Question 769570

{{{2x+y+3=0}}}......eq.1
{{{x^2+y^2=5}}}.....eq.2
__________________________

{{{2x+y+3=0}}}......eq.1.....solve for {{{y}}}

{{{y=-2x-3}}}....eq.1a....substitute in eq.2

{{{x^2+(-2x-3)^2=5}}}.....eq.2...solve for {{{x}}}


{{{x^2+((-2x)^2-2x(-3)+3^2)=5}}}

{{{x^2+4x^2+12x+9=5}}}

{{{5x^2+12x+9-5=0}}}

{{{5x^2+12x+4=0}}}....use quadratic formula to find {{{x}}}

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

{{{x = (-12+- sqrt( 12^2-4*5*4 ))/(2*5) }}}

{{{x = (-12 +- sqrt(144 -80))/10 }}}

{{{x = (-12 +- sqrt(64))/10 }}}

{{{x = (-12 +- 8)/10 }}}

solutions:

{{{x = (-12 + 8)/10 }}}

{{{x = -4/10 }}}

{{{x = -2/5 }}}

{{{x = -0.4 }}}

or

{{{x = (-12 - 8)/10 }}}

{{{x = -20/10 }}}

{{{x = -2 }}}

now go to eq.1a and find {{{y}}}

{{{y=-2x-3}}}....eq.1a ...plug in {{{x = -0.4 }}}

{{{y=-2(-0.4)-3}}}

{{{y=0.8-3}}}

{{{y=-2.2}}}

or

{{{y=-2x-3}}}....plug in {{{x = -2 }}}

{{{y=-2(-2)-3}}}

{{{y=4-3}}}

{{{y=1}}}

solution pairs are:

{{{x = -0.4 }}} and {{{y=-2.2}}}
{{{x = -2 }}} and {{{y=1}}}