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Question 231058: Assume that the denominator x of a fraction is greater than the numerator, 2x-5. Find all such fractions for which both the numerator and the denominator are positive integers.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! numerator is 2x-5
denominator is x
your fraction is 2x-5/x
you have:
2x - 5 >= 0 (equation 1)
and you have:
x > 2x-5 (equation 2)
You need to find a common solution to both of these equations.
Solve equation 1 first.
2x - 5 >= 0 (equation 1)
Add 5 to both sides to get:
2x >= 5
Divide both sides by 2 to get:
x >= 5/2
Since x has to be an integer, this makes:
x >= 3 (the next highest integer).
Next solve equation 2.
x > 2x-5 (equation 2)
Add 5 to both sides and subtract x from both sides to get:
5 > x
This is the same as:
x < 5
We have:
x >= 3
and:
x < 5
This means that x can be 3 or 4 only.
We test this out to confirm.
We take values of x = 2,3,4,5 and solve the original equation:
When x = 2, (2x-5)/x = -1/2.
Since this is a negative number, this answer is NOT good which is ok since it's out of the limits we set up.
When x = 3, (2x-5)/x = 1/3
Numerator is less than denominator and both numerator and denominator are positive so this IS good which is ok since it's within the limits we set up.
When x = 4, (2x-5)/x = 3/4
Numerator is less than denominator and both numerator and denominator are positive so this IS good which is ok since it's within the limits we set up.
When x = 5, (2x-5)/x = 5/5
Numerator is NOT less than denominator so this is NOT good which is ok since it's outside of the limits we set up.
We'll take two more just to confirm.
When x = 1, (2x-5)/x = -3/1 which is NOT good as expected.
When x = 9, (2x-5)/x = 13/9 which is NOT good as expected.
Looks like your answer is:
x = 3 and 4 only.
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