Question 238478
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
:
{{{(x+4)/((x+1)+4)}}} = {{{x/(x+1)}}} + {{{1/8}}}
which is:
{{{(x+4)/(x+5)}}} = {{{x/(x+1)}}} + {{{1/8}}}
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: {{{3/4}}}
:
Check solution
{{{(3+4)/(4+4)}}} = {{{3/4}}} + {{{1/8}}}
{{{7/8}}} = {{{6/8}}} + {{{1/8}}}
;
:
But what about the other solution x=-9, that fraction = {{{(-9)/(-8)}}} = {{{9/8}}} is the fraction then
:
{{{(9+4)/(8+4)}}} = {{{9/8)}}} + {{{1/8}}}
{{{13/12}}} = {{{9/8}}} + {{{1/8}}}
Common denominator of 24
{{{26/24}}} = {{{27/24)}}} + {{{3/24}}}; this solution does not work