Here’s a little maths trick I have for some reason carried around in my brain for years:
Dividing through by the common
Why this is possible is left as an excercise for the reader 🙂
Filed under geekery
Line 3: (a^2)-(b^2) = ab-(b^2)
How do you get:
(a+b)(a-b) = b(a-b)
From line 3? If you divide ‘b’ from the right side of the equation, you must do the same on the left, giving you:
( (a^2)-(b^2) /b) = b(a-b)
simplified: (a^2)/b) – b = a – b
Eg replacing with #’s
4/2 – 2 = 2 – 2
Was the answer faulty math? XD
Had to check wikipedia for this one
very interesting though
Ah-hah – so that’s what they call it 🙂
on line 3, I rewrote both sides of the equation, but didn’t actually perform any calculations.
On the left, (a^2)-(b^2) can be written as (a+b)(a-b), since if you multiply that out you get (a^2) -ab + ab -(b^2). -ab + ab is 0, so you get (a^2)-(b^2)
On the right, ab-(b^2) has b common to both ab and b^2, so the b can be factored out and the line written as b(a-b). If you multiply b(a-b) out you get ab – (b^2).
In both cases the lines are rewritten in a different form without any calculation being done.
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