Question 604144
{{{y= x^2-6x+10}}}


{{{y+x=4}}}
{{{y = 4 - x}}}
substitute this into first equation:
{{{y= x^2-6x+10}}}
{{{4 - x= x^2-6x+10}}}
{{{0= x^2-6x+10-4+x}}}
{{{0= x^2-5x+6}}}
{{{0=(x - 3)*(x - 2)}}}
x - 3 = 0 or x - 2= 0
x = 3 or x = 2


if x = 3 then y = 4 - x = 4 - 3 = 1
if x = 2 then y = 4 - y = 4 - 2 = 2


so, the solutions are:
{(3,1), (2,2)}