Question 967820
{{{x^2+y^2=25}}}......eq.1
{{{x+y=7}}}......eq.2
_____________________

{{{x+y=7}}}......eq.2......solve for {{{y}}}

{{{y=7-x}}}.....substitute in eq.1


{{{x^2+(7-x)^2=25}}}......eq.1...solve for {{{x}}}

{{{x^2+7^2-14x+x^2=25}}}

{{{2x^2-14x+49=25}}}

{{{2x^2-14x+49-25=0}}}

{{{2x^2-14x+24=0}}}...simplify, both sides divide by {{{2}}}

{{{x^2-7x+12=0}}}

{{{x^2-4x-3x+12=0}}}

{{{(x^2-4x)-(3x-12)=0}}}

{{{x(x-4)-3(x-4)=0}}}

{{{(x-3)(x-4)=0}}}

solutions:

if {{{(x-3)=0}}}=>{{{x=3}}}
if {{{(x-4)=0}}}=>{{{x=4}}}

substitute in {{{y=7-x}}} and find {{{y}}}

{{{y=7-x}}} if {{{x=3}}}
{{{y=7-3}}}
{{{y=4}}}

{{{y=7-x}}} if {{{x=4}}}
{{{y=7-4}}}
{{{y=3}}}

so, your solutions are:

{{{x=3}}} and {{{y=4}}}

or 

{{{x=4}}}  and {{{y=3}}}


{{{drawing( 600, 600, -7, 7, -7, 7,
circle(3,4,.1),circle(4,3,.1),
locate(3,4,p(3,4)),locate(4,3,p(4,3)),
 graph( 600, 600, -7, 7, -7, 7, 7-x,sqrt(25- x^2),-sqrt(25- x^2))) }}}