Question

Solve the following system of linear equations graphically:
$2x-3y-17=0,4x+y-13=0.$
Shade the region between the lines and the x-axis.

Answer 1

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 3y − 17 = 0

2x − 3y − 17 = 0
⇒ 3y = (2x − 17)
∴ $y=\frac{2x-17}{3}$ ...........(i)
Putting x = 1, we get y = −5.
Putting x = 4, we get y = −3.
Putting x = 7, we get y = −1.
Thus, we have the following table for the equation 2x − 3y − 17 = 0.

x	1	4	7
y	−5	−3	−1

Now, plot the points A(1, −5), B( 4, −3) and C(7, −1) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x − 3y − 17 = 0.

Graph of 4x + y − 13 = 0
4x + y − 13 = 0
⇒ y = (−4x + 13) ...........(ii)
Putting x = 4, we get y = −3.
Putting x = 2, we get y = 5.
Putting x = 3, we get y = 1.
Thus, we have the following table for the equation 4x + y − 13 = 0.

x	4	2	3
y	−3	5	1

Now, plot the points P(2, 5) and Q(3, 1). The point B(4, −3) has already been plotted. Join PQ and QB to get the graph line PB. Extend it on both ways.
Then, PB is the graph of the equation 4x + y − 13 = 0.

The two graph lines intersect at B(4, −3).
∴ The solution of the given system of equations is x = 4 and y = −3.
These graph lines intersect the x-axis at R and S.
Hence, the region bounded by these lines and the x-axis has been shaded.