Question 683653
x is numerator
y is denominator
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y = 2x + 1
{{{(x + 4)/(y-4) = y/x}}}
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Cross Multiply
x(x+4) = y(y-4)
{{{x^2 + 4x = y^2 - 4y}}}
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{{{x^2 + 4x - y^2 + 4y  = 0}}}
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{{{x^2 + 4x -(2x+1)^2 + 4(2x+1) = 0}}} ==> substitution
{{{x^2 + 4x -(4x^2 + 4x + 1) + 8x + 4 = 0}}}
{{{x^2 + 4x - 4x^2 - 4x - 1 + 8x + 4 = 0}}}
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{{{-3x^2 + 8x + 3 = 0}}}
{{{-1(3x^2 - 8x - 3)=0}}}
(3x+1)(x-3)=0
3x+1=0
3x=-1
x = {{{-1/3}}}
x-3=0
x = 3
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Using x = {{{-1/3}}}
y = {{{2x+1 = 2(-1/3) + 1 = -2/3 + 1 = -2/3 + 3/3 = 1/3}}}
Original fraction
{{{highlight_green((-1/3)/(1/3))}}} which is {{{(-1/3)(3) = -1}}}
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Using x = 3
y = 2x + 1 = 2(3) + 1 = 6 +1 = 7
Original fraction
{{{highlight_green(3/7)}}}
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Testing both originals
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Checking {{{(-1/3)/(1/3)}}} which is -1
{{{(-1+4)/(1-4) = 3/-3 = -1}}}
it works! {{{-1/1}}} and {{{1/-1}}} are reciprocals
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checking {{{3/7}}}
{{{(3+4)/(7-4) = 7/3}}}
so
{{{3/7}}} and {{{7/3}}} are reciprocals
it is correct!
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Ultimately, there are two original fractions that qualify
{{{(-1/3)/(1/3)}}} and {{{3/7}}}
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