Question 931592
{{{4x^2+y^2=64}}}....eq.1
{{{x^2+y^2=52}}}.....eq.2
___________________________subtract eq.2 from eq.1

{{{4x^2+y^2-(x^2+y^2)=64-52}}}

{{{4x^2+cross(y^2)-x^2-cross(y^2)=12}}}

{{{3x^2=12}}}

{{{x^2=12/3}}}

{{{x^2=4}}}

{{{x=sqrt(4)}}}

solutions: {{{x=2}}} or {{{x=-2}}}

substitute these solutions in eq.1 or eq.2 and solve for {{{y}}}

{{{2^2+y^2=52}}}.....eq.2

{{{4+y^2=52}}}

{{{y^2=52-4}}}

{{{y^2=48}}}

{{{y=sqrt(48)}}}

solutions: {{{y=6.93}}} or {{{y=-6.93}}}

system solutions:

{{{x=2}}} and {{{y=6.93}}}
{{{x=2}}} and {{{y=-6.93}}}
or 
{{{x=-2}}} and {{{y=6.93}}}
{{{x=-2}}} and {{{y=-6.93}}}

{{{drawing( 600, 600, -10, 10, -10, 10,circle(-2,6.93,.14),locate(-2,6.93,p(-2,6.93)),
circle(2,-6.93,.14),locate(2,-6.93,p(2,-6.93)),circle(2,6.93,.14),locate(2,6.93,p(2,6.93)),circle(-2,-6.93,.14),locate(-2,-6.93,p(-2,-6.93)), graph( 600, 600, -10, 10, -10, 10,-sqrt(52-x^2),sqrt(52-x^2),-sqrt(64-4x^2), sqrt(64-4x^2))) }}}