System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Thank you for your time
 A: Subtract the first row from the last row of the system: then you get
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1  -16 a_2 -64 &=0
\end{split}
$$
Then multiply by 8 the second row and subtract it again from the third row:
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
64a_1  -16 a_2 -64 &=0
\end{split}\implies
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
56a_1 -8a_3+64 &=0
\end{split}
$$
Then you can proceed by substitution and do the following steps
$$
\begin{split}
a_1a_3-a_2^2 &=0\\
a_1+a_3-2a_2-16&=0\\
a_1 &=\frac{1}{7}(a_3-8)
\end{split}\implies
\begin{split}
\frac{1}{7}a_3^2-\frac{8}{7}a_3-a_2^2 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1&=\frac{1}{7}(a_3-8)
\end{split}
$$
and finally we have 
$$
\begin{split}
9a_3^2-424a_3-3600 &=0\\
a_2&=\frac{4}{7}(a_3-15)\\
a_1  &=\frac{1}{7}(a_3-8)
\end{split}
$$
where the first equation is a quadratic equation respect to the single $a_3$.
In sum, a nice way to proceed for such system is to see if you can subtract a multiple of one row from another, in order to simplify the structure and possibly arrive to a system where all the equations are linear except one, which however contains only one of the variables.
A: As long as $a_1a_3-a_2^2= 0$ we have
$$
a_1+a_3-2a_2= 16\\
64a_1 -16 a_2 =64
$$
or
$$
a_1+a_3-2a_2=16\\
4a_1 - a_2 =4
$$
solving for $a_1,a_2$ we have
$$
a_1 = \frac 17(a_3-8)\\
a_2 = \frac 47(a_3-15)
$$
substituting now into $a_1a_3-a_2^2= 0$ we have
$$
9a_3^2-424a_3-3600=0
$$
and solving gives
$$
a_3 = \left\{\frac{100}{9},\ 36\right\}
$$
etc.
A: By the second equation we get
$$a_3=2a_2+16-a_1$$ so we get in (1)
$$2a_1a_2+16a_1-a_1^2-a_2^2=0$$ (I)
and in (3)
$$2a_1a_2+16a_1-a_1^2+64a_1-a_2^2-16a_2-64=0$$(II)
with $$2a_1a_2=a_1^2-16a_1+a_2^2$$ we get in (II):
We get
$$4a_1-a_2=4$$
Now e can eliminate $a_2$
$$2a_1(4a_1-4)+16a_1-a_1^2-(4a_1-4)^2=0$$
Now you can solve this equation for $a_1$.
A: I'm going to use $x=a_1, \ y=a_2, \ z=a_3$
$$xz - y^2=0 \tag A$$
$$x + z - 2y - 16 = 0 \tag B$$
$$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$
Substitute $xz = y^2$ from (A) into (C)
\begin{align}
   xz + 64x - y^2 - 16y - 64 &= 0 \\
   y^2 + 64x - y^2 - 16y - 64 &= 0 \\
   64x - 16y - 64 &= 0 \\
   y &= 4x - 4 \tag D
\end{align}
Substitute $y = (4x - 4)$ from (D) into (B)
\begin{align}
   x + z - 2y - 16 = 0 \\
      x + z - 2(4x - 4) - 16 = 0 \\
      -7x + z - 8 &= 0 \\
      z &= (7x + 8) \tag E
\end{align}
Substitute (D) and (E) into (A)
\begin{align}
   xz - y^2 &= 0 \\
   x(7x + 8) - (4x - 4)^2 &= 0 \\
   7x^2 + 8x - 16x^2 + 32x - 16 &= 0 \\
   -9x^2 + 40x - 16 &= 0 \\
   -(9x - 4)(x - 4) &= 0 \\
   x &\in \left\{ \dfrac 49, 4 \right\}
\end{align}
We get $$
   (a_1,a_2,a_3) = \left(\dfrac 49, -\dfrac{20}{9}, \dfrac{100}{9}  \right)
   \ \text{and} \ 
   (a_1,a_2,a_3) = (4, 12, 36)$$
