MHB Restrictions on Variables for Simplified Expression

[mathdrama]

I'm not really sure where to put this, it's from an MCR3U course(Advanced Functions):

Simplify and state the restrictions on the variables.
a) (3x)2/4xy * 16y/36x

I got as far as
= 12xy2/36x2y
= Xy2/3x2y
= 3x-1y

but I don't really know how to state the restrictions. Thanks in advance.

 

[Nono713]

I'm not really sure where to put this, it's from an MCR3U course(Advanced Functions):

Simplify and state the restrictions on the variables.
a) (3x)2/4xy * 16y/36x

I got as far as
= 12xy2/36x2y
= Xy2/3x2y
= 3x-1y

but I don't really know how to state the restrictions. Thanks in advance.

Please note that the original and the simplified expression are not the same! For instance, the simplified expression has no restrictions on the two variables $x$ and $y$ (it is defined for any two real numbers), however the original expression has some discontinuities (look at divisions by zero), and at precisely these points, it is not defined (and since the simplified expression is, they are not equivalent!). In order to make the statement "(3x)2/4xy * 16y/36x = 3x-1y" true, you have to impose restrictions on the variables $x$ and $y$ such that:

- if one expression is defined at $(x, y)$, then the other is also defined at $(x, y)$ and they are equal at that point
- if one expression is not defined at $(x, y)$, then the other is not defined either

In this case, the simplified expression is defined everywhere, but the original expression is not defined as specific points $(x, y)$ (can you find them?), so you need to define the simplified expression to exclude these points, and then the equality will be true. If you're wondering how to write them down, this would do:

... = 3x - 1y for (x, y) not equal to [...]

(usually you don't have to list every single pair (x, y), you can often simplify them into a smaller set of equivalent restrictions)

However I should note that you made a mistake in your simplification. But just to be sure, did you mean:

$$3x \frac{2}{4xy} \times \frac{16y}{36x}$$

 

[mathdrama]

Please note that the original and the simplified expression are not the same! For instance, the simplified expression has no restrictions on the two variables $x$ and $y$ (it is defined for any two real numbers), however the original expression has some discontinuities (look at divisions by zero), and at precisely these points, it is not defined (and since the simplified expression is, they are not equivalent!). In order to make the statement "(3x)2/4xy * 16y/36x = 3x-1y" true, you have to impose restrictions on the variables $x$ and $y$ such that:

- if one expression is defined at $(x, y)$, then the other is also defined at $(x, y)$ and they are equal at that point
- if one expression is not defined at $(x, y)$, then the other is not defined either

In this case, the simplified expression is defined everywhere, but the original expression is not defined as specific points $(x, y)$ (can you find them?), so you need to define the simplified expression to exclude these points, and then the equality will be true. If you're wondering how to write them down, this would do:

... = 3x - 1y for (x, y) not equal to [...]

(usually you don't have to list every single pair (x, y), you can often simplify them into a smaller set of equivalent restrictions)

However I should note that you made a mistake in your simplification. But just to be sure, did you mean:

$$3x \frac{2}{4xy} \times \frac{16y}{36x}$$

Sorry, I'm not sure how to describe the expression on this forum. Does this link help? http://i57.tinypic.com/t8105t.png

"Please note that the original and the simplified expression are not the same! " When you say this, is this because there was a mistake in the simplification, or is that how it works?

 

[Nono713]

Sorry, I'm not sure how to describe the expression on this forum. Does this link help? http://i57.tinypic.com/t8105t.png

[snip]

"Please note that the original and the simplified expression are not the same! " When you say this, is this because there was a mistake in the simplification, or is that how it works?

Ah, ok, with the $3x$ raised to a power it would make a difference. Haven't checked, but I'll assume the expression is right.

The simplification you gave in your opening post is technically "wrong", in the sense that the two are not equivalent, as I've said above. However they are "almost" the same, they only differ at a few points (due to you implicitly making assumptions on the possible values of $x$ and $y$ while manipulating the expression, e.g. division by zero) which is exactly why you need to put restrictions on the variables to make it correct. Once you add the restrictions, the original and simplified expression are the same, which is the point of the exercise :p

 

[I like Serena]

Simplify and state the restrictions on the variables.
a) (3x)2/4xy * 16y/36x

I got as far as
= 12xy2/36x2y
= Xy2/3x2y
= 3x-1y

Welcome to MHB, mathdrama!

