Question 107477
Let x=first number, y=second number


Since one number (say the first number) is 3 times the other, then


{{{x=3y}}}


Now the statement "..the sum of their reciprocals
Is 1/6" translates to:


{{{1/x+1/y=1/6}}}



{{{1/(3y)+1/y=1/6}}} Now plug in {{{x=3y}}}



{{{1/(3y)+(1/y)(3/3)=1/6}}} Now multiply the second fraction by {{{3/3}}} to get to the common denominator {{{3y}}}



{{{1/(3y)+(3)/(3y)=1/6}}} Multiply 



{{{(1+3)/(3y)=1/6}}} Combine the fractions



{{{4/(3y)=1/6}}} Combine the numerators




{{{4=(3y)/6}}} Multiply both sides by 3y



{{{24=3y}}} Multiply both sides by 6



{{{8=y}}} Divide both sides by 3 to solve for y



So one number is 8



Now multiply 8 by 3 to find the first number


{{{x=3*8=24}}}



So another number is 24




Answer:


So our answer is 

{{{x=8}}} and {{{y=24}}}




Check:


Remember their reciprocals should add to {{{1/6}}}



{{{1/8+1/24}}} Turn 8 into 1/8 and 24 into 1/24



={{{3/24+1/24}}} Multiply 1/8 by 3/3




={{{4/24}}} Add the fractions 



={{{1/6}}} Reduce




So our answer is verified



note: the previous answer is incorrect. When you check by adding the reciprocal of 3/2 and 9/2 you get {{{2/3+2/9=6/9+2/9=8/9}}} which is not {{{1/6}}}