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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 $x=6$, $y \not= 0$.

 Dec 2, 2018
 #1
avatar+4221 
+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. 

 Dec 2, 2018
 #2
avatar+5073 
+2

deleted

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 Dec 2, 2018
edited by Rom  Dec 3, 2018
edited by Rom  Dec 3, 2018
 #3
avatar+4221 
+2

Not quite, Rom. Continuing with my method and plugging the values of b, we get b=6, so the answer is \(\boxed{\frac{5}{6}}.\) Can someone verify this?

tertre  Dec 3, 2018
 #4
avatar+100513 
+1

9/11 = .8181.....

5/6 = .8333....

11/13 = .846....

 

Mmmmm.....it looks like you could  be correct, tertre.....

 

 

cool cool cool

CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
 #5
avatar+100513 
+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   !!!!

 

 

 

cool cool cool

 Dec 3, 2018
edited by CPhill  Dec 3, 2018
 #6
avatar+4221 
+1

Thank you, CPhill! Great solution!

tertre  Dec 3, 2018

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