How do you solve the system $ - 6x + 6y = 6$ and $ - 6x + 3y = - 12\,$?

Hint: Let us solve the system of equations by using elimination method.
There are three ways to solve systems of linear equations in two variables: graphing, substitution and elimination method.
In this system of equations multiplication is not necessary to eliminate a variable.
In the elimination method, you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.
The elimination method for solving systems of linear equations uses the additional property of equality.

Complete step-by-step solution:
First mark the given two equations as $\left( 1 \right)$ and $\left( 2 \right)$ .
The given equations are
$ - 6x + 6y = 6$ …. $\left( 1 \right)$
$ - 6x + 3y = - 12\,$ ….. $\left( 2 \right)$
 Solve one of the equations for one of the variables. Let’s solve the first equation for y.
Note that the coefficients of $x$ in both equations are numerically equal.
So, we can eliminate $x$ easily in the system of equations.
Now, adding the two equations we obtain an equation,
\[\begin{array}{*{20}{c}}
  { - 6x + 6y = 6} \\
  { - 6x + 3y = - 12} \\
  { + {\text{ }} - {\text{ + }}} \\
  {\overline {{\text{ }}0 + 3y = 18} }
\end{array}\]
$3y = 18$
$y = \dfrac{{18}}{3}$
 That is, $y = 6$
Now, we substitute $y = 6$ in any one of the above equations to solve for x.
Substituting $y = 6$ in equation $\left( 1 \right)$
We obtain,
$ - 6x + 6\left( 6 \right) = 6$
$ - 6x + 36 = 6$
Shifting the constants to the right-hand side,
$ - 6x = 6 - 36$
$ - 6x = - 30$
Shifting the variable to the other side,
$ - x = \dfrac{{ - 30}}{6}$
$ - x = - 5$
$ \Rightarrow x = 5.$
Now, $\left( {5,6} \right)$ is a solution to the given system of equations.
Because the two given equations are true when $x = 5$ and $y = 6$ as from the given two equations.
Substitute $x = 5$ and $y = 6$ in equation $\left( 1 \right)$ and $\left( 2 \right)$ , we get,
$ - 6\left( 5 \right) + 6\left( 6 \right) = 6$
$ - 30 + 36 = 6$
And, $ - 6x + 3y = - 12\,$
$ - 6\left( 5 \right) + 3\left( 6 \right) = - 12$
$ - 30 + 18 = - 12\,$

Note: We can also solve this system of equations by cross multiplication method.
The given system of equation can be written in the form as
$ - 6x + 6y - 6 = 0$
$ - 6x + 3y + 12 = 0$
For the cross-multiplication method, we write the coefficient as
$\begin{array}{*{20}{c}}
  {}&x&{}&y&{}&1&{} \\
  6&{}&{ - 6}&{}&{ - 6}&{}&6 \\
  3&{}&{12}&{}&{ - 6}&{}&3
\end{array}$
Hence, we get,
$\dfrac{x}{{\left( 6 \right)\left( {12} \right) - \left( 3 \right)\left( { - 6} \right)}} = \dfrac{y}{{\left( { - 6} \right)\left( { - 6} \right) - \left( {12} \right)\left( { - 6} \right)}} = \dfrac{1}{{\left( { - 6} \right)\left( 3 \right) - \left( { - 6} \right)\left( 6 \right)}}$
$\dfrac{x}{{72 + 18}} = \dfrac{y}{{36 + 72}} = \dfrac{1}{{ - 18 + 36}}$
$\dfrac{x}{{90}} = \dfrac{y}{{108}} = \dfrac{1}{{18}}$
$\dfrac{x}{{90}} = \dfrac{1}{{18}}$
$x = \dfrac{{90}}{{18}}$
$x = 5$
$\dfrac{y}{{108}} = \dfrac{1}{{18}}$
$y = \dfrac{{108}}{{18}}$
$y = 6$
Hence, the solution is
$\left( {5,6} \right)$