The numberator of a fraction exceeds the denominator by 3. If 3 is added to the numerator and 3 is subtracted from the denominator, the resulting fraction is equal to 3/2. Find the original fraction.
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Previous response:
http://www.jiskha.com/display.cgi?id=1313109184
13/17
To solve this problem, let's assume that the original fraction is x/y, where x is the numerator and y is the denominator.
We are given two conditions:
1. The numerator (x) exceeds the denominator (y) by 3:
x = y + 3
2. When 3 is added to the numerator (x + 3) and 3 is subtracted from the denominator (y - 3), the resulting fraction is equal to 3/2:
(x + 3) / (y - 3) = 3/2
Now we will substitute the value of x from the first condition into the second condition:
(y + 3 + 3) / (y - 3) = 3/2
(y + 6) / (y - 3) = 3/2
Now, cross-multiply to get rid of the fractions:
2(y + 6) = 3(y - 3)
2y + 12 = 3y - 9
Simplify the equation:
12 + 9 = 3y - 2y
21 = y
Now that we know the value of y, we can substitute it back into the first condition to find x:
x = y + 3
x = 21 + 3
x = 24
Therefore, the original fraction is 24/21, which can be simplified as 8/7.