Question 93084
Given:
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{{{(2x-1)/(x+3) - (1-x)/(x+3)}}}
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To perform the operations and simplify completely you first notice that both fractions have
the denominator {{{(x+3)}}} in common. Therefore their numerators can be written over the
single common denominator to get:
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{{{((2x-1)-(1-x))/(x+3)}}}
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In the numerator the first quantity in parentheses is (2x-1). Since its set of parentheses 
is not preceded by a minus sign, you can just remove those parentheses without doing
anything to the terms inside. When you do the problem becomes:
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{{{(2x-1-(1-x))/(x+3)}}}
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Notice now the next set of parentheses in the numerator. The terms inside this set of
parentheses are +1 and -x. Since the parentheses are preceded by a minus sign, when you
remove the parentheses you change the sign of all the terms inside. When you do that the 
problem then becomes:
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{{{(2x-1-1+x)/(x+3)}}}
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Then you can combine the like terms in the numerator. The +2x and the +x combine to
give +3x and the -1 and -1 combine to give -2. This makes the problem:
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{{{(3x - 2)/(x+3)}}}
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That's the answer to the problem. There are no other cancellations or combinations
that can be made. If you like you can long divide x + 3 into 3x - 2 and the answer you
will get is:
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{{{3 -11/(x+3)}}}
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But that is the same as the answer:
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{{{(3x - 2)/(x+3)}}}
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It's just in a different form.
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Hope this helps you to understand what the problem is asking for you to do, and provides
you with some insight into how to do the problem too.