Question 684378
if  ({{{-1}}},{{{ -3}}})
{{{-3=(-1)^2+4(-1)}}}
{{{-3=1-4}}}
{{{-3=-3}}}

{{{2y-y=1}}}
{{{2(-3)-(-3)=1}}}
{{{-6+3=1}}}
{{{-3<>1}}}


so, ({{{-1}}},{{{ -3}}}) is not common solution


than common solution will be:

for {{{y=x^2+4x}}} and 

{{{2y-y=1}}}...=>...{{{y=1}}}...plug it in {{{y=x^2+4x}}} and solve for {{{x}}}


{{{1=x^2+4x}}}

{{{0=x^2+4x-1}}}

or

{{{x^2+4x-1=0}}}....use quadratic formula


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


{{{x = (-4 +- sqrt( 4^2-4*1*(-1) ))/(2*1) }}}


{{{x = (-4 +- sqrt( 16+4 ))/2 }}}


{{{x = (-4 +- sqrt( 20 ))/2 }}}


{{{x = (-4 +- 4.47)/2 }}}


solutions:

{{{x = (-4 +4.47)/2 }}}

{{{x = (0.47)/2 }}}

{{{x = 0.235}}}

or

{{{x = (-4 -4.47)/2 }}}

{{{x = -8.47/2 }}}

{{{x = -4.235}}}


{{{drawing(600,600,-10,10,-10,10,grid(1),circle(0.235,1,0.2),circle(-4.235,1,0.2),graph(600,600,-10,10,-10,10,x^2+4x,1))}}}


 common solution ({{{0.235}}},{{{1}}}) and ({{{-4.235}}},{{{1}}})