How to these cancel down

T

thickmax

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May 6, 2021
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I am working through some resolutions that I am just not understanding,

? /1 − 2? = 2/2? 1 − ? works out to ?(1 − ?) = 4?(1 − 2?)


but I cannot figure this out,

I think I am missing a method here, but without knowing what it's called, I cannot teach myself it.

Please help me understand how this cancels, or tell me what this process is called
 
thickmax said:
I am working through some resolutions that I am just not understanding,

? /1 − 2? = 2/2? 1 − ? works out to ?(1 − ?) = 4?(1 − 2?)


but I cannot figure this out,

I think I am missing a method here, but without knowing what it's called, I cannot teach myself it.

Please help me understand how this cancels, or tell me what this process is called
Click to expand...
I cannot figure-out the first statement. As posted - it does not make sense. You need to use parentheses () to indicate grouping. A photo of the original assignment will be very helpful.
 
your syntax is not clear

is x/1 - 2x supposed to be [MATH]\frac{x}{1-2x}[/MATH] or something else?

… please clarify your expressions
 
Subhotosh Khan said:
I cannot figure-out the first statement. As posted - it does not make sense. You need to use parentheses () to indicate grouping. A photo of the original assignment will be very helpful.
Click to expand...
I disagree with the student showing a photo. I understand that it would help the helpers know what the problem is, but I feel that the student needs to know how to write the problem. That is, I think that the 1st step in helping the student is for the student to know how to write the problem. Seriously, this is how I feel and it is not about the friendly war which we have.
 
It is clear that the op does not know that the 2 on the right side is being multiplied by what comes after that 2.

Can you please tell me what you would get for [math]2(\dfrac{5}{7})[/math]? Thank you.
 
2 x 5/7 =10/7

i get the 1-2*x=1-2x


Thanks for the reply, but I'm still struggling to understand what you are trying to tell me with the common denominator...


The mistake was not putting an extra 2, after the first 2. Simply only seeing a 2, but no counting for all of the 2's. If the second 2 was a different number, I'd like to have thought I would have detected my mistake.
 
Try to factor 1-2x (you can't so you have 1-2x)
Try to factor 1-x (you can't so you have 1-x)
Now we try to get the common denominator.
Did you write down 1-2x yet (for the common denominator)? Of course not, you haven't written down anything yet for the common denominator. So write down (1-2x). Look at the next factor, (1-x). Have you written it down yet? If yes, then do not write it down (next to (1-2x)). If no, then write it down next to (1-2x).

Since we do not have a factor of (1-x) written down yet, we write it down next to (1-2x). Since there are no other factors we are done.
The denominator we want is (1-2x)(1-x)
 
Jomo said:
Try to factor 1-2x (you can't so you have 1-2x)
Try to factor 1-x (you can't so you have 1-x)
Now we try to get the common denominator.
Did you write down 1-2x yet (for the common denominator)? Of course not, you haven't written down anything yet for the common denominator. So write down (1-2x). Look at the next factor, (1-x). Have you written it down yet? If yes, then do not write it down (next to (1-2x)). If no, then write it down next to (1-2x).

Since we do not have a factor of (1-x) written down yet, we write it down next to (1-2x). Since there are no other factors we are done.
The denominator we want is (1-2x)(1-x)
Click to expand...
Ok, think I get the factoring now, thank you. But how do the bracketed terms swap equations?
 
Jomo said:
I disagree with the student showing a photo. I understand that it would help the helpers know what the problem is, but I feel that the student needs to know how to write the problem. That is, I think that the 1st step in helping the student is for the student to know how to write the problem. Seriously, this is how I feel and it is not about the friendly war which we have.
Click to expand...
I totally agree with "the student to know how to write the problem". However the way the problem was stated, I could not figure out anything about the "format" of the problem.
 
thickmax said:
Ok, think I get the factoring now, thank you. But how do the bracketed terms swap equations?
Click to expand...
There are two ways to think of it. One is called "cross multiplication", which says that the equation `a/b=c/d` is equivalent (if b and d are nonzero) to `ad=bc`. Have you heard of that?

The more general way is to multiply both sides of the equation by the common denominator. In the case of `a/b=c/d`, that is `bd`, and multiplying makes `a/b (bd)/1=c/d (db)/1`, which cancels to leave `ad=bc`.
 
I can understand
a/b=c/d is equivalent to ad=bc

but i still do not get the last bit you have said.

Can you write it before you've cancelled it please?
 
thickmax said:
I can understand
a/b=c/d is equivalent to ad=bc

but i still do not get the last bit you have said.

Can you write it before you've cancelled it please?
Click to expand...

But that was the last bit I said. I'm not sure what you don't get.

Here's the whole process:

[MATH]\frac{a}{b}=\frac{c}{d}[/MATH]​
​
[MATH]\frac{a}{b}\cdot\frac{bd}{1}=\frac{c}{d}\cdot\frac{bd}{1}[/MATH]​
​
[MATH]\frac{a}{\cancel{b}}\cdot\frac{\cancel{b}d}{1}=\frac{c}{\cancel{d}}\cdot\frac{b\cancel{d}}{1}[/MATH]​
​
[MATH]\frac{a}{1}\cdot\frac{d}{1}=\frac{c}{1}\cdot\frac{b}{1}[/MATH]​
​
[MATH]\frac{ad}{1}=\frac{bc}{1}[/MATH]​
​
[MATH]ad=bc[/MATH]​
 
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