Today our topic of discussion is Solving by using Graphical Method.
Solving by using Graphical Method
Solving by using Graphical Method
In a simple equation in two variables, the relation of existing variables x and y and can be expressed by picture. This picture is called the graph of that relation. In the graph of such equation, there exist infinite number of points. Plotting a few such points and joining them with each other, we get the graph.
Each of a simple simultaneous equations has infinite number of solutions. Graph of each equation is a straight line. Coordinates of each point of the straight line satisfies the equation. To indicate a graph, two or more than two points are necessary.
Now we shall try to solve graphically the following system of equations:
2x + y =3…(1)
4x + 2y =6…(2)
From equation (1) we get, y = 3 – 2x
Taking some values of z in the equation, we find the corresponding values of y I and make the adjoining table:
∴ three points on the graph of the equation are: (- 1, 5), (0, 3)(3, – 3) .
Again, from equation (2) we get, 2y = 6 – 4x or, y = (6 – 4x)/2
Taking some values of x in the equation, we find the corresponding values of y and make the adjoining table:
∴ three points on the graph of the equation are: (-2,7), (0,3) (6,-9).
In a graph paper let XOX’ and YOY’ be respectively the X – axis and Y-axis and O is the origin. We take each side of the smallest squares of the graph paper as unit along with both axes. Now, we plot the points (-1,5), (0,3) and (3,-3) obtained from equation (1) and join them each other. The graph is a straight line.
Again, we plot the points (-2,7), (0,3) (6,-9) obtained from equation (2) and join them each other. In this case also the graph is a straight line.
But we observe that the two straight lines coincide and they have turned into the one straight line. Again, if both sides of equation (2) are divided by 2, we get he equation (1). That is why the graphs of the two equations coincide.
Here, the system of equations,
4x+2y=6… (2), are consistent and mutually dependent. Such system of equations have infinite number of solutions and its graph is a straight line.
2x – y =4…(1)
4x – 2y =12…(2)
From equation (1) we get, y = 2x – 4
Taking some values of x in the equation, we find the corresponding values of y and make the adjoining table:
[[x, – 1, 0, 4], [y, – 6, – 4, 4]]
∴ three points on the graph of the equation are: (-1,-6), (0,-4), (4, 4).
Again, from equation (2) we get,
4x – 2y = 12
or ,2x-y=6 [dividing both sides by 2]
or, y = 2x – 6
Taking some values of x in the equation, we find the corresponding values of y and make the adjoining table:
∴ three points on the graph of the equation are: (0, -6), (3,0), (6,6). In a graph paper let XOX’ and YOY’ be respectively the X – axis and Y axis and O is the origin.
We take each side of smallest squares of the graph paper as unit along with both axes. Now, we plot the points (-1, -6), (0,-4) (4, 4) obtained from equation (1) and join them each other. The graph is a straight line.
= integrate 1/2 * integrate 1/2 dx from 0 to ∞ dx from 0 to ∞ (6, 9)
Again, we plot the points (0,-6), (3,0), (6,6) obtained from equation (2) and join them each other. In this case also the graph is a straight line.
We observe in the graph, though each of the given equations has separately infinite number of solutions, they have no common solution as system of simultaneous equations. Further, we observe that the graphs of the two equations are straight lines parallel to each other. That is, the lines will never intersect each other.
Therefore, there will be no common point of the lines. In this case we say, such system of equations has no solution. We know, such system of equations is inconsistent and independent of each other. Now, we shall solve the system of two consistent and independent equations by graphs. Graphs of two such equations in two variables intersect at a point.
Both the equations will be satisfied by the coordinates of that point. The very coordinates of the point of intersection will be the solution of the two equations.
Example 8. Solve and show the solution in graph:
2x + y = 8
3x – 2y = 5
Solution: Given equations are:
2x + y – 8 =0…(1)
3x – 2y – 5 =0…(2)
By the method of cross-multiplication we get,
x/(1(- 5) – (- 2)(- 8)) = y/((- 8) * 3 – (- 5) * 2) = 1/(2(- 2) – 3 * 1)
or, x/(- 5 – 16) = y/(- 24 + 10) = 1/(- 4 – 3)
or, x/- 21 = y/- 14 = 1/- 7
or, x/21 = y/14 = 1/7
x/21 = 1/7
Again, x = 21/7 = 3 y/14 = 1/7 y = 14/7 = 2
∴ solution: (x, y) = (3, 2)
Let XOX’ and YOY’ be X-axis and Y-axis respectively and O be the origin. Taking each two sides of the smallest squares along with both axes of the graph paper as one unit, we plot the point (3,2).
