Math Help - Reducing Algebra Fractions by -1

1. Reducing Algebra Fractions by -1

The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$

1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\frac{y-x}{-(-x+y)}$

2.Here $\frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\frac{y-x}{-(y-x)}$

3.Next we have $\frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.

Does my logic up to 3's question look correct also?

2. Hi
Originally Posted by cmf0106
The book is trying to show me that reducing a fraction or adding two fractions sometimes only requires that -1 be factored from one or more denominators. Here is the example they give:
$\frac{y-x}{x-y}=\frac{y-x}{-(-x+y)}=\frac{y-x}{-(y-x)}=\frac{1}{-1}=-1$

1.In the first step here I understand factoring out the "-" (the -1 value) because if the variable does not have a number coefficient we can assume its one so this part is fine.
$\frac{y-x}{-(-x+y)}$

2.Here $\frac{y-x}{-(-x+y)}$
once we distribute the "-" we get a negative y and a positive x, therefore we can rearrange the denominator to match the numerator in $\frac{y-x}{-(y-x)}$
What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x.

3.Next we have $\frac{1}{-1}$ Now here is where I have a problem how can we know the values of both of these are 1 and -1? I am not certain how this is accomplished.
Relevant question for someone who begins ^^ It's so good seeing you trying to understand !

We have $\frac{y-x}{-(y-x)}$. This can be rewritten this way :

$\frac{1*{\color{red}(y-x)}}{(-1)*{\color{red}(y-x)}}$ and hence the result. Does it look correct to you ?

When you'll advance into maths lessons, maybe you will learn that most of these steps are only here to guide you to the solution

3. Thanks Moo! I will chew on that for awhile and let you know what the outcome is, I think it makes sense let me try a few more practice problems and we will see.

as for

"What you said for 1. is not wrong. But I'd say they made these two transformations to show you that there is the common factor y-x.
But what you said in 2. looks incorrect to me. They don't distribute - again, they just use the commutativity of the addition : a+b=b+a. Here : -x+y=y+(-x)=y-x."

Yah sorry I didnt mean to phrase it that way, it wouldnt make sense anyways because we factor out the "-", if we factored it out one step and then the next we redistributed it that would defeat the purpose of factoring

4. Alright moo update, this stuff is really putting me in a rut.

$\frac{3}{y-x}+\frac{x}{x-y}$

1.The book solves it $\frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-y+x)}+\frac{x}{x-y}=\frac{3}{-(x-y)}+\frac{x}{x-y}=\frac{-3}{x-y}+\frac{x}{x-y}=\frac{-3+x}{x-y}$

2. I started solving it like this, which looks different from the book $\frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$

I used $\frac{3}{-(-x+y)}$ so it would yield the other fractions denominator $\frac{x}{x-y}$

Here $\frac{3+x}{(x-y)(x-y)}$ the two $(x-y)$ can merge into one but Im not understanding why the 3 is supposed to be -3 here, this is assuming my way of solving for it is correct in the first place.

5. Originally Posted by cmf0106
Alright moo update, this stuff is really putting me in a rut.

$\frac{3}{y-x}+\frac{x}{x-y}$

2. I started solving it like this, which looks different from the book $\frac{3}{y-x}+\frac{x}{x-y}=\frac{3}{-(-x+y)}+\frac{x}{x-y}=\frac{3+x}{(x-y)(x-y)}$
First problem in the first equality : y-x=-x+y, it's not equal to -(-x+y)

The purpose of this is to make appear the other fraction's denominator x-y. You can see that -(x-y)=-x+y=y-x.
So $\frac{3}{y-x}=\frac{3}{-(x-y)}$

Now if you want to know why it becomes $\frac{-3}{x-y}$ :

$\frac{3}{-(x-y)}=\frac{3}{-1} \cdot \frac{1}{x-y}=-3 \cdot \frac{1}{x-y}=\frac{-3}{x-y}$

See ?

Now, there is also a problem with the second equality. You should remember that :
$\frac ac+\frac bc=\frac{a+b}{c}$ and there is no multiplication involved. The multiplication occurs if there is not the same denominator

6. But I could chose to make either both denominators appear as $y-x$ or $x-y$ correct?

7. Originally Posted by cmf0106
But I could chose to make either both denominators appear as $y-x$ or $x-y$ correct?
Correct

It'd be nice if you could show how you would have done for y-x (unless you have many more exercises ^^)

8. Originally Posted by Moo
Correct

It'd be nice if you could show how you would have done for y-x (unless you have many more exercises ^^)
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot

so $\frac{3}{y-x}+\frac{x}{x-y}$

Lets Change $\frac{3}{y-x}=-(y-x)=-y+x$ this equals our other denominator $x-y$

next $=\frac{3}{-(y-x)}=\frac{-3+x}{x-y}$

Probably wrong lol.

9. Originally Posted by cmf0106
Man this is probably the hardest thing Ive encountered so far, hard for me to wrap my brain around. Anyways lets give the other denominator a shot

so $\frac{3}{y-x}+\frac{x}{x-y}$

Lets Change ${\color{red}\frac{3}{y-x}=\frac{3}{-(y-x)}=\frac{3}{-y+x}}$ this equals our other denominator $x-y$

next $=\frac{3}{-(y-x)}=\frac{-3+x}{x-y}={\color{red}\frac{x-3}{x-y}}$

Probably wrong lol.
Let's clean this up just a little bit so that your equalities make sense, and your conclusion looks fine. Great job!