# Lesson: QC Strategy - Matching Operations

## Comment on QC Strategy - Matching Operations

### On the next to last example

On the next to last example why are you plugging in -1 for X when you said only multiplying by a positive is acceptable? Furthermore, in the follow up example where 2x is subtracted from both quantities, the answer is the same as D; so couldn't you say both methods were valid?﻿ ### Plugging in a negative value

Plugging in a negative value for a variable is not the same as multiplying both quantities by a negative number.﻿

### with due respect, why not to

with due respect, why not to replace 'w' by '2' (as w>0 ) to get answer very fast ( either quantity A or B is greater )..!? ### That approach can get you

That approach can get you into trouble. If you use the Plugging In Numbers approach, you can't just stop at plugging in one value, since there might be other values of w such that the two quantities are equal or Quantity B is greater (for more on this, watch the part that starts at 2:50 of this video: https://www.greenlighttestprep.com/module/gre-quantitative-comparison/vi...

For example, let's say that we're told that w > 0 and we have the following:
QUANTITY A: w
QUANTITY B: w^2

Using your approach, we might plug in w = 2 (since we're told that w > 0, and 2 > 0). When we do this, we get:
QUANTITY A: 2
QUANTITY B: 4
Here, Quantity B is greater. Does this mean the correct answer is B?
No.

If we plug in w = 1, we get:
QUANTITY A: 1
QUANTITY B: 1
Here, the two quantities are equal.

If we plug in w = 1/2, we get:
QUANTITY A: 1/2
QUANTITY B: 1/4
Here, Quantity A is greater.

So, be careful; plugging in ONE value will not always yield the correct answer.

### Also, why not to cancel 'y-3x

Also, why not to cancel 'y-3x' from both sides in first question to get answer more quickly ? ### We are, indeed, "canceling" y

We are, indeed, "canceling" y - 3x^2 from both quantities. This video is meant to show HOW that works. We do so, by performing the same operation to both quantities.

The term "cancel" can sometimes be misinterpreted. It doesn't mean that, if both quantities share the same algebraic expression, then we can just eliminate those expressions from both quantities.

For example, if we have:
QUANTITY A: 3/x
QUANTITY B: 2/x
We can't just say "Since both expressions have an x term, we just cancel them to get:
QUANTITY A: 3
QUANTITY B: 2
This will yield an incorrect conclusion.

So, we need to be careful to follow the rules/guidelines outlined in the video.

### I guess I am confused by the

I guess I am confused by the way the problem 7w + 4 and w-2 is explained. It's stated that in order to solve, we have to get the variable on one side and the non variables are on the other side. Is that in regard to the w>0? I don't mean to be dumb, because I understand the way the problem is solved, I'm just not sure I understand the initial thought process, and if I could look at a problem and accomplish just that. ### Good question.

You're referring to the question that appears at 1:37 in the above video:

Given: w > 0
QUANTITY A: 7w + 4
QUANTITY B: w - 2

For Quantitative Comparison questions involving variables, it's often useful to get all of the variables to just one of the quantities. In most cases, that will make it easier to compare the quantities.

Cheers,
Brent

### So could I say it is

So could I say it is appropriate to set them equal to each other in that case? ### Are you asking whether we

Are you asking whether we should create the equation 7w + 4 = w - 2?

If so, then the answer is no.

The primary goal of Quantitative Comparison questions is to determine the relationship between the two quantities (Is Quantity A greater? Is Quantity B greater? Do they have equal value?). So, we can never assume that the quantities are of equal value.

Does that help?

Cheers,
Brent

### In the reinforcement question

In the reinforcement question 150-159 range can you please explain me the second last one by plugging values.. Mp=p.. Though I got your explanation but just wanted to know if it can be solved by plugging in values too.
thanks ### I'm happy to help!

I'm happy to help!

Here's my solution: https://greprepclub.com/forum/m-p-and-x-are-positive-integers-3017.html#...

