Page No 3.30
Q.23. Solve the following system of linear equation graphically and shade the region between the two lines and x-axis:
(i) 2x + 3y = 12,
x − y = 1
(ii) 3x + 2y − 4 = 0,
2x − 3y − 7 = 0
(iii) 3x + 2y − 11 = 0
2x − 3y + 10 = 0
Ans. (i) The given equations are:
2x + 3y = 12 .....(i)
x - y = 1 .....(ii)
Putting x = 0 in equation (i), we get:
⇒ 2 x 0 + 3y = 12
⇒ y = 4
x = 0, y = 4
Putting y = 0 in equation (i) we get:
⇒ 2x + 3 x 0 = 12
⇒ x =6
x = 6, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points A(0,4), B(6,0) from table
x - y = 1 ....(ii)
Putting x = 0 in equation (ii) we get:
⇒ 0 - y = 1
⇒ y = - 1
x = 0, y = -1
Putting y = 0 in equation (ii), we get:
⇒ x - 0 = 1
⇒ x = 1
x = 1, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points C(0,- 1), D(1,0) from table.
The two lines intersect at P(3,2). The region enclosed by the lines represented by the given equations and x−axis are shown in the above figure
Hence, x = 3 and y = 2 is the solution.
(ii) The given equations are:
3x + 2y - 4 = 0 ....(i)
2x - 3y - 7 = 0 ....(ii)
Putting x = 0 in equation (i) we get:
⇒ 3 x 0 + 2y = 4
⇒ y = 2
x = 0, y = 2
Putting y = 0 in equation (i) we get:
⇒ 3x + 2 x 0 = 4
⇒ x = 4/3
x = 4/3, y = 0
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the two points A(0,2), B(4/3,0)
2x - 3y - 7 = 0 .......(ii)
Putting x = 0 in equation (ii) we get:
⇒ 2 x 0 - 3y = 7
⇒ y = -7/3
x = 0, y = -7/3
Putting y = 0 in equation (ii), we get:
⇒ 2x - 3 x 0 = 7
⇒ x = 7/2
x = 7/2, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points C(0,-7/3),D(7/2,0) from table.
The two lines intersect at P(2, - 1).The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, x = 2 and y = - 1 is the solution.
(iii) The given equations are:
3x + 2y - 11 = 0 .....(i)
2x - 3y + 10 = 0 .....(ii)
Putting x = 0 in equation (i) we get:
⇒ 3 x 0 + 2y = 11
⇒ y = 11/2
x = 0, y = 11/2
Putting y = 0 in equation (i) we get:
⇒ 3x + 2 x 0 = 11
⇒ x = 11/3
x = 11/3, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points A(0,11/2), B(11/3,0) from table
2x - 3y + 10 = 0 .......(ii)
putting x = 0 in equation (ii) we get:
⇒ 2 x 0 - 3y = - 10
⇒ y =10/3
x = 0, y = 10/3
Putting y = 0 in equation (ii) we get:
⇒ 2x - 3 x 0 = - 10
⇒ x = -5
x = -5, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points C(0,10/3), D(-5,0) from table.
The two lines intersect at P(1,4).The area enclosed by the lines represented by the given equations and the coordinates x−axis and shaded the area in graph.
Hence, x = 1 and y = 4 is the solution.
Q.24. Draw the graphs of the following equations on the same graph paper.
2x + 3y = 12,
x − y = 1
Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis.
Ans. The given equations are
2x + 3y = 12 .....(i)
x - y = 1 .....(ii)
Putting x = 0 in equation (i), we get:
⇒ 2 x 0 + 3y = 12
⇒ y = 4
x = 0, y = 4
Putting y = 0 in equation (i), we get:
⇒ x + 3 x 0 = 6
⇒ x = 6
x = 6, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points A(0,2) and B(6,0) from table.
x - y = 1 .....(ii)
Putting x = 0 in equation (ii) we get:
⇒ 0 - y = 1
⇒ y = - 1
x = 0, y = - 1
Putting y = 0 in equation (ii), we get:
⇒ x - 0 = 1
⇒ x = 1
x = 1, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points C(0, - 1), D(1, 0) from table.
Draw the graph by plotting the two points from table.
The intersection point is P(3, 2)
Three points of the triangle are A(0,4),C(0,-1),and P(3,2)
Hence the value of x = 3 and y = 2.
Q.25. Draw the graphs of x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular area. Calculate the area bounded by these lines and x-axis.
Ans. The given equations are
x - y + 1 = 0 ....(i)
3x + 2y - 12 = 0 ....(ii)
Putting x = 0 in equation (i) we get:
⇒ 0 - y = -1
⇒ y = 1
x = 0, y = 1
Putting y = 0 in equation (i) we get:
⇒ x - 0 = - 1
⇒ x = - 1
x = - 1, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points A(0,1), B(-1,0) from table.
