Question 669201


let's two natural no. be {{{x}}} and {{{y}}}

 if difference of two natural no. are {{{4}}}, we have

{{{x-y=4}}}........eq. 1

 and, if the sum of their square is {{{58}}} we have

{{{x^2+y^2=58}}}........eq. 2

solve the system:

{{{x-y=4}}}........eq. 1
{{{x^2+y^2=58}}}........eq. 2
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{{{x-y=4}}}........eq. 1..solve for {{{x}}}

{{{x=y+4}}}.....substitute in 2

{{{(y+4)^2+y^2=58}}}.....solve for {{{y}}}

{{{y^2+8y+16+y^2=58}}}

{{{2y^2+8y+16=58}}}

{{{2y^2+8y+16-58=0}}}

{{{2y^2+8y-42=0}}}.....solve quadratic

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y = (-8 +- sqrt( 8^2-4*2*(-42) ))/(2*2) }}}

{{{y = (-8 +- sqrt( 64+336 ))/4 }}}

{{{y = (-8 +- sqrt( 400))/4 }}}

{{{y = (-8 +- 20)/4 }}}

solutions:

{{{y = (-8 + 20)/4 }}}

{{{y =12/4 }}}

{{{highlight(y =3)}}}

and

{{{y = (-8 -20)/4 }}}

{{{y =-28/4 }}}

{{{highlight(y =-7)}}}


now find {{{x}}}:

{{{x=y+4}}}

{{{x=3+4}}}

{{{highlight(x=7)}}}

and

{{{x=y+4}}}

{{{x=-7+4}}}

{{{highlight(x=-3)}}}


so, you have two pairs of solutions:

{{{highlight(x=7)}}} and {{{highlight(y =3)}}}

or

{{{highlight(x=-3)}}} and {{{highlight(y =-7)}}}