SOLUTION: here is my question.
The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more
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Question 238478: here is my question.
The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fraction. Find the original fraction.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The numerator of a fraction is 1 less than the denominator.
If the numerator and the denominator are both increased by 4,
the new fraction will be 1/8 more than the original fraction.
Find the original fraction.
;
Let x = the numerator
then
(x+1) = the denominator
:
= +
which is:
= +
Multiply equation by 8(x+5)(x+1), results:
:
8(x+1)(x+4) = 8(x+5)*x + (x+1)(x+5)
FOIL
8(x^2 + 5x + 4) = 8(x^2 + 5x) + x^2 + 6x + 5
:
8x^2 + 40x + 32 = 8x^2 + 40x + x^2 + 6x + 5
:
Group like terms on the left
8x^2 - 8x^2 - x^2 + 40x - 40x - 6x + 32 - 5 = 0
:
-x^2 - 6x + 27 = 0
Multiply by -1, change the signs
x^2 + 6x - 27 = 0
Factor
(x+9)(x-3) = 0
Two solutions
x = -9
x= +3
:
The original fraction:
:
Check solution
= +
= +
;
:
But what about the other solution x=-9, that fraction = = is the fraction then
:
= +
= +
Common denominator of 24
= + ; this solution does not work
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