Lesson: Properties of Fractions - Part II

Comment on Properties of Fractions - Part II

Dear Brent,

in the fifth problem I am having troubles with the expression's simplification. More specifically, where does that 9 come from? Which property of fractions has been used?

One approach is to simplify the expression.
(27x + 23y)/(3x + 2y) = (27x + 18y + 5y)/(3x + 2y)
= (27x + 18y)/(3x + 2y) + (5y)/(3x + 2y)
= 9 + (5y)/(3x + 2y)

Thank you very much!
greenlight-admin's picture

Question link: http://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-320...

I'm using the RULE that says: (a + b)/c = a/c + b/c

So, in my solution, I first rewrote 23y as (18y + 5y)

This allowed me to take 27x + 18y + 5y and add brackets as follows to get the equivalent form: (27x + 18y) + 5y

So, we get: (27x + 23y)/(3x + 2y) = [27x + 18y + 5y]/(3x + 2y)

= [(27x + 18y) + 5y]/(3x + 2y)

= (27x + 18y)/(3x + 2y) + 5y/(3x + 2y) {I applied the RULE at the top}

= 9 + [5y/(3x + 2y)]

How did I get 9?
Well 27x + 18y = 9(3x + 2y)
So, (27x + 18y)/(3x + 2y) = (9)(3x + 2y)/(3x + 2y)
= 9 {since (3x + 2y)/(3x + 2y) = 1}

Does that help?

Thanks for explaining. I too got confused by 9.

Now let's take a closer look at (5y)/(3x + 2y)
Notice that (5y)/(3y + 2y) = 5y/5y = 1 [since the numerator and denominator are EQUAL]
However, since we're told that y < x, we know that 3y + 2y < 3y + 2x
This means that (5y)/(3x + 2y) < 1, [since the numerator is LESS THAN the denominator]

If (5y)/(3x + 2y) < 1, then we can conclude that 9 + (5y)/(3x + 2y) < 10

I did not understand this part of the solution. Can you please explain?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-32...

If (5y)/(3x + 2y) < 1, then we can add 9 to both sides of the inequality to get:

(5y)/(3x + 2y) + 9 < 10

Does that help?


I just assumed x to be 3 and y to be 2. When i do this, it makes sense. But is that how you concluded? By assuming values of x and y?
greenlight-admin's picture

x = 3 and y = 2 is ONE pair of values that satisfy the given restriction that 0 < y < x

However, there are infinitely many possible pairs of values for x and y that satisfy the given restriction.

On the third reinforcement activity for this video the question is as follows "If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of n/m?" - when coming to the final answer to the question, how is the numerator 1? Shouldn't it be x^2/2(3)? Please let me know! Thank you!
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-m-is-the-product-of-all-the-integers...

n = (4)(5).....(10)(11)
m = (2)(3)(4)(5).....(10)(11)

Notice that every value in m is also in n
So, all of m's values will pair with some of n's values to get many 1's
That is: 4/4 = 1, 5/5 = 1, 6/6 = 1, . . . 10/10 = 1 and 11/11 = 1
All that remains, is (2)(3) in the denominator.

Another approach is to first notice that m = (2)(3)(n)
So, n/m = n/(2)(3)(n) = 1/(2)(3) = 1/6

Does that help?


Hi there,

I am struggling with the second problem in the practice list.
When using the property (x+y)/z=x/z+y/z

If b/(a+b) = 7/12, then we can also say that (a + b)/b = 12/7

What I don't understand is why we would add 1 next.

the explanation suggest:
Simplify: a/b + 1 = 12/7
Subtract 1 from both sides to get: a/b = 12/7 - 1
Rewrite 1 as follows: a/b = 12/7 - 7/7
Evaluate: a/b = 5/7

Please help break this process down for me! I appreciate your help!

Thank you in advance
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-b-a-b-8385.html

We know that: (a + b)/b = 12/7
Applying the fraction property, we get: a/b + b/b = 12/7
Since b/b = 1, we can write: a/b + 1 = 12/7

Does that help?


understood thank you!


Hi Brent, I've gone over this problem a few times, but I seem to not understand it. Could you please explain this one again.

Also, for comparison questions, are there certain criteria of a question that would be able to tell me that plugging in numbers could solve the question easily?

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/compare-for-x-2155.html

I posted a 3rd solution here: https://gre.myprepclub.com/forum/compare-for-x-2155.html#p28345
Please let me know if that helps.

Plugging in values has its drawbacks. The biggest problem is that, unless we're able to show conflicting outcomes (meaning the correct answer is D), we can never be 100% certain of the correct answer.

That said, I'd say that plugging in numbers is a great approach when you're not sure how to proceed with a question.


Thank you so much. That helped Brent!

Can you give me the answer break down for this problem?
greenlight-admin's picture

Hey Brent, quick question on this one. https://gre.myprepclub.com/forum/if-0-x-y-then-which-of-the-following-must-be-true-3579.html

For A of your answer, wouldn't dividing by 2 yield
y>x (which is still a true statement)
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-0-x-y-then-which-of-the-following-mu...

You're absolutely right. Good catch - thanks!
I have edited my solution accordingly.



How come we are able to conclude that 3x+2y > 3y+2y? And therefore, the value of 3x+2y will be less than 1?

Thanks Bretnt for the help!
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-32...

Think of it this way:
We're told that: y < x
Multiply both sides by 3 to get: 3y < 3x
Add 2y to both sides to get: 3y + 2y < 3x + 2y

Does that help?


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