Question 864850
If the equation is {{{x/3 = 12/(x+5)}}}, then...



{{{x/3 = 12/(x+5)}}}



{{{x(x+5) = 3*12}}}



{{{x^2+5x = 36}}}



{{{x^2+5x-36=0}}}



{{{(x+9)(x-4)=0}}}



{{{x+9=0}}} or {{{x-4=0}}}



{{{x=-9}}} or {{{x=4}}}



If the equation is {{{x/3 = 12/(x+5)}}}, then the two solutions are {{{x=-9}}} or {{{x=4}}}

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If the equation is {{{x/3 = 12/x + 5}}}, then...



{{{x/3 = 12/x + 5}}}



{{{x/3-12/x = 5}}}



{{{(x^2)/(3x)-12/x = 5}}}



{{{(x^2)/(3x)-36/(3x) = 5}}}



{{{(x^2-36)/(3x) = 5}}}



{{{x^2-36 = 5(3x)}}}



{{{x^2-36 = 15x}}}



{{{x^2-15x-36 = 0}}}



Then you'd use the quadratic formula to solve for x. Since these solutions are much more complicated (and involve radicals), I have a feeling the equation is originally {{{x/3 = 12/(x+5)}}}