2}{3}$

So $- \frac{2}{3} < x < \frac{3}{2}$