Question

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.

Answer 1

The given equations are:

2x − 3y + 6 = 0 .........(i)

2x + 3y − 18 = 0 .........(ii)

Putting x = 0 in equation (i) we get

`=> 2 xx 0 - 3y = -6`

=> y = 2

x = 0, y = 0

Putting y = 0 in equation (i) we get

`=> 2x - 3 xx 0 = -6`

=> x  = -3

x = -3, y = 0

Use the following table to draw the graph

x	0	-3
y	2	0

Draw the graph by plotting the two points A(0, 2), B(-3, 0) from table.

2x + 3y - 18 = 0 ........(ii

Putting x = 0 in equation (ii) we get

`=> 2 xx 0 + 3y = 18`

=> y = 6

x = 0, y = 6

Putting y = 0 in equation (ii) we get

`=> 2x + 3 xx 0 = 18`

=> x = 9

x = 9, y = 0

Use the following table to draw the graph.

x	0	9
y	6	0

Draw the graph by plotting the two points C(0,6), D(9,0) from a table.

The two lines intersect at P(3,4).

Hence x= 3, y = 4 is the solution of the given equations.

The area enclosed by the lines represented by the given equations and the y−axis

Now,

Required area = Area of PCA

Required area = 1/2 (base x height)

Required area = 1/2 (CA x PM)

Required area = 1/2 (4 x 3) sq.units

Hence the required area is 6 sq.units