I need help... I did most of my hw, but these just can't? IDK. but...

Among all fractions $x$ that have a positive integer numerator and denominator and satisfy $$\frac{9}{11} \le x \le \frac{11}{13},$$ which fraction has the smallest denominator?

AND

Evaluate: $\dfrac{(x+y)^2 - (x-y)^2}{y}$ for $x=6$, $y \not= 0$.

Guest Dec 2, 2018

#1**+1 **

1. First, try to change the x in the middle, to a fraction. Let's say x/y. We then multiply both sides by b, to get 9b/11 and 11b/13. Now, the trick here is to find a positive integer that satisfies and is less than 1, since 11/13 is less than 1.

tertre Dec 2, 2018

#2

#5**+3 **

Here might be another way to look at this

Notice that the fractions seem to have the form n / [ n + 2] where n is an integer

Suppose that there exists a fraction such that

9/11 < n / [n + 2] < 11/13

We can split this into two inequalities

Looking at the inequality on the left....we have

9[n + 2] < 11n

9n + 18 < 11n

18 < 2n

9 < n or

n > 9

Looking at the inequality on the right.... we have

13(n) < 11[n + 2]

13n < 11n + 22

2n < 22

n < 11

This implies that

9 < n < 11

So...n = 10 is the only integer that satisfies this

And n + 2 = 12

So....the fraction is 10/12 = 5/6 ......just as tertre found!!!

Good job, tertre !!!!

CPhill Dec 3, 2018