Question 258000
<pre><font size = 4 color = "indigo"><b>

Note: The other two tutors' solutions are both incorrect.


{{{system(x-2y-8=0, x^2+y^2-12x+6y+29=0)}}}

Solve the first equation for x

{{{x-2y-8=0}}}
{{{x=2y+8}}}

Substitute {{{(2y+8)}}} for x in the second equation:

{{{x^2+y^2-12x+6y+29=0}}}

{{{(2y+8)^2+y^2-12(2y+8)+6y+29=0}}}

{{{(2y+8)(2y+8)+y^2-12(2y+8)+6y+29=0}}}

{{{(4y^2+16y+16y+64)+y^2-12(2y+8)+6y+29=0}}}

{{{4y^2+32y+64+y^2-24y-96+6y+29=0}}}

{{{5y^2+14y-3=0}}}

Factor:

{{{(5y-1)(y+3)=0}}}

Use zero-factor principle:

{{{5y-1=0}}} gives solution {{{y=1/5}}}

{{{y+3=0}}} gives solution {{{y=-3}}}

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

For {{{y=1/5}}} we substitute {{{(1/5)}}} for y in 

{{{x=2y+8}}}

{{{x=2(1/5)+8}}}

{{{x=2/5+8}}}

{{{x=2/5+40/5}}}

{{{x=42/5) 

So one solution is (x,y) = ({{{42/5}}},{{{1/5}}})

or if you prefer, (x,y) = (8.4,.2)

For {{{y=-3}}} we substitute {{{(-3)}}} for y in 

{{{x=2y+8}}}

{{{x=2(-3)+8}}}

{{{x=-6+8}}}

{{{x=2}}}

So the other solution is (x,y) = (2,-3)

Edwin</pre>