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 $x=6$, $y \not= 0$.
+4622 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.
+130070 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 !!!!
