Mathematics
posted by David .
Find a fraction which reduces to 3/4 when the numerator and denominator are each decreased by 1, and which reduces to 4/5 when the numerator and denominator are each increased by 1.

Let the original fraction be x/y
equation #1 : (x1)/(y1) = 3/4
4x  4 = 3y  3
4x  3y = 1
equation #2 : (x+1)/*(y+1) = 4/5
5x + 5 = 4y + 4
5x  4y = 1
#1 times 4 :16x  12y = 4
#2 times 3 : 15x  12y = 3
subtract them ...
x = 7
back in #1
28  3y = 1
3y = 27
y = 9
the fraction was 7/9
check:
6/8 = 3/4 ✔
8/10 = 4/5 ✔
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