Agree, but math isn't high tech. Most people understand when you want to find the answer to a problem, certain basic steps should be followed. Let me give you an example:
[(3x^2-27)divided by 4)] times[ 8x^2) divided by(9-3x)]divided by [(x^2+3x) divided by 6]
Ok, to find the answer we can invert the third fraction then multiply straight across like this
(3x^2-27)(8x^2) 6
-------------------------------------
4(9-3x) (x^2+3x)
but first we can factor and cross cancel
(3(x-3)(x+3)(8x^2) 6
--------------------------------------...
-12(x-3) x (x+3)
after we cross cancel we get
=-12x
Or, to make it simpler, we can start by inverting the last fraction and multiplying:
3x² - 27 .. 8x² ....... 6
----------- * -------- * ----------
.... 4 ...... 9 - 3x . x² + 3x
Now you can do some factoring and canceling:
3(x² - 9) .. 8x² ...... 6
----------- * -------- * ----------
.... 4 ..... 3(3 - x) . x(x + 3)
You can cancel the 4 with the 8, remove the 3's and the x's.
(x² - 9) .... 2x ........ 6
----------- * -------- * ----------
.... 1 ...... (3 - x) . (x + 3)
Now x²-9 is a difference of squares which can be rewritten as (x - 3)(x + 3):
(x-3)(x+3) .... 2x ........ 6
-------------- * -------- * ----------
....... 1 ...... (3 - x) . (x + 3)
Cancel the x+3:
(x - 3) .. 2x .... 6
------- * -------- * ---
...1 ... (3 - x) .. 1
Now notice if you factor out a -1, you have 3-x which will cancel:
-1 .. 2x .. 6
--- * ---- * ---
.1 ... 1 ... 1
Multiply across and you get:
-12x
So you see, there isn't really anything high tech about it. It's just math and it's just plain fun.![]()
It's a 'nothing' thread.
Agree, but math isn't high tech. Most people understand when you want to find the answer to a problem, certain basic steps should be followed. Let me give you an example:
[(3x^2-27)divided by 4)] times[ 8x^2) divided by(9-3x)]divided by [(x^2+3x) divided by 6]
Ok, to find the answer we can invert the third fraction then multiply straight across like this
(3x^2-27)(8x^2) 6
-------------------------------------
4(9-3x) (x^2+3x)
but first we can factor and cross cancel
(3(x-3)(x+3)(8x^2) 6
--------------------------------------...
-12(x-3) x (x+3)
after we cross cancel we get
=-12x
Or, to make it simpler, we can start by inverting the last fraction and multiplying:
3x² - 27 .. 8x² ....... 6
----------- * -------- * ----------
.... 4 ...... 9 - 3x . x² + 3x
Now you can do some factoring and canceling:
3(x² - 9) .. 8x² ...... 6
----------- * -------- * ----------
.... 4 ..... 3(3 - x) . x(x + 3)
You can cancel the 4 with the 8, remove the 3's and the x's.
(x² - 9) .... 2x ........ 6
----------- * -------- * ----------
.... 1 ...... (3 - x) . (x + 3)
Now x²-9 is a difference of squares which can be rewritten as (x - 3)(x + 3):
(x-3)(x+3) .... 2x ........ 6
-------------- * -------- * ----------
....... 1 ...... (3 - x) . (x + 3)
Cancel the x+3:
(x - 3) .. 2x .... 6
------- * -------- * ---
...1 ... (3 - x) .. 1
Now notice if you factor out a -1, you have 3-x which will cancel:
-1 .. 2x .. 6
--- * ---- * ---
.1 ... 1 ... 1
Multiply across and you get:
-12x
So you see, there isn't really anything high tech about it. It's just math and it's just plain fun.![]()