SOLUTION: Please help me solving this quadratic equation system: {{{ x^2+2*x-y+4=0 }}} {{{ 3*x-y+6=0 }}}

 ------- eq (i)

3x - y + 6 = 0 ======> y = 3x + 6 ------ eq (ii)

 -------- Substituting 3x + 6 for y in eq (i)





(x - 2)(x + 1) = 0

x = 2      OR      x = - 1



y = 3(2) + 6 ------- Substituting 2 for x in eq (ii)

y = 6 + 6, or 12





y = 3(- 1) + 6 ------- Substituting - 1 for x in eq (ii)

y = - 3 + 6, or 3