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

Aug 31, 2015

$x \in \left(- \frac{2}{3} , 0\right) \cup \left(0 , \frac{3}{2}\right)$

#### Explanation:

Right from the start, you know that $x$ cannot be equal to zero, since it is the denominator of a fraction. So you need to have $x \ne 0$.

With that in mind, multiply the left-hand side of the inequality by $1 = \frac{x}{x}$ to get rid of the denominator

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

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

Next, add $- 6$ to both sides of the inequality

$6 {x}^{2} - 5 x - 6 < \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}$

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

To help you determine the intervals on which this quadratic function is smaller than zero, you need to first determine its root by using the quadratic formula

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

${x}_{1 , 2} = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \cdot 6 \cdot \left(- 6\right)}}{2 \cdot 6}$

${x}_{1 , 2} = \frac{5 \pm \sqrt{169}}{12}$

${x}_{1 , 2} = \frac{5 \pm 13}{12} = \left\{\begin{matrix}{x}_{1} = \frac{5 + 13}{12} = \frac{3}{2} \\ {x}_{2} = \frac{5 - 13}{12} = - \frac{2}{3}\end{matrix}\right.$

You can thus rewrite the quadratic as

$6 \left(x - \frac{3}{2}\right) \left(x + \frac{2}{3}\right) = 0$

So, you need this expression to be negative, which implies that $\left(x - \frac{3}{2}\right)$ must be positive whenever $\left(x + \frac{2}{3}\right)$ is negative, and vice versa.

For $x > - \frac{2}{3}$ and $x < \frac{3}{2}$ you get

$\left\{\begin{matrix}x - \frac{3}{2} < 0 \\ x + \frac{2}{3} > 0\end{matrix}\right. \implies \left(x - \frac{3}{2}\right) \left(x + \frac{2}{3}\right) < 0$

Any value of $x > \frac{3}{2}$ will make both terms positive, and any value of $x < - \frac{2}{3}$ will make both terms negative. Keeping in mind that you also need $x \ne 0$, the solution set for this inequality will be $x \in \left(- \frac{2}{3} , 0\right) \cup \left(0 , \frac{3}{2}\right)$.