# Lesson: Properties of Fractions - Part II

## Comment on Properties of Fractions - Part II

### Dear Brent,

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! 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

Thanks for explaining. I too got confused by 9.

### Now let's take a closer look

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? 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?

Cheers,
Brent

### I just assumed x to be 3 and

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? ### x = 3 and y = 2 is ONE pair

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 second reinforcement

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! 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?

Cheers,
Brent

### Hi there,

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 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?

Cheers,
Brent

### understood thank you!

understood thank you!

### https://greprepclub.com/forum

https://greprepclub.com/forum/compare-for-x-2155.html

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? I posted a 3rd solution here: https://greprepclub.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.

Cheers,
Brent

### Thank you so much. That

Thank you so much. That helped Brent!

### https://greprepclub.com/forum

https://greprepclub.com/forum/qotd-21-if-bc-0-and-3b-2c-18-then-which-of-the-2748.html
Can you give me the answer break down for this problem? ### Hey Brent, quick question on

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

y>x (which is still a true statement) You're absolutely right. Good catch - thanks!
I have edited my solution accordingly.

Cheers,
Brent

### https://greprepclub.com/forum

https://greprepclub.com/forum/if-0-y-x-then-which-of-the-following-3202.html

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! 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?

Cheers,
Brent