Question

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

Answer 1

Not too hard...

Explanation:

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

`(3x)/(6 + sqrt(7)) = 6 - sqrt(7)`

...now multiply both sides by `6 + sqrt(7)` giving:

`3x = (6-sqrt(7))(6 + sqrt(7))`

...and divide by 3 on both sides:

`x = ((6-sqrt(7))(6 + sqrt(7)))/3`

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

`x = (36 + 6sqrt(7) - 6sqrt(7) - 7)/3`
...gives
`x = 29/3`

GOOD LUCK

Answer 2

`x=29/3`

Explanation:

`x/(6-sqrt7)=(6+sqrt7)/3`
`x=((6+sqrt7)*(6-sqrt7))/3`
Numerator is in the form `(a+b)(a-b)=a^2-b^2`
Hence `x=(6^2-sqrt7^2)/3=(36-7)/3=29/3`