Question 769223
Let the positive integer to be x whereas the number to be y.
{{{x=2y-6}}} ---> Equation 1 ... Substitute into Equation 2
{{{x^2+y^2=164}}} ---> Equation 2
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{{{(2y-6)^2+y^2=164}}}
{{{(4y^2-24y+36)+y^2=164}}}
{{{4y^2-24y+36+y^2=164}}}
{{{4y^2+y^2-24y+36-164=0}}}
{{{5y^2-24y-128=0}}}
{{{(5y+16)(y-8)=0}}}
{{{y=-16/5}}} or {{{y=8}}}
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If {{{y=-16/5}}}, {{{x=2(-16/5)-6=-12&2/5}}}
* {{{x=-12&2/5}}} is not the answer as x must be a positive integer.
If {{{y=8}}}, {{{x=2(8)-6=10}}}
* {{{x=10}}} is the answer as x is a positive integer.
Thus, the integer is 10.