Graphically,solve the following pair of equations:
$2x + y = 6$
$2x - y + 2 = 0$
Find the ratio of the areas of the two triangles formed by the lines representing these equations with the $x$-axis and the lines with the $y$-axis.

(4:1) Given equations are $2x + y = 6$ and $2x - y + 2 = 0$.
Table for equation $2x + y = 6$:

$x$	$0$	$3$
$y$	$6$	$0$
Points	$B$	$A$

Table for equation $2x - y + 2 = 0$:

$x$	$0$	$-1$
$y$	$2$	$0$
Points	$D$	$C$

Let $A_1$ and $A_2$ represent the areas of $\triangle ACE$ and $\triangle BDE$,respectively.
Now,$A_1 = \text{Area of } \triangle ACE = \frac{1}{2} \times AC \times PE = \frac{1}{2} \times 4 \times 4 = 8$.
And $A_2 = \text{Area of } \triangle BDE = \frac{1}{2} \times BD \times QE = \frac{1}{2} \times 4 \times 1 = 2$.
Therefore,$A_1 : A_2 = 8 : 2 = 4 : 1$.
Hence,the pair of equations intersect graphically at point $E(1, 4)$,i.e.,$x = 1$ and $y = 4$.