I need to solve the system of equations
$$6x+y=3\tag 1 $$
$$x^2+y^2=16\tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This I need to solve by completing the square.
On
Exercise: Solve $37x^2-36x-7=0$ by the method of completing the square.
There is an alternate form of completing the square which can be useful when the coefficient of $x^2$ is large. Although the method avoids the use of fractions the trade-off is that it usually involves rather large numbers. The steps (using the present equation to illustrate) are
Given $ax^2+bx+c=0$,
From there the solution $x=\dfrac{18\pm\sqrt{583}}{37}$ is straightforward.
On
$ 37 x^2 - 36 x - 7 = 0 $
Factor out the $37$ from the first two terms
$ 37 ( x^2 - \dfrac{36}{37} x ) - 7 = 0$
Take half of $\dfrac{36}{37}$ which is $\dfrac{18}{37} $ and add it with its sign to $ x $ then square both as follows
$ 37 ( x - \dfrac{18}{37} )^2 - 7 = 37 ( \dfrac{18}{37} )^2$
Multiply through by $37^2$
$ 37 ( 37 x - 18)^2 - 37^2 (7) = 37 (18)^2 $
Divide through by $37$
$ (37 x - 18 )^2 - 7(37) = (18)^2 = 324 $
So,
$ (37 x - 18)^2 = 324 + 7(37) = 583 $
Therefore,
$ x = \dfrac{1}{37} \left ( 18 \pm \sqrt{ 583 } \right)$
To complete the square for something like this, the best way is to start by making the first coefficient a square. So we begin by multiplying both sides by $37$ to get $$37^2x^2-36\times37x-7\times37=0\ .$$ To avoid messing around with fractions, we would like the coefficient of $x$ to be even. Here, this is already the case. So completeing the square gives $$(37x-18)^2=7\times37+18^2\ .$$ I'm sure you can finish it from here.