Question 257977
<pre><font size = 4 color = "indigo"><b>
{{{system(y=x+3, x^2+y^2+4x-8y+11=0)}}}

Substitute {{{(x+3)}}} for y in the second equation:

{{{x^2+y^2+4x-8y+11=0}}}

{{{x^2+(x+3)^2+4x-8(x+3)+11=0}}}

{{{x^2+(x+3)(x+3)+4x-8x-24+11=0}}}

{{{x^2+(x^2+3x+3x+9)+4x-8x-24+11=0}}}

{{{x^2+(x^2+6x+9)+4x-8x-24+11=0}}}

{{{x^2+x^2+6x+9+4x-8x-24+11=0}}}

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

Divide every term by 2

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

Factor:

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

Use zero-factor principle:

{{{x+2=0}}} gives solution {{{x=-2}}}

{{{x-1=0}}} gives solution {{{x=1}}}

Now we must find the value of y for each of these
two values for x.

For {{{x=-2}}} we substitute -2 for x in 

{{{y=x+3}}}

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

{{{y=1}}}

So one solution is (x,y) = (-2,1)

For {{{x=1}}} we substitute 1 for x in 

{{{y=x+3}}}

{{{y=(1)+3}}}

{{{y=4}}}

So the other solution is (x,y) = (1,4)

Graphically we have this

{{{drawing(200,200,-6,4,-2,8,
locate(1,4,"(1,4)"), locate(-4,2,"(-2,1)"),
graph(200,200,-6,4,-2,8,x+3), circle(-2,4,3) )}}}

Edwin</pre>