Solve the following system of equations graphically:
$2x\ -\ 3y\ +\ 6\ =\ 0$
$2x\ +\ 3y\ -\ 18\ =\ 0$
Also, find the area of the region bounded by these two lines and y-axis.

 Given:

The given equations are:

$2x\ -\ 3y\ +\ 6\ =\ 0$

$2x\ +\ 3y\ -\ 18\ =\ 0$

To do:

We have to solve the given system of equations and calculate the area formed by the given lines and y-axis.

Solution:

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation $2x-3y+6=0$,

$3y=2x+6$

$y=\frac{2x+6}{3}$

If $x=0$ then $y=\frac{2(0)+6}{3}=\frac{6}{3}=2$

If $x=3$ then $y=\frac{2(3)+6}{3}=\frac{12}{3}=4$

$x$	$0$	$3$
$y$	$2$	$4$

For equation $2x+3y-18=0$,

$3y=18-2x$

$y=\frac{18-2x}{3}$

If $x=0$ then $y=\frac{18-2(0)}{3}=\frac{18}{3}=6$

If $x=3$ then $y=\frac{18-2(3)}{3}=\frac{12}{3}=4$

$x$	$0$	$3$
$y$	$6$	$4$

The equation of y-axis is $x=0$.

The above situation can be plotted graphically as below:

 

The lines AB, CD and AC represent the equations $2x-3y+6=0$, $2x+3y-18=0$ and y-axis respectively.

As we can see, the points of intersection of the lines AB, CD and AC taken in pairs are the vertices of the given triangle.

Hence, the vertices of the given triangle are $(0,2), (3,4)$ and $(0,6)$. 

We know that,

Area of a triangle$=\frac{1}{2}bh$

In the graph, the height of the triangle is the distance between point B and AC.

Height of the triangle$=3$ units.

Base of the triangle$=$Distance between the points A and C.

Base of the triangle$=6-2=4$ units.

Area of the triangle formed by the given lines and y-axis$=\frac{1}{2}\times3\times4$

$=6$ sq. units. 

The area bounded by the given lines and y-axis is $6$ sq. units.