Question 136648
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.
:
Write an equation for what it says:
{{{x/((x+1))}}} + {{{1/8}}} = {{{((x+4))/((x+1+4))}}}
:
{{{x/((x+1))}}} + {{{1/8}}} = {{{((x+4))/((x+5))}}}
:
The common denominator will be 8(x+1)(x+5)
Multiply each term by this, cancel out the denominators, leaving us with:
x(8(x+5) + (x+1)(x+5) = 8(x+4)(x+1)
:
8x^2 + 40x + x^2 + 6x + 5 = 8(x^2 + 5x + 4)
:
8x^2 + 40x + x^2 + 6x + 5 = 8x^2 + 40x + 32
Combine on the left:
8x^2 + x^2 - 8x^2 + 40x + 6x - 40x + 5 - 32 = 0
A quadratic equation
x^2 + 6x - 27 = 0
Factors to:
(x+9)(x-3) = 0
We want the positive solution:
x = 3
:
Check solution in original equation
{{{3/((4))}}} + {{{1/8}}} = {{{((7))/((8))}}}
:
Actually the negative solution x = -9 will work also:
{{{(-9)/((-8))}}} + {{{1/8}}} = {{{((-5))/((-5))}}}
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Did I make this understandable? Any questions?