Question 927944
{{{x=y+3}}}.....eq.1
{{{x=y^2+1}}}...eq.2
______________

{{{x=y^2+1}}}...eq.2 substitute {{{x=y+3}}}

{{{y+3=y^2+1}}}...solve for {{{y}}}

{{{0=y^2-y-3+1}}}

{{{y^2-y-2=0}}}

{{{y^2+y-2y-2=0}}}

{{{(y^2+y)-(2y+2)=0}}}

{{{y(y+1)-2(y+1)=0}}}

{{{(y-2)(y+1)=0}}}

solutions:

if {{{(y-2)=0}}}=> {{{y=2}}}

if {{{(y+1)=0}}}=> {{{y=-1}}}

now find {{{x}}}:

{{{x=y+3}}}.....eq.1 if {{{y=2}}}

{{{x=2+3}}}

{{{x=5}}}

{{{x=y+3}}}.....eq.1 if {{{y=-1}}}

{{{x=-1+3}}}

{{{x=2}}}

so, your solutions are:

({{{5}}},{{{2}}})

and

({{{2}}},{{{-1}}})


{{{drawing( 600, 600, -10, 10, -10, 10,circle(2,-1,.12),circle(5,2,.12),locate(2,-1,p(2,-1)),locate(5,2,p(5,2)), graph( 600, 600, -10, 10, -10, 10, x-3,sqrt(x-1),-sqrt(x-1))) }}}