Class 10

### Topic Covered

♦ Introduction
♦ Pair of Linear Equations in Two Variables

### Introduction

● You must have come across situations like the one given below :

● Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. If each ride costs Rs 3, and a game of Hoopla costs Rs 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent Rs 20. May be you will try it by considering different cases. If she has one ride, is it possible? Is it possible to have two rides?

● Let us try this approach, Denote the number of rides that Akhila had by x, and the number of times she
played Hoopla by y. Now the situation can be represented by the two equations:

y = 1/2 x ............  (1)

3x + 4y = 20............. (2)

● There are several ways to find the solutions of this pair of equations, which we will study in this chapter.

### Pair of Linear Equations in Two Variables

● As the following are examples of linear equations in two variables:

2x + 3y = 5

x – 2y – 3 = 0

and x – 0.y = 2, i.e., x = 2

● You also know that an equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers, and a and b are not both zero, is called a linear equation in two variables x and y. (We often denote the condition a and b are not both zero by a^2 + b^2 ≠ 0).

=> For example, let us substitute x = 1 and y = 1 in the left hand side (LHS) of the equation 2x + 3y = 5. Then

LHS = 2(1) + 3(1) = 2 + 3 = 5, which is equal to the right hand side (RHS) of the equation.

● Therefore, x = 1 and y = 1 is a solution of the equation 2x + 3y = 5.

Now let us substitute x = 1 and y = 7 in the equation 2x + 3y = 5. Then,

LHS = 2(1) + 3(7) = 2 + 21 = 23 which is not equal to the RHS.

● Therefore, x = 1 and y = 7 is not a solution of the equation.
● Geometrically, It means that the point (1, 1) lies on the line representing the equation 2x + 3y = 5, and the point (1, 7) does not lie on it. So, every solution of the equation is a point on the line representing it.

● In fact, this is true for any linear equation, that is, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa.

Now, consider Equations (1) and (2) given above. These equations, taken together, represent the information we have about Akhila at the fair.

● These two linear equations are in the same two variables x and y. Equations like these are called a pair of linear equations in two variables.

=> Let us see what such pairs look like algebraically.

● The general form for a pair of linear equations in two variables x and y is

a_(1) x + b_(1) y + c_1 = 0

and a_(2) x + b_(2) y + c_2 = 0,

where a_1, b_1, c_1, a_2, b_2, c_2 are all real numbers and a_(1)^2 + b_(1)^2 ≠ 0 , a_(2)^2 + b_(2)^2 ≠ 0 .

● Some examples of pair of linear equations in two variables are:

2x + 3y – 7 = 0 and 9x – 2y + 8 = 0

5x = y and –7x + 2y + 3 = 0

x + y = 7 and 17 = y

● You have also studied in Class IX that given two lines in a plane, only one of the following three possibilities can happen:

(i) The two lines will intersect at one point.

(ii) The two lines will not intersect, i.e., they are parallel.

(iii) The two lines will be coincident.

We show all these possibilities in Fig. 3.1:

In Fig. 3.1 (a), they intersect.
In Fig. 3.1 (b), they are parallel.
In Fig. 3.1 (c), they are coincident.

=> Both ways of representing a pair of linear equations go hand-in-hand—the algebraic and the geometric ways. Let us consider some examples.
Q 3149167913

Let us take the example given in Section 3.1. Akhila goes to a fair with
Rs 20 and wants to have rides on the Giant Wheel and play Hoopla. Represent this
situation algebraically and graphically (geometrically).
Class 10 Chapter 3 Example 1
Solution:

The pair of equations formed is :

 y = 1/2 x

i.e., x-2y =0 (1)

3x + 4y = 20 (2)

Let us represent these equations graphically. For this, we need at least two
solutions for each equation. We give these solutions in Table 3.1.

Recall from Class IX that there are infinitely many solutions of each linear
equation. So each of you can choose any two values, which may not be the ones we
have chosen. Can you guess why we have chosen x = 0 in the first equation and in the
second equation? When one of the variables is zero, the equation reduces to a linear

equation in one variable, which can be solved easily. For instance, putting x = 0 in
Equation (2), we get 4y = 20, i.e., y = 5. Similarly, putting y = 0 in Equation (2), we get

3x =20 , i.e., x = 20/3 . But as 20/3 is

not an integer, it will not be easy to
plot exactly on the graph paper. So,
we choose y = 2 which gives x = 4,
an integral value.

Plot the points A(0, 0), B(2, 1)
and P(0, 5), Q(4, 2), corresponding
to the solutions in Table 3.1. Now
draw the lines AB and PQ,
representing the equations
x – 2y = 0 and 3x + 4y = 20, as
shown in Fig. 3.2.

In Fig. 3.2, observe that the two lines representing the two equations are
intersecting at the point (4, 2). We shall discuss what this means in the next section.
Q 3159167914

Romila went to a stationery shop and purchased 2 pencils and 3 erasers
for Rs. 9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and
she also bought 4 pencils and 6 erasers of the same kind for Rs. 18. Represent this
situation algebraically and graphically.
Class 10 Chapter 3 Example 2
Solution:

Let us denote the cost of 1 pencil by Rs. x and one eraser by Rs. y. Then the
algebraic representation is given by the following equations:

2x + 3y = 9 (1)

4x + 6y = 18 (2)

To obtain the equivalent geometric representation, we find two points on the line
representing each equation. That is, we find two solutions of each equation.

These solutions are given below in Table 3.2.

We plot these points in a graph
paper and draw the lines. We find that
both the lines coincide (see Fig. 3.3).
This is so, because, both the
equations are equivalent, i.e., one can
be derived from the other .
Q 3179167916

Two rails are represented by the equations
x + 2y – 4 = 0 and 2x + 4y – 12 = 0.
Represent this situation geometrically.
Class 10 Chapter 3 Example 3
Solution:

Two solutions of each of
the equations :

x + 2y – 4 = 0 (1)

2x + 4y – 12 = 0 (2)

are given in Table 3.3

To represent the equations graphically, we plot the points R(0, 2) and S(4, 0), to
get the line RS and the points P(0, 3) and Q(6, 0) to get the line PQ.

We observe in Fig. 3.4, that the
lines do not intersect anywhere, i.e.,
they are parallel.

So, we have seen several
situations which can be represented
by a pair of linear equations. We
have seen their algebraic and
geometric representations. In the
next few sections, we will discuss
how these representations can be
used to look for solutions of the pair
of linear equations.