Solving System of Equations using Substitution or Elimination

[Hazel]

x^2-13xy+12y^2=0 (1)
x^2+xy=156 (2)

What I have so far:
x^2+xy=156
xy=156-x^2
y=(156-x^2)/x)

Plugged y=(156-x^2)/x) into (1):
x^2-13x(156-x^2)/x)+12(156-x^2)/x)^2=0

For 1st half I Multiplied x to -13x in order to get the same denominator so I can multiply it to (156-x^2)/x):
(-13x^2)/x)(156-x^2)/x)
=(-2028x^2+13x^4)/x^2
=-2028+13x^2

For 2nd half I squared (156-x^2)/x) and factored:
12(156-x^2)/x)(156-x^2)/x)
=12(24336-156x^2-156x^2+x^4)/x^2)
=12(24336-312x^2+x^4)/x^2)

Next, I multiplied x^2 to 12 in order to get the same denominator so I can multiply it to (24336-312x^2+x^4)/x^2):
(12x^2)x^2)(24336-312x^2+x^4)/X^2)

I got for 2nd half:
(292032x^2-3744x^4+12x^6)/x^4)

So all together I got:
x^2-2028+13x^2+(292032x^2-3744x^4+12x^6/x^4)=0

Well apparently I went wrong somewhere in the 2nd half because MyMathLab "Help me Solve this" feature is telling me it suppose to had been:
x^2-2028+13x^2+(292032-3744x^2+12x^4/x^2)=0 Then went on to solving the problem.

I'm stuck here. I would like to know where I went wrong. Can you please tell me?

 

[MarkFL]

Hello and welcome to MHB! :D

I think you are off to a good start by solving the second equation for $y$:

\(\displaystyle y=\frac{156-x^2}{x}\)

What I suggest doing next is factoring the first equation. What do you find?

 

[Hazel]

x^2-13xy+12y^2=0
(x-12y)(x-y)=0
x=12y x=y

 

[MarkFL]

x^2-13xy+12y^2=0
(x-12y)(x-y)=0
x=12y x=y

Good! (Yes) Now you have two cases to consider:

$x=y$ and $x=12y$

Substitute each in turn into the other equation, and solve for $y$, and then you will know $x$ as well from the substitution. You should get 4 points in total. What do you find?

 

[Hazel]

Using x^2+xy=156?
x=12y x=y
So 1st it will be:
(12y)^2+(12y)y=156?
Then:
(y)^2+(y)y=156?

 

[Hazel]

(12y)^2+(12y)y=156
144y^2+12y^2=156
156y^2=156
y^2=1
y=\pm\sqrt{1}
y=\pm1

x^2+x(x)=156
x^2+x^2=156
2x^2=156
x^2=78
x=\pm\sqrt{78}

 

[MarkFL]

(12y)^2+(12y)y=156
144y^2+12y^2=156
156y^2=156
y^2=1
y=\pm\sqrt{1}
y=\pm1

x^2+x(x)=156
x^2+x^2=156
2x^2=156
x^2=78
x=\pm\sqrt{78}

Yes, so what are your 4 points? :D

 

[Hazel]

Right?
(1,12) (-1,-12) (\sqrt{78},\sqrt{78}) (-\sqrt{78},-\sqrt{78})

 

[Hazel]

Actually it's:
(12,1) (-12,-1) (\sqrt{78},\sqrt{78}) (-\sqrt{78},-\sqrt{78})

My mistake

 

[MarkFL]

Actually it's:
(12,1) (-12,-1) (\sqrt{78},\sqrt{78}) (-\sqrt{78},-\sqrt{78})

My mistake

Yes, that's correct. :D

 

[Hazel]

Thanks! Solved!