♦ Graphical Method of Solution of a Pair of Linear Equations

Let us look at the earlier examples one by one.

♦ In the situation of Example 1, find out how many rides on the Giant Wheel Akhila had, and how many times she played Hoopla.

In Fig. 3.2, you noted that the equations representing the situation are geometrically shown by two lines intersecting at the point `(4, 2).` Therefore, the point `(4, 2)` lies on the lines represented by both the equations `x – 2y = 0` and `3x + 4y = 20.` And this is the only common point.

Let us verify algebraically that x = 4, y = 2 is a solution of the given pair of equations. Substituting the values of x and y in each equation, we get `4 – 2 × 2 = 0` and `3(4) + 4(2) = 20.` So, we have verified that `x = 4, y = 2` is a solution of both the equations.

`=> "Since (4, 2) is the only common point on both the lines, "`

`"there is one and only one solution for this pair of linear equations in two variables."`

Thus, the number of rides Akhila had on Giant Wheel is `4` and the number of times she played Hoopla is `2.`

♦ In the situation of Example 2, can you find the cost of each pencil and each eraser?

`=>` In Fig. 3.3, the situation is geometrically shown by a pair of coincident lines. The solutions of the equations are given by the common points.

`=>` From the graph, we observe that every point on the line is a common solution to both the equations. So, the equations `2x + 3y = 9` and `4x + 6y = 18` have infinitely many solutions. This should not surprise us, because if we divide the equation 4x + 6y = 18 by 2 , we get `2x + 3y = 9,` which is the same as Equation (1). That is, both the equations are equivalent. From the graph, we see that any point on the line gives us a possible cost of each pencil and eraser. For instance, each pencil and eraser can cost Rs. 3 and Rs. 1 respectively. Or, each pencil can cost Rs. 3.75 and eraser can cost Rs. 0.50, and so on.

♦ In the situation of Example 3, can the two rails cross each other?

`=>` In Fig. 3.4, the situation is represented geometrically by two parallel lines. Since the lines do not intersect at all, the rails do not cross. This also means that the equations have no common solution.

`=>` A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.

`=>` We can now summaries the behavior of lines representing a pair of linear equations in two variables and the existence of solutions as follows :

(i) the lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations).

(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).

(iii) the lines may be coincident. In this case, the equations have infinitely many

solutions [dependent (consistent) pair of equations].

`=>` Let us now go back to the pairs of linear equations formed in Examples 1, 2, and 3, and note down what kind of pair they are geometrically.

(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)

(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)

(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)

`=>` Let us now write down, and compare, the values of ` (a_1)/(a_2) , (b_1)/(b_2) ` and ` (c_1)/(C_2)` in all the three examples. Here, `a_1, b_1, c_1` and `a_2, b_2, c_2` denote the coefficents of equations given in the general form in Section 3.2.

`=>` From the table above, you can observe that if the lines represented by the equation

`a_1 x + b_1 y + c_1 = 0`

and `a_2 x + b_2 y + c_2 = 0`

are (i) intersecting, then `color{orange}{a_1/a_2 ≠ b_1/b_2}` .

(ii) coincident, then ` color{orange}{a_1/a_2 = b_1/b_2 =c_1/c_2}` .

(iii) parallel, then ` color{orange}{a_1/a_2 = b_1/b_2 ≠ c_1/c_2}`.

In fact, the converse is also true for any pair of lines. You can verify them by considering some more examples by yourself.

♦ In the situation of Example 1, find out how many rides on the Giant Wheel Akhila had, and how many times she played Hoopla.

In Fig. 3.2, you noted that the equations representing the situation are geometrically shown by two lines intersecting at the point `(4, 2).` Therefore, the point `(4, 2)` lies on the lines represented by both the equations `x – 2y = 0` and `3x + 4y = 20.` And this is the only common point.

Let us verify algebraically that x = 4, y = 2 is a solution of the given pair of equations. Substituting the values of x and y in each equation, we get `4 – 2 × 2 = 0` and `3(4) + 4(2) = 20.` So, we have verified that `x = 4, y = 2` is a solution of both the equations.

`=> "Since (4, 2) is the only common point on both the lines, "`

`"there is one and only one solution for this pair of linear equations in two variables."`

Thus, the number of rides Akhila had on Giant Wheel is `4` and the number of times she played Hoopla is `2.`

♦ In the situation of Example 2, can you find the cost of each pencil and each eraser?

`=>` In Fig. 3.3, the situation is geometrically shown by a pair of coincident lines. The solutions of the equations are given by the common points.

`=>` From the graph, we observe that every point on the line is a common solution to both the equations. So, the equations `2x + 3y = 9` and `4x + 6y = 18` have infinitely many solutions. This should not surprise us, because if we divide the equation 4x + 6y = 18 by 2 , we get `2x + 3y = 9,` which is the same as Equation (1). That is, both the equations are equivalent. From the graph, we see that any point on the line gives us a possible cost of each pencil and eraser. For instance, each pencil and eraser can cost Rs. 3 and Rs. 1 respectively. Or, each pencil can cost Rs. 3.75 and eraser can cost Rs. 0.50, and so on.

