Lesson: The Something Method

Comment on The Something Method

outbox thinking :d﻿

outbox thinking :d﻿

Thanks!

How would one know if this

How would one know if this technique should be used and will yield an answer ?

You can use it in situations

You can use it in situations where an algebraic expression is grouped together.

Two rudimentary examples:

(2)(x² - x - 9) = 6
This tells us that (x² - x - 9) = 3

12/(x² - x - 9) = 3
This tells us that (x² - x - 9) = 4

I get these 2 examples for

I get these 2 examples for applying the *something* method. For the 1st: If 2 multiplied by *something* = 6; then that *something* must be equal to 3. However, if the *something* is “x-squared - x - 9” (which is equal to 3), then the value of x still needs to be solved if the question is asking for the value of x. Is this correct? Thanks.

That's correct. We'd know

That's correct. We'd know that (x² - x - 9) = 3, but we would still need to solve an equation for x.

This is a good method but any

This is a good method but any other method to solve this?

We can just apply basic

We can just apply basic equation-solving techniques.

Given: (12y)/(5y + 1/x) = 2

Multiply both sides by (5y + 1/x) to get: (12y) = 2(5y + 1/x)

Expand: 12y = 10y + 2/x

Subtract 10y from both sides: 2y = 2/x

Multiply both sides by x to get: 2xy = 2

Divide both sides by 2 to get: xy = 1

Voila!

I'm confused with this

I'm confused with this question. At 00:56, if '12y/6y = 2'; then y = 1?

As such, the best answer, in evaluating the entire equation (including the numerator), is: 'x + y = 1' (or y + x = 1)?

Thanks.

Sorry but I'm not too sure

Sorry but I'm not too sure what you're asking.
At that point in the video we know that 12y/something = 2

Since we know that 12y/6y = 2, we can then conclude that: something = 6y

It's important to note that 12y/6y = 2 for ALL non-zero values of y.

For example, if y = 10, then 12y/6y = 12(10)/6(10) = 120/60 = 2
Likewise, if y = 7.2, then 12y/6y = 12(7.2)/6(7.2) = 86.4/43.2 = 2
Etc.

Does that help?

QUESTION #2: As such, the best answer, in evaluating the entire equation (including the numerator), is: 'x + y = 1' (or y + x = 1)?
Sorry can you elaborate on your question?
I'm not sure what you mean by "evaluate the entire equation."

Cheers,
Brent

I just watched the video

I just watched the video again. From my POV, it seems that in getting to the answer 'xy = 1', only the expression '5y + 1/x' (the denominator) was evaluated?

I'm referring to the numerator part ('12 y'), as well as the right hand side of the equation, which is = '2' when I said 'entire equation.'

Perhaps I'm missing something?

Thanks.

I'm having trouble with some

I'm having trouble with some of your terminology.

When we EVALUATE an algebraic expression, we do so with given values.
For example, on its own, the expression 3x² + x has no value. However we can evaluate (assign a value to) that expression for various values of x.
For example, if x = 6, we get: 3x² + x = 3(5²) + 5 = 80
So when x = 6, the expression 3x² + x evaluates to equal 80.

So, when you say "getting to the answer 'xy = 1', only the expression '5y + 1/x' (the denominator) was evaluated?" I'm not sure what you're referring to when you use the word "evaluated"

The main idea here is that, if 12y/something = 2, then something = 6y
In other words, 5y + 1/x = 6y

Next, if 5y + SOMETHING = 6y, then SOMETHING = y
In other words, 1/x = y
From here, we can multiply both sides of the equation by x to get: 1 = xy

Does that help?

Cheers,
Brent

To clarify: when we get to

To clarify: when we get to *5y + 1/x = 6y*; y = 1/x (or 1), and x is also equal to 1, since xy = 1? And if the question asks for *x + y*, the answer is 2?

Thanks.

You're correct to say that y

You're correct to say that y = 1/x
However, we cannot then conclude that it must be the case that y = 1 and x = 1.
The fact of the matter is that the equation y = 1/x has infinitely many solutions.

For example it could be the case that x = 3 and y = 1/3
This satisfies the condition that y = 1/x

Likewise it could be the case that, x = -0.25 and y = -4
This also satisfies the condition that y = 1/x

And so on.

Cheers,
Brent

I just got confused. I saw a

I just got confused. I saw a reply to an earlier question, and I think that reply works for me: Please see below.

“Given: (12y)/(5y + 1/x) = 2
Multiply both sides by (5y + 1/x) to get: (12y) = 2(5y + 1/x)
Expand: 12y = 10y + 2/x
Subtract 10y from both sides: 2y = 2/x
Multiply both sides by x to get: 2xy = 2
Divide both sides by 2 to get: xy = 1”

Now that solving this problem is clear to me. My new questions are:

1. Does the *something* method always work better than the *basic equation solving technique*?

2. Could you please provide examples of when thie *something* method is the *must” technique to be used?

Thanks!

That (algebraic) approach

That (algebraic) approach definitely works.

1. No, the Something Method does not always work better than general algebraic equation solving.

2. There are no instances in which an equation only be solved by applying the Something Method. So if you are uncomfortable with the Something Method, you can just stick with general algebraic equation solving.

Cheers,
Brent

That's really interesting.

That's really interesting.

Oh wow...brilliant something

Oh wow...brilliant something method ever...thanks Brent!!