Fairly basic algebra question - but it's bugging me now

  • #1
kel
62
0
Hi,
I'm just going through some notes on linear interpolation, but this bit has got me confused - I think it's an easy work through on gradients etc, but my maths seems to be failing me (mature student here so go easy on me :shy: )

Ok here goes (see attached images - not sure about latex yet. A look at image 2 first, it just shows the gradient equation)

My question is: How can e.g. (x2-x) become (x-x2), I was always taught that when you move the lower part of a division/fraction to the other side of the equals sign, you don't change it! So, how does this work?

Is there some trick that occurs when you rearrange an equation or is there some maths rule that I was never taught.

I know this is probably a solve in 2 seconds type question, but it's starting to bug me now !

Cheer
Kel
 

Attachments

  • Image2.gif
    Image2.gif
    966 bytes · Views: 535
  • Image1.gif
    Image1.gif
    4.3 KB · Views: 585
Physics news on Phys.org
  • #2
Do not post your questions in the tutorial section!
A mentor will move this thread to where it belongs.
 
  • #3
Whoops! Sorry, forgot which area I was in. o:)

Please moderator (any one of you) please could you move this to the relevant area.

Thankyou!
 
  • #4
kel said:
Hi,
I'm just going through some notes on linear interpolation, but this bit has got me confused - I think it's an easy work through on gradients etc, but my maths seems to be failing me (mature student here so go easy on me :shy: )

Ok here goes (see attached images - not sure about latex yet. A look at image 2 first, it just shows the gradient equation)

My question is: How can e.g. (x2-x) become (x-x2), I was always taught that when you move the lower part of a division/fraction to the other side of the equals sign, you don't change it! So, how does this work?

Is there some trick that occurs when you rearrange an equation or is there some maths rule that I was never taught.
Well, there is NO (x2-x) that becomes (x-x2)! You mean "How can (x2-x1) become (x1-x2).
You have
[tex]y= \frac{y_2-y_1}{x_2-x_1}(x- x_1)+ y_2[/tex]
and it becomes
[tex]y= y_1\frac{x-x_1}{x_1-x_2}+ y_2\frac{x-x_1}{x_2-x_1}[/tex]
Actually that is wrong but not because of the switch of (x2-x1) to (x1-x2).
What you do have is
[tex]\frac{y_2-y_1}{x_2-x_1}(x- x_1)= y2\frac{x-x_1}{x_2-x_1}- y_1\frac{x-x_1}{x_2-x_1}[/itex]
Do you see that it is negative y1 times the fraction? Take that negative into the denominator of the fraction:
[tex]-y_1\frac{x-x_1}{x_2-x_1}= y_1\frac{x-x_1}{x_1-x_2}[/tex]

What is wrong is that the last "y2[/sup]" has been dropped! did you mean
[tex]y- y_2= y_1\frac{x-x_1}{x_1-x_2}+ y_2\frac{x-x_1}{x_2-x_1}[/tex]
[tex]y- y_2= y_1A_1(x)+ y_2A_2(x)[/tex]?


I know this is probably a solve in 2 seconds type question, but it's starting to bug me now !

Cheer
Kel
 
  • #5
Actually, I just lifted it straight from the course text (pdf file). So any errors will be in the original teaching notes.

Cheers
kel
 
  • #6
So basically I can take the minus sign down to the denominator, but then I have to switch the numbers in the demominator to keep the value of the overall multiplication the same ie -2(1/3-1) = 2(1/1-3), I worked these both to be -1, so I guess it works.

Can I also ask, how does this work when you have a long denominator ie -2(1/(2+3-5)*2) can you still move the minus sign down? do I just have to change the +'s to -'s or would I have to rearrange.

Thanks again
Kel
 
  • #7
kel said:
So basically I can take the minus sign down to the denominator, but then I have to switch the numbers in the demominator to keep the value of the overall multiplication the same ie -2(1/3-1) = 2(1/1-3), I worked these both to be -1, so I guess it works.

Can I also ask, how does this work when you have a long denominator ie -2(1/(2+3-5)*2) can you still move the minus sign down? do I just have to change the +'s to -'s or would I have to rearrange.

Thanks again
Kel
You can "move" a minus sign wherever you want:
[tex]-\frac{a}{b}= \frac{-a}{b}= \frac{a}{-b}[/tex]
In the example you give
[tex]-2\frac{1}{(2+ 3-5)^2}= \frac{2}{-(2+3-5)^2}= \frac{-2}{(2+3-5)^2}[/itex]
Of course, sinced 2+3-5= 0 that fraction doesn't exist wherever the "-" is! (Were you trying to catch me?)
 
  • #8
Nope, just made that question up on the spot.

Thanks for your help
Kel
 

Similar threads

Replies
4
Views
2K
Replies
16
Views
2K
Replies
5
Views
7K
Replies
13
Views
3K
Replies
13
Views
9K
Replies
1
Views
1K
Replies
11
Views
2K
Replies
4
Views
1K
Back
Top