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The given equations are:

1. $x + 3y = 6$
2. $2x - 3y = 12$

Comparing these equations with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we get the coefficients:
For equation (1): $a_1 = 1$, $b_1 = 3$, $c_1 = -6$
For equation (2): $a_2 = 2$, $b_2 = -3$, $c_2 = -12$

Now, find the ratios of the coefficients:

$$\frac{a_1}{a_2} = \frac{1}{2}$$ $$\frac{b_1}{b_2} = \frac{3}{-3} = -1$$ $$\frac{c_1}{c_2} = \frac{-6}{-12} = \frac{1}{2}$$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (as $\frac{1}{2} \neq -1$), the lines intersect at a unique point. Therefore, the pair of equations is consistent.

For the equation $x + 3y = 6$, we find a few points:

• If $x = 0$, then $3y = 6 \Rightarrow y = 2$. Point: $(0, 2)$
• If $y = 0$, then $x = 6$. Point: $(6, 0)$
• If $x = 3$, then $3 + 3y = 6 \Rightarrow 3y = 3 \Rightarrow y = 1$. Point: $(3, 1)$

For the equation $2x - 3y = 12$, we find a few points:

• If $x = 0$, then $-3y = 12 \Rightarrow y = -4$. Point: $(0, -4)$
• If $y = 0$, then $2x = 12 \Rightarrow x = 6$. Point: $(6, 0)$
• If $x = 3$, then $2(3) - 3y = 12 \Rightarrow 6 - 3y = 12 \Rightarrow -3y = 6 \Rightarrow y = -2$. Point: $(3, -2)$

When you plot these points on a graph and draw the lines, you will observe that both lines intersect at the point $(6, 0)$. This is the unique solution to the pair of equations.
Thus, the solution is:

$$x = 6, y = 0$$