Let's see...

$$ \frac{(3x)^2}{4xy} \cdot \frac{16y}{36x} $$

$$ =\frac{3^2x^2 \cdot 16y}{ 4xy \cdot 36x} $$

$$ =\frac{9x^2 \cdot \cancel 4 \cdot 4y}{\cancel 4xy \cdot 36x} $$

$$ =\frac{9x^2 \cdot 4y}{xy \cdot 36x} $$

$$ =\frac{36x^2y}{36x^2y} $$

Hmm... that does not look the same as what you have...

 

[mathdrama]

Welcome to MHB, mathdrama!

Let's see...

$$ \frac{(3x)^2}{4xy} \cdot \frac{16y}{36x} $$

$$ =\frac{3^2x^2 \cdot 16y}{ 4xy \cdot 36x} $$

$$ =\frac{9x^2 \cdot \cancel 4 \cdot 4y}{\cancel 4xy \cdot 36x} $$

$$ =\frac{9x^2 \cdot 4y}{xy \cdot 36x} $$

$$ =\frac{36x^2y}{36x^2y} $$

Hmm... that does not look the same as what you have...

Thank you. To make sure I understand this, the way to find the restrictions is to make the common denominator = 0, right?

In the simplified expression, I would have to make 36x^2y = 0, right?

The thing is, I don't know how to factor 36x^2y.

Is it (x + 6)(x + 6)y = 0

 

[MarkFL]

In general, you want to look at the original expression, at least before any factors are divided out or singularities removed to determine the restrictions.

The original form is:

$$\frac{(3x)^2}{4xy} \cdot \frac{16y}{36x}$$

we could first find the product:

$$\frac{144x^2y}{144x^2y}$$

Now, before you reduce, look at what can cause the denominator to be zero. When the denominator is a set of factors, then we can use the zero-factor property, which essentially states that when we equate the product of some factors to zero, then if anyone of those factors is zero, the product is zero, since zero times any finite value is zero.

We know 144 can never be zero, so we look at:

$$x^2=0\implies x=0$$

$$y=0$$

So, we cannot have those values for $x$ and $y$. Now we may simplify:

$$\frac{\bcancel{144x^2y}}{\bcancel{144x^2y}}=1$$

Hence, the original expression is equivalent to 1 where $x\ne0$ and $y\ne0$.

 

[I like Serena]

Thank you. To make sure I understand this, the way to find the restrictions is to make the common denominator = 0, right?

Yes.

In the simplified expression, I would have to make 36x^2y = 0, right?

As Mark already remarked, no, you have to check the original expression.
Each individual denominator that can be zero causes a restriction.

The thing is, I don't know how to factor 36x^2y.

Is it (x + 6)(x + 6)y = 0

It is already factorized.
More specifically:
$$36x^2y = 36 \cdot x^2 \cdot y = 36 \cdot x \cdot x \cdot y$$

 

[mathdrama]

In general, you want to look at the original expression, at least before any factors are divided out or singularities removed to determine the restrictions.

The original form is:

$$\frac{(3x)^2}{4xy} \cdot \frac{16y}{36x}$$

we could first find the product:

$$\frac{144x^2y}{144x^2y}$$

Now, before you reduce, look at what can cause the denominator to be zero. When the denominator is a set of factors, then we can use the zero-factor property, which essentially states that when we equate the product of some factors to zero, then if anyone of those factors is zero, the product is zero, since zero times any finite value is zero.

We know 144 can never be zero, so we look at:

$$x^2=0\implies x=0$$

$$y=0$$

So, we cannot have those values for $x$ and $y$. Now we may simplify:

$$\frac{\bcancel{144x^2y}}{\bcancel{144x^2y}}=1$$

Hence, the original expression is equivalent to 1 where $x\ne0$ and $y\ne0$.

Oh, now I finally understand.

Thank you very much kind MHB folks.