Question 1132097

you are right, it is  ordered pair ({{{1}}},{{{2}}})

explanation:

{{{x=(4-x)/(y^2-x)}}}

{{{x(y^2-x)=(4-x)}}}

{{{(y^2-x)=(4-x)/x}}}

{{{y^2-x=(4-x)/x}}}

{{{y^2=(4-x)/x+x}}}

{{{y^2=(4-x)/x+x^2/x}}}

{{{y^2=(x^2-x+4)/x}}}

{{{y=sqrt((x^2-x+4)/x)}}}

{{{y=sqrt(x^2-x+4)/sqrt(x)}}}=>{{{x<>0}}} and {{{x-4<>0}}}

since {{{x<>0}}}, lowest integer we can choose for {{{x}}} is {{{ 1}}}, then {{{y}}} is

{{{y=sqrt(1^2-1+4)/sqrt(1)}}}

{{{y=sqrt(4)/1}}}

{{{y }}}= ±{{{ 2}}}=>since we need  ordered pairs of {{{positive}}} integers, we will use only {{{y=2}}}

so, we have one ordered pair of positive integers ({{{x}}},{{{y}}}) satisfy the equation {{{x=(4-x)/(y^2-x)}}} and it is  ({{{1}}},{{{2}}})