We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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**+2 **

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