Dividing with AlgebraDate: 3/25/96 at 13:10:53 From: Carl Stromberg Subject: Math problem I have a big problem: I shall factorize and reduce this problem: (3x-9) / (2x-6)= and (x2+2xy) / (3x+6y)= I hope for answer! Sincerely. Carl Christian Stromberg, Norway Date: 3/26/96 at 15:56:19 From: Doctor Patrick Subject: Re: Math problem Hi Carl! When you want to solve a problem like this, you need to look for two things. The first thing to do is to is to look for numbers that will divide all the terms within one set of parentheses, and then a number that will divide the terms in the other set of parentheses. The second thing we need to do is to choose from the possible factors (if there are more than one) and find which one will give us a common factor in the parentheses. For example, in the first problem you asked about, (3x-9)/(2x-6), we can factor a 3 out of the first set of terms, since both 3x and 9 are divisible by three. This gives us 3(x-3). The second set of terms can be factored by 2, giving us 2(x-3). Do you see how that works? The problem now is (3(x-3))/(2(x-3)). Do you see any common term we can cancel out? How about (x-3)? If we divide the (x-3) in the numerator by the (x-3) in the denominator, we get 3/2, a much nicer number then what we started with. I'll let you try the second problem on your own. You can solve it in the same way. First find numbers that can be factored out of the original terms. Then divide out any common terms in the numerator and denominator. Good luck -Doctor Patrick, The Math Forum |
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