Question 937348

A)
you have here a system of two equations and you need to find {{{x}}} and {{{y}}} that satisfy each equation

{{{x+y=5}}}........eq.1
{{{x^2+y^2=25}}}........eq.2
-------------------------

start with

{{{x+y=5}}}........eq.1...solve for {{{x}}}

{{{x=5-y}}}........eq.1a....substitute this in eq.2

{{{(5-y)^2+y^2=25}}}........eq.2

{{{25-10y+y^2+y^2=25}}}

{{{2y^2-10y+25=25}}}

{{{2y^2-10y=25-25}}}

{{{2y(y-5)=0}}}

solutions:

if {{{2y=0}}} => {{{y=0}}}

if {{{(y-5)=0}}} => {{{y=5}}}

now find {{{x}}}

{{{x=5-y}}}........eq.1a if {{{highlight(y=0)}}}
{{{x=5-0}}}
{{{highlight(x=5)}}}


{{{x=5-y}}}........eq.1a if {{{highlight(y=5)}}}
{{{x=5-5}}}
{{{highlight(x=0)}}}

 Answer is: ({{{0}}},{{{5}}}) ({{{5}}},{{{0}}})


B)
{{{3x-4y=25 }}} 
{{{x^2+y^2=25}}}
_________________

{{{3x=4y+25 }}}

{{{x=(4/3)y+25/3 }}}


{{{x^2+y^2=25}}}

{{{((4/3)y+25/3)^2 +y^2=25}}}

{{{(16/9)y^2+2(4/3)(25/3)y+9y^2/9=25}}}

{{{(25/9)y^2+(8/3)(8*25/9)y=25}}}

{{{25/9(y^2+8y+25)=25}}}

{{{cross(25)(y^2+8y+25)=cross(25)*9}}}

{{{y^2+8y+25=9}}}

{{{y^2+8y+25-9=0}}}

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

{{{(y+4)^2=0

{{{highlight(y = -4)}}}


now find {{{x}}}:

{{{x=(4/3)y+25/3 }}}

{{{x=(4/3)(-4)+25/3 }}}

{{{x=-16/3+25/3 }}}

{{{x=(25-16)/3 }}}

{{{x=9/3 }}}

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


 Answer is: ({{{3}}},{{{-4}}})