Cheers,
Brent

### I have a question about

I have a question about multiplying and dividing quantities on both sides by a positive value.....

Example:
given: x < 0
Quant A: x^3
Quant B: x^5

Can't I divide by x^3 on both sides leaving me with

Quant A: 1
Quant B: x^2

Aren't I dividing by a negative on both sides?
Should i not do this? ### I address this concept at 2

I address this concept at 2:45 in the above video.

However, I'll show you why dividing both quantities by a negative can lead to false conclusions.

Let's say we're asked to compare the following:
QUANTITY A: -6
QUANTITY B: 12

Clearly Quantity B is greater. In fact, if we follow any of the allowable operations, Quantity B will always be greater.

However, if we break one of the rules and divide both quantities by -3, we get:
QUANTITY A: 2
QUANTITY B: -4
Now, Quantity A is greater.

I hope that helps.

Cheers,
Brent

### I understand...

I understand...
So the safe method in the example I provided would be to divide by x^2 instead of x or x^3. ### Yes, that would be a safer

Yes, that would be a safer approach.

If we divide both quantities by x², we get:
Quant A: x
Quant B: x³

Now let's test some cases....
CASE I: If x = -1, then we get:
Quant A: x = -1
Quant B: x³ = (-1)³ = -1
In this case, the quantities are equal

CASE II: If x = -2, then we get:
Quant A: x = -2
Quant B: x³ = (-2)³ = -8
In this case, Quantitt A is greater

Cheers,
Brent

### So i solved this question can

So I solved this question can you confirm if my explanation is right

x - y=5
A: x^2-y^2
B: 5

X^2 - Y^2 can be written as (X-Y)(X+Y), and X-Y = 5. So we get
A: 5(X+Y)
B: 5

Dividing both sides by 5 we get
A: X+Y
B: 1

Let's just deal with one variable so we rewrite the given equation as X = Y + 5 and substitute in A to get
A:5+2Y
B: 1

We do not know whats the true value of Y it can be negative or positive so answer is D. ### That's a perfectly valid

That's a perfectly valid solution. Nice work!

### Hey Brent, at the 3:20

Hey Brent, at the 3:20 minute mark I may be overthinking, but how are you getting the values for dividing by 3 and dividing by -2. A bit lost how you're getting those values ### With each step in the process

With each step in the process, I'm manipulating the most recent values.
QUANTITY A: 2
QUANTITY B: 5

When we add 8 to both quantities we get:
QUANTITY A: 2 + 8 = 10
QUANTITY B: 5 + 8 = 13

Next, we subtract 4 from both quantities to get:
QUANTITY A: 10 - 4 = 6
QUANTITY B: 13 - 4 = 9

Next, we divide both quantities by 3to get:
QUANTITY A: 6 ÷ 3 = 2
QUANTITY B: 9 ÷ 3 = 3

And so on.

Does that help?

### Haha, Brent you're the best.

Haha, Brent you're the best. Extremely slow moment for me. thank you as always!

### Hello Brent,

Hello Brent,

Is there any limitation to applying this strategy because I tried applying it unfortunately it didn't work?

here below is the link to the question
https://greprepclub.com/forum/0-t-7960.html The strategy won't always work, especially in situations in which we don't know whether the variable(s) is/are negative or positive.
That said, it's a very effective strategy and should be considered for a lot of questions.

The question you noted can be solved using Matching Operations.
Here's my solution: https://greprepclub.com/forum/0-t-7960.html#p14521

### I have a question when and

I have a question when and how do we know when to apply approximation and when to match operations....like is there a preference? ### There's no guaranteed rule

There's no guaranteed rule for determining the best strategy.
However, if we're given variables in both quantities, then matching operations is a good strategy to start with.
If both quantities consist of numbers and operations only, approximation is a good place to start.
Some other useful strategies include:
- moving the terms with variables to one side
- arranging the values so that one quantity equals 0
- arranging the values so that one quantity equals 1