How do you solve for x in x=18/(x+3)?

Jun 23, 2016

$x = - 6 , x = 3.$

Explanation:

$x = \frac{18}{x + 3}$
$\Rightarrow x \left(x + 3\right) = 18.$
$\Rightarrow {x}^{2} + 3 x - 18 = 0.$
$\Rightarrow {x}^{2} + 6 x - 3 x - 18 = 0 , \ldots \ldots \ldots \left[6 \times 3 = 18 , 6 - 3 = 3.\right]$
$\Rightarrow x \left(x + 6\right) - 3 \left(x + 6\right) = 0.$
$\Rightarrow \left(x + 6\right) \left(x - 3\right) = 0.$
$\Rightarrow x = - 6 , x = 3.$

Clearly, these satisfy the given eqn. hence, the roots are confirmed.

Jun 23, 2016

So solutions are $x = 3 \text{ and } x = - 6$

Explanation:

Multiply both sides by $\left(x + 3\right)$

$x \left(x + 3\right) = 18 \times \frac{x + 3}{x + 3}$

But $\frac{x + 3}{x + 3} = 1$ giving;

$x \left(x + 3\right) = 18$

Subtract $18$ from both sides

$x \left(x + 3\right) - 18 = 18 - 18$

$x \left(x + 3\right) - 18 = 0$

Expand the brackets

${x}^{2} + 3 x - 18 = 0 \leftarrow \text{ solve as a quadratic}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note that $\left(- 3\right) \times 6 = - 18$ and that $- 3 + 6 = + 3$

$\implies \left(x - 3\right) \left(x + 6\right) = 0$

So solutions are $x = 3 \text{ and } x = - 6$
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