# How do you solve x/9 + 5/x = 2?

Oct 1, 2016

$x = 3$ or $x = 15$

#### Explanation:

As $\frac{x}{9} + \frac{5}{x} = 2$

we have $\frac{x \times x + 5 \times 9}{9 x} = 2$

or $\frac{{x}^{2} + 45}{9 x} = 2$

Assuming $x \ne 0$ and multiplying both sides by $9 x$

${x}^{2} + 45 = 18 x$ or ${x}^{2} - 18 x + 45 = 0$

or ${x}^{2} - 3 x - 15 x + 45 = 0$

or $x \left(x - 3\right) - 15 \left(x - 3\right) = 0$

or $\left(x - 15\right) \left(x - 3\right) = 0$

i.e. either $x - 3 = 0$ i.e. $x = 3$

or $x - 15 = 0$ i.e. $x = 15$

Oct 1, 2016

$x = 15 \text{ or } x = 3$

#### Explanation:

If you have an equation which has fractions, you can get rid of the fractions by multiplying every term by the LCD.

In $\frac{x}{9} + \frac{5}{x} = 2 \text{ "larr" } L C D = \textcolor{red}{9 x}$

$\frac{\textcolor{red}{9 x} \times x}{9} + \frac{\textcolor{red}{9 x} \times 5}{x} = \textcolor{red}{9 x} \times 2 \text{ } \leftarrow$ cancel the denominators.

$\frac{\textcolor{red}{\cancel{9} x} \times x}{\cancel{9}} + \frac{\textcolor{red}{9 \cancel{x}} \times 5}{\cancel{x}} = \textcolor{red}{9 x} \times 2$

${x}^{2} + 45 = 18 x \text{ } \leftarrow$ make the quadratic = 0

${x}^{2} - 18 x + 45 = 0$

$\left(x - 15\right) \left(x - 3\right) = 0$

Making each factor = 0 gives,
$x = 15 \text{ or } x = 3$