My Math Forum multiply the fractions and make the answer as easy as possible (almost done)

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 May 20th, 2015, 05:01 AM #1 Member   Joined: Jan 2015 From: Norway Posts: 95 Thanks: 1 multiply the fractions and make the answer as easy as possible (almost done) 1.32 D) a^2-b^2/2x^2-2y^2 * x+y/3a-3b = a^2-b^2*x+y/2x^2-2y^2*3a*3b What happens now is that you shorten the fraction, you shorten a and b to remove the 3a and 3b to be 3 and 3. You shorten the fraction of x and y to remove them and make 2x^2 - 2y^2 to be 2x - 2y. What happens now is this a-b/2x-2y*3*3 What you do now is multiply 2x with 3 and multiply 2y with 3 = a+b/2*3*x-2*3*y Making it = a+b/6x-6y. To end it with a+b/6(x-y). I am done with the task, but there are one thing I don't understand. Why and how did a-b change into a+b? After I know this, I understand the whole task. Last edited by skipjack; May 20th, 2015 at 12:19 PM.
 May 20th, 2015, 10:01 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 You've been told a few times to use BRACKETS when necessary; do you not understand what that means? For example, your "a^2-b^2/2x^2-2y^2 * x+y/3a-3b" should be: (a^2 - b^2) / (2x^2 - 2y^2) * (x + y) / (3a-3b) Make SURE you understand why...else you'll keep being in trouble... The (a - b) did NOT change into (a + b). What happened is the (a - b)'s all cancelled out, but the (a + b) didn't.
May 21st, 2015, 02:18 AM   #3
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Joined: Jan 2015
From: Norway

Posts: 95
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Quote:
 Originally Posted by Denis You've been told a few times to use BRACKETS when necessary; do you not understand what that means? For example, your "a^2-b^2/2x^2-2y^2 * x+y/3a-3b" should be: (a^2 - b^2) / (2x^2 - 2y^2) * (x + y) / (3a-3b) Make SURE you understand why...else you'll keep being in trouble... The (a - b) did NOT change into (a + b). What happened is the (a - b)'s all cancelled out, but the (a + b) didn't.
ah, thanks, i understand brackets now.

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