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Algebra

by heshamelaziry » Mon Nov 23, 2009 5:04 pm
2a/2a + 3b - (3b/2a + 3b -1) =


Note: 1 is subtracted from the preceding fraction


[spoiler]OA: 4a/2a + 3b[/spoiler]

[spoiler]I tried to solve the between parenthesis first and add this to the preceding fraction and got 0, which is wrong. Why ?[/spoiler]
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by Stuart@KaplanGMAT » Mon Nov 23, 2009 6:42 pm
heshamelaziry wrote:2a/2a + 3b - (3b/2a + 3b -1) =


Note: 1 is subtracted from the preceding fraction


[spoiler]OA: 4a/2a + 3b[/spoiler]

[spoiler]I tried to solve the between parenthesis first and add this to the preceding fraction and got 0, which is wrong. Why ?[/spoiler]
Assuming the original question is:

2a/(2a+3b) - (3b/(2a+3b) - 1)

= 2a/(2a+3b) - 3b/(2a+3b) + 1

= (2a-3b)/(2a+3b) + 1

= (2a-3b)/(2a+3b) + (2a+3b)/(2a+3b)

= ((2a-3b) + (2a+3b))/(2a+3b)

= 4a/(2a + 3b)
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