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

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Answer 1 · 2017-10-12T19:02:46

Not too hard...

...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 · 2017-10-12T19:05:15

#x=29/3#

#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#