Question 205861
The numerator of a fraction is one more than the denominator.
 If the numerator and the denominator are both increased by 2, the new fraction
 will be one fourth less than the original fraction.
 What is the original fraction?
:
Looks like you are on the right track, here's how I would do it
:
Let x = the denominator
"The numerator of a fraction is one more than the denominator." therefore:
(x+1) = the numerator
;
" If the numerator and the denominator are both increased by 2, the new fraction will be one fourth less than the original fraction."
{{{(x+1+2)/(x+2)}}} = {{{(x+1)/x}}} - {{{1/4}}}
:
{{{(x+3)/(x+2)}}} = {{{(x+1)/x}}} - {{{1/4}}}
Multiply equation by 4x(x+2)
4x(x+2)*{{{(x+3)/(x+2)}}} = 4x(x+2)*{{{(x+1)/x}}} - 4x(x+2)*{{{1/4}}}
Cancel the denominators:
4x(x+3) = 4(x+2)(x+1) - x(x+2)
:
4x^2 + 12x = 4(x^2 + 3x + 2) - x^2 - 2x
:
4x^2 + 12x = 4x^2 + 12x + 8 - x^2 - 2x
Arrange as a quadratic equation on the left
4x^2 - 4x^2 + x^2 + 12x - 12x + 2x - 8 = 0
:
x^2 + 2x - 8 = 0
Factors to:
(x+4)(x-2) = 0
Positive solution
x = 2 is the denominator
then
2 + 1 = 3 is the numerator
:
 What is the original fraction? {{{3/2}}} 
:
:
Check solution in the statement:
if the numerator and the denominator are both increased by 2, the new fraction will be one fourth less than the original fraction."
{{{5/4}}} = {{{3/2}}} - {{{1/4}}}
{{{5/4}}} = {{{6/4}}} - {{{1/4}}}