Question 925853
{{{2x-y=0}}}....eq.1
{{{2xy=4 }}}....eq.2
________________

start with {{{2x-y=0}}}....eq.1, solve for {{{x}}}

{{{2x=y}}}

{{{x=y/2}}}...substitute in eq.2

{{{2(y/2)y=4 }}}....eq.2

{{{cross(2)(y/cross(2))y=4 }}}

{{{y*y=4 }}}

{{{y^2=4 }}}

{{{y=sqrt(4) }}}

solutions:

{{{y=2}}} or {{{y=-2}}}

go to eq.1 or 2, substitute {{{y}}} 


{{{2x-y=0}}}....eq.1 if {{{y=2}}}

{{{2x-2=0}}}
{{{2x=2}}}
{{{x=2/2}}}
{{{x=1}}}

or

if {{{y=-2}}}

{{{2x-(-2)=0}}}
{{{2x+2=0}}}
{{{2x=-2}}}
{{{x=-2/2}}}
{{{x=-1}}}

solutions to the system:

{{{x=1}}},{{{y=2}}}
or
{{{x=-1}}},{{{y=-2}}}

{{{2x-y=0}}}....eq.1
{{{2xy=4 }}}....eq.2

{{{drawing( 600, 600, -10, 10, -10, 10,locate(1,2,p(1,2)),locate(-1,-2,p(-1,-2)),circle(1,2,.12),circle(-1,-2,.12), graph( 600, 600, -10, 10, -10, 10, 2x, 4/(2x))) }}}