Example 9. Solve with the help of graphs:
3x – y = 3
5x + y = 21
Solution: Given equations are:
3x – y =3…(1)
5x + y = 21 (2) From equation (1) we get, 3x – y = 3, or , y = 3x – 3
Taking some values of x in the equation, we find the corresponding values of y and make the adjoining table:
∴ three points on the graph of the equation are: (-1,-6), (0,-3), (3,6)
Again, from equation (2) we get, 5x + y = 21 , or, y = 21 – 5x
Taking some values of x in the equation, we find the corresponding values of y and make the adjoining table:
∴ three points on the graph of the equation are: (3, 6), (4, 1), (5,-4). In a graph paper let XOX’ and YOY’ be respectively the X-axis and Y-axis and O is the origin.
We take each side of the smallest squares of the graph paper as unit along with both axes. Now, we plot the points (-1,-6), (0,-3), (3,6) obtained from equation (1) and join them each other. The graph is a straight line.
Again, we plot the points (3, 6), (4, 1), (5,-4) obtained from equation (2) and join them each other. In this case also the graph is a straight line.
Let the two straight lines intersect each other at
- It is seen from the picture that the coordinates of P are (3,6).
∴ solution: (x, y) = (3, 6)
Example 10. Solve by graphical method:
2x + 5y = – 14
4x – 5y = 17
Solution: Given equations are:
2x + 5y =-14…(1)
4x – 5y =17…(2)
y = (- 2x – 14)/5 Taking some convenient values of z in the equation, we find the corresponding values of y and make the adjoining table:
From equation (1) we get, 5y = – 14 – 2x
three points on the graph of the equation are: (3,-4), (1/2, – 3) , (-2,-2).
y = (4x – 17)/5
Taking some convenient values of in the equation, we find the corresponding values of y and make the adjoining table:
Again, from equation (2) we get, 5y = 4x – 17 , or,
∴ three points on the graph of the equation are: (3,-1), (1/2, – 3), (- 2, – 5)
In a graph paper let XOX’ and YOY’ be respectively the X – axis and Y-axis and O is the origin. We take two sides of the smallest squares of the graph paper as unit along with both axes. (1/2, – 3)
Now, we plot the points (3,-4), (-2,-2) obtained from equation (1) and join them each other. The graph is a straight line.
Again, we plot the points (3,-1), (1/2, – 3),(-2,-5) obtained from equation (2) and join them each other. In this case also the graph is a straight line.
Let the two straight lines intersect each other at P. It is seen from the picture that the coordinates of P are (1/2, – 3)
solution: (x, y) = (1/2, – 3)
Example 11. Solve with the help of graphs:
3 – 3/2 * x = 8 – 4x
Solution: Given equation is 3 – 3/2 * x = 8 – 4x
3 Let, y=3x-8-4x
y=3 – 3/2 * x …(1) and y=8 – 4x …(2)
Taking some values of r in equation (1), we find the corresponding values of y and make the adjoining table:
∴ three points on the graph of the equation are: (-2, 6), (0,3), (2, 0).
Again, taking some values of x in equation (2), we find the corresponding values of y and make the adjoining table:
In a graph paper let XOX’ and YOY’ be respectively the X-axis and Y axis and O is the origin.
We take each side of the smallest squares of the graph paper as unit along with both axes. Now, we plot the points (-2,6), (0,3), (2, 0) obtained from equation (1) and join them each other. The graph is a straight line.
Again, we plot the points (1,4), (2, 0), (3,-4) obtained from equation (2) and join them each other. In this case also the graph is a straight line.
Let the two straight lines intersect each other at P. It is seen from the picture that the coordinates of P are (2,0).
∴ solution: x = 2
Work: Find four points on the graph of the equation 2x – y – 3 = 0 in terms of a table. Then, taking unit of a fixed length on the graph paper, plot the points and join them each other. Is the graph a straight line?
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