Asked 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.
Answers
Answered by
Reiny
Let the original fraction be x/y
equation #1 : (x-1)/(y-1) = 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 ✔
equation #1 : (x-1)/(y-1) = 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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