# How do you solve \frac { 3} { 6+ \sqrt { 7} } = \frac { 6- \sqrt { 7} } { x }?

Oct 12, 2017

Not too hard...

#### Explanation:

...multiply both sides of your initial equation by $x$...

$\frac{3 x}{6 + \sqrt{7}} = 6 - \sqrt{7}$

...now multiply both sides by $6 + \sqrt{7}$ giving:

$3 x = \left(6 - \sqrt{7}\right) \left(6 + \sqrt{7}\right)$

...and divide by 3 on both sides:

$x = \frac{\left(6 - \sqrt{7}\right) \left(6 + \sqrt{7}\right)}{3}$

...which is your answer, but I'm guessing your instructor would like to see it simplified. If you multiply out the numerator...

$x = \frac{36 + 6 \sqrt{7} - 6 \sqrt{7} - 7}{3}$
...gives
$x = \frac{29}{3}$

GOOD LUCK

Oct 12, 2017

$x = \frac{29}{3}$

#### Explanation:

$\frac{x}{6 - \sqrt{7}} = \frac{6 + \sqrt{7}}{3}$
$x = \frac{\left(6 + \sqrt{7}\right) \cdot \left(6 - \sqrt{7}\right)}{3}$
Numerator is in the form $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
Hence $x = \frac{{6}^{2} - {\sqrt{7}}^{2}}{3} = \frac{36 - 7}{3} = \frac{29}{3}$