SOLUTION: Let x be a positive number and y be its reciprocal. Compute 1/(x+1) + 1/(y+1)

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Question 238037: Let x be a positive number and y be its reciprocal. Compute
1/(x+1) + 1/(y+1)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Since the reciprocal of x is 1/x, y = 1/x. Substituting this into the expression for y we get:
1%2F%28x%2B1%29++%2B++1%2F%281%2Fx%2B1%29
Now we simplify. We'll start with the second term by eliminating the fractions within the "big" fraction. This is done by multiplying the numerator and denominator of the "big" fraction by the Lowest Common Denominator (LCD) of the "little" fractions. Since there is only one "little" fraction, finding the LCD is simple. It is x.
1%2F%28x%2B1%29++%2B++%281%2F%281%2Fx%2B1%29%29%28x%2Fx%29
1%2F%28x%2B1%29++%2B++x%2F%28%281%2Fx%29%2Ax%2B1%2Ax%29%29%29
1%2F%28x%2B1%29++%2B++x%2F%281%2Bx%29
Not only have we eliminated the fraction with a fraction, we have coincidentally gotten the denominators to match (since 1+x=x+1 by the Commutative Property). So we can go ahead and add the two fractions:
%281%2Bx%29%2F%281%2Bx%29
And finally, as usual with answers that are fractions, we try to reduce the fraction. This is very easy and obvious, I hope.
cross%281%2Bx%29%2Fcross%281%2Bx%29
1%2F1
which simplifies to
1
The x's have disappeared. This means that 1%2F%28x%2B1%29++%2B++1%2F%28y%2B1%29 is equal to 1 no matter what number you use for x (except x cannot be zero because zero does not have a reciprocal).