♦ In the situation of Example 3, can the two rails cross each other?

`=>` In Fig. 3.4, the situation is represented geometrically by two parallel lines. Since the lines do not intersect at all, the rails do not cross. This also means that the equations have no common solution.

`=>` A pair of linear equations which has no solution, is called an inconsistent pair of linear equations. A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations. A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.

`=>` We can now summaries the behavior of lines representing a pair of linear equations in two variables and the existence of solutions as follows :

(i) the lines may intersect in a single point. In this case, the pair of equations has a unique solution (consistent pair of equations).

(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).

(iii) the lines may be coincident. In this case, the equations have infinitely many

solutions [dependent (consistent) pair of equations].

`=>` Let us now go back to the pairs of linear equations formed in Examples 1, 2, and 3, and note down what kind of pair they are geometrically.

(i) x – 2y = 0 and 3x + 4y – 20 = 0 (The lines intersect)

(ii) 2x + 3y – 9 = 0 and 4x + 6y – 18 = 0 (The lines coincide)

(iii) x + 2y – 4 = 0 and 2x + 4y – 12 = 0 (The lines are parallel)

`=>` Let us now write down, and compare, the values of ` (a_1)/(a_2) , (b_1)/(b_2) ` and ` (c_1)/(C_2)` in all the three examples. Here, `a_1, b_1, c_1` and `a_2, b_2, c_2` denote the coefficents of equations given in the general form in Section 3.2.

`=>` From the table above, you can observe that if the lines represented by the equation

`a_1 x + b_1 y + c_1 = 0`

and `a_2 x + b_2 y + c_2 = 0`

are (i) intersecting, then `color{orange}{a_1/a_2 ≠ b_1/b_2}` .

(ii) coincident, then ` color{orange}{a_1/a_2 = b_1/b_2 =c_1/c_2}` .

(iii) parallel, then ` color{orange}{a_1/a_2 = b_1/b_2 ≠ c_1/c_2}`.

In fact, the converse is also true for any pair of lines. You can verify them by considering some more examples by yourself.

Q 3109178018

Check graphically whether the pair of equations

`x + 3y = 6` (1)

and `2x – 3y = 12` (2)

is consistent. If so, solve them graphically.

Class 10 Chapter 3 Example 4

`x + 3y = 6` (1)

and `2x – 3y = 12` (2)

is consistent. If so, solve them graphically.

Class 10 Chapter 3 Example 4

Let us draw the graphs of the Equations (1) and (2). For this, we find two

solutions of each of the equations, which are given in Table 3.5

Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph

paper, and join the points to form the lines AB and PQ as shown in Fig. 3.5.

We observe that there is a point B (6, 0) common to both the lines

AB and PQ. So, the solution of the pair of linear equations is x = 6 and

y = 0, i.e., the given pair of equations is consistent.

Q 3119278110

Graphically, find whether the following pair of equations has no solution,

unique solution or infinitely many solutions:

`5x - 8 y +1 = 0` (1)

`3x - 24/5 y + 3/5 = 0` (2)

Class 10 Chapter 3 Example 5

unique solution or infinitely many solutions:

`5x - 8 y +1 = 0` (1)

`3x - 24/5 y + 3/5 = 0` (2)

Class 10 Chapter 3 Example 5

Multiplying Equation (2) by `5/3` , we get

`5x - 8 y +1 = 0`

But, this is the same as Equation (1). Hence the lines represented by Equations (1)

and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions.

Plot few points on the graph and verify it yourself.

Q 3139278112

Champa went to a ‘Sale’ to purchase some pants and skirts. When her

friends asked her how many of each she had bought, she answered, “The number of

skirts is two less than twice the number of pants purchased. Also, the number of skirts

is four less than four times the number of pants purchased”. Help her friends to find

how many pants and skirts Champa bought.

Class 10 Chapter 3 Example 6

friends asked her how many of each she had bought, she answered, “The number of

skirts is two less than twice the number of pants purchased. Also, the number of skirts

is four less than four times the number of pants purchased”. Help her friends to find

how many pants and skirts Champa bought.

Class 10 Chapter 3 Example 6

Let us denote the number of pants by x and the number of skirts by y. Then

the equations formed are :

`y = 2x – 2` (1)

and `y = 4x – 4` (2)

Let us draw the graphs of

Equations (1) and (2) by finding two

solutions for each of the equations.

They are given in Table 3.6.

Plot the points and draw the lines passing through them to represent the equations,

as shown in Fig. 3.6.

The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution

of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did

not buy any skirt.

Verify the answer by checking whether it satisfies the conditions of the given

problem.