3x + 2y = 12 ....(i)
Putting x = 0 in equation (ii) we get:
⇒ 3 x 0 + 2y = 12
⇒ y = 6
x = 0, y = 6
Putting y = 0 in equation (ii) we get:
⇒ 3x + 2 x 0 = 12
⇒ x = 4
x = 4, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points C(0,6), D(4,0) from table.
The two lines intersect at P(2,3)
Now, Required area = Area of shaded region
⇒ Required area = Area of PBD
⇒ Required area = 1/2(base x height)
⇒ Required area = 1/2(BD x PM)
⇒ Required area = 1/2(5 x 3)sq. units
Hence the area = 7.5 sq.units
Q.26. Solve graphically the system of linear equations:
4x − 3y + 4 = 0
4x + 3y − 20 = 0
Find the area bounded by these lines and x-axis.
Ans. The given equations are
4x − 3y + 4 = 0 .....(i)
4x + 3y − 20 = 0 .....(ii)
Putting x = 0 in equation (i), we get:
⇒ 4 x 0 - 3y = - 4
⇒ y = 4/3
x = 0, y = 4/3
Putting y = 0 in equation (i), we get:
⇒ 4x - 3 x 0 = -4
⇒ x = - 1
x = - 1, y = 0
Use the following table to draw the graph.
The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).
4x + 3y = 20 .....(i)
Putting x = 0 in equation in equation (ii) we get:
⇒ 4 x 0 + 3y = 20
⇒ y = 20/3
x = 0, y = 20/3
Putting y = 0 in equation (ii), we get:
⇒ 4x + 3 x 0 = 20
⇒ x = 5
x = 5, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at P(2,4).
Hence x = 2, y =4 is the solution of the given equations.
Now,
⇒ Required area = Area of PBD
⇒ Required area = 1/2(base x height)
⇒ Required area = 1/2(BD x PM)
⇒ Required area = 1/2(6 x 4)
Hence, the area = 12 sq. units
Q.27. Solve the following system of linear equations graphically :
3x + y − 11 = 0, x − y − 1 = 0.
Shade the region bounded by these lines and y-axis. Also, find the area of the region bounded by the these lines and y-axis.
Ans. The given equations are
3x + y - 11 = 0 .....(i)
x - y - 1 = 0 .....(ii)
Putting x = 0 in equation (i), we get:
⇒ 3 x 0 + y = 11
⇒ y = 11
x = 0, y = 11
Putting y = 0 in equation (i), we get:
⇒ 3x + 0 = 11
⇒ x = 11/3
x = 11/3, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
x - y = 1 ....(i)
Putting x = 0 in equation (ii) we get:
⇒ 0 - y = 1
⇒ y = -1
x = 0, y = - 1
Putting y = 0 in equation (ii), we get:
⇒ x - 0 = 1
⇒ x = 1
x = 1, y = 0
Use the following table to draw the graph.
Draw the graph by plotting the two points from table.
The two lines intersect at P(3,2)
Hence x = 3, y = 2 is the solution of the given equations
The area enclosed by the lines represented by the given equations and the y−axis is shaded region in the figure
Now, Required area = Area of shaded region
⇒ Required area = Area of PAC
⇒ Required area = 1/2 (base x height)
⇒ Required area = 1/2 (AC x PM)
⇒ Required area = 1/2 (12 x 3) sq. units
Hence the required area is 18 sq. units.
Q.28. Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system.
(i) 2x + y = 6
x − 2y = −2
(ii) 2x − y = 2
4x − y = 8
(iii) x + 2y = 5
2x − 3y = −4
(iv) 2x + 3y = 8
x − 2y = −3
Ans. (i) The given equations are
2x + y = 6 ....(i)
x - 2y = - 2 ....(ii)
The two points satisfying (i) can be listed in a table as,
The two points satisfying (ii) can be listed in a table as,
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 2, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are (3, 0) and (-2, 0) respectively.
(ii) The given equations are
2x - y = 2 ......(i)
4x - y = 8 ......(ii)
The two points satisfying (i) can be listed in a table as,
The two points satisfying (ii) can be listed in a table as,
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 3, y = 4.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are respectively.
(iii) The given equations are
x + 2y = 5 .....(i)
2x - 3y = -4 .....(ii)
The two points satisfying (i) can be listed in a table as,
The two points satisfying (ii) can be listed in a table as,
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are (5, 0) and (-2, 0) respectively.
(iv) The given equations are
2x + 3y = 8 .....(i)
x - 2y = - 3 .....(ii)
The two points satisfying (i) can be listed in a table as,
The two points satisfying (ii) can be listed in a table as,
Now, graph of equations (i) and (ii) can be drawn as,
It is seen that the solution of the given system of equations is given by x = 1, y = 2.
Also, it is observed that the coordinates of the points where the lines (i) and (ii) meet the x-axis are (4,0) and (-3,0) respectively.
