Introduction to Pair of Linear Equations in Two Variables

Linear equations in two variables are equations which can be

expressed as ax + by + c = 0, where a, b and c are real numbers and

both a, and b are not zero. The solution of such equations is a pair of

values for x and y which makes both sides of the equation equal. 

Introduction

Let’s look at the solutions of some linear equations in two variables.

Consider the equation 2x + 3y = 5. There are two variables in this

equation, x and y.

● Scenario 1: Let’s substitute x = 1 and y = 1 in the Left Hand

Side (LHS) of the equation. Hence, 2(1) + 3(1) = 2 + 3 = 5 =

RHS (Right Hand Side). Hence, we can conclude that x = 1 and

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

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

● Scenario 2: Let’s substitute x = 1 and y = 7 in the LHS of the

equation. Hence, 2(1) + 3(7) = 2 + 21 = 23 ≠ RHS. Therefore, x

= 1 and y = 7 is not a solution of the equation 2x + 3y = 5.

Geometrically, this means that the point (1, 1) lies on the line

representing the equation 2x + 3y = 5. Also, the point (1, 7) does not

lie on this line. In simple words, every solution of the equation is a

point on the line representing it.

To generalize, 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.

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Pair of Linear Equations in Two Variables

Here is a situation: The number of times Ram eats a mango is half the

number of rides he eats an apple. He goes to the market and spends

Rs. 20. If one mango costs Rs.3 and one apple costs Rs.4, then how

many mangoes and apples did Ram eat?

Let’s say that the number of apples that Ram ate is y and the number

of mangoes is x. Now, the situation can be represented as follows:

y = (½)x … {since he ate mangoes (x) which were half the number of

apples (y)}

3x + 4y = 20 … {since each apple (y) costs Rs.4 and mango (x) costs

Rs.3}

Both these equation together represent the information about the

situation. Also, these two linear equations are in the same variables, x

and y. These are known as a ‘Pair of Linear Equations in Two

Variables’.

To generalize them, a pair of linear equations in two variables x and y

is:

a1x + b1 y + c1 = 0 and a2x + b2 y + c2 = 0.

Where a1, b1, c1, a2, b2, c2 are all real numbers and a12+ b12 ≠ 0, a22+

b22 ≠ 0. Some examples of a pair of linear equations in two variables

are:

● 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0

● 5x = y and –7x + 2y + 3 = 0

Geometric Representation of a Pair of Linear Equations in Two Variables

By now, we know that the geometrical or graphical representation of

linear equations in two variables is a straight line. Hence, a pair of

linear equations in two variables will be two straight lines which are

considered together. We also know that when there are two lines in a

plane:

● The two lines will intersect at one point. {Fig.1 (a)}

● They will not intersect, i.e., they are parallel. {Fig.1 (b)}

● The two lines will be coincident. {Fig.1 (c)}

Fig. 1

Linear equations can be represented both algebraically and

geometrically. Let’s see how. Going back to our earlier example of

Ram, let’s try to represent the situation both algebraically and

geometrically.

Solution: Algebraic Representation

The pair of equations formed is: y = (1/2)x

So, 2y = x

Hence, x – 2y = 0 … (1)

3x + 4y = 20 … (2)

Geometric Representation

To represent these equations graphically we need at least two

solutions for each equation. These solutions are listed in the tables

below:

x 0 2

y = (1/2)x 0 1

x 0 4

y = (20 – 3x)/4 5 2

Now, we take a graph paper and plot the points A(0, 0), B(2, 1) and

P(0, 5), Q(4, 2), corresponding to the solutions in tables above. Next,

we draw the lines AB and PQ as shown below.

Fig. 2

This is the geometric representation of the equations x – 2y = 0 and 3x

+ 4y = 20.

Solved Examples for You

Question: Aftab tells his daughter, “Seven years ago, I was seven

times as old as you were then. Also, three years from now, I shall be

three times as old as you will be.” (Isn’t this interesting?) Represent

this situation algebraically.

Solution: Let Aftab’s present age be x and his daughter’s present age

by y.

7 years ago, Aftab’s age = x – 7

His daughter’s age = y – 7

Also, according to the question,

(x – 7) = 7(y – 7)

So, x – 7 = 7y – 49

Hence, x – 7y = 42

Now, three years later,

Aftab’s age = x + 3

His daughter’s age = y + 3

Also, according to the question,

(x + 3) = 3(y + 3)

So, x + 3 = 3y + 9

Hence, x – 3y = 6

Therefore, the situation can be algebraically represented as:

● x – 7y = 42

● x – 3y = 6

Solution of Linear Equations in Two Variables

Now that we know what Linear Equations are and the ways of

converting a statement in the form of the linear equation and the

various terminologies associated with it. We can discuss the methods

of solving linear equations for finding the required solution. Solving

linear equations is very simple. Let us get to know the methods to do

it! 

Methods for Solving Linear Equations

Linear equations can be solved graphically as well as algebraically.

Let us learn about both of them.

Graphical Method

In Graphical method, we draw the lines for the given pair of equations

with possible satisfying values on a graph and find out the case being

satisfied i.e., whether the drawn lines are intersecting at a point

(consistent solution) or are parallel with each other (inconsistent

solution) or are coincident (dependent solution).

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Algebraic Methods

● Substitution Method: We substitute one of the given equations

in another one by substituting one variable in the form of the

other. Now, the equation will contain only one variable and

then solve it accordingly to get the desired result.

● Elimination Method: As the name suggests, in the elimination

method, we try to eliminate one of the variables from the given

set of equations. Solving it will give us the desired result.

● Cross-Multiplication Method: The general form of a pair of

linear equations in two variables is:

a1x1 + b1y1 = c1 … (i)

a2x2 + b2y2 = c2 … (ii)

In this method, we multiply equation (i) by the coefficient of y2 (or,

x2) i.e., b2 (or, a2) & equation (ii) by that of y1 (or, x1) i.e., b1 (or, a1)

& eliminate one of the variables and solve accordingly. The name is

given as the multiplication in the equations with the coefficients of the

variables are done in a cross fashion.

Check out our detailed article on Linear Equations with 2 variables

here.

Solved Linear Equations Examples

Solve by the Graphical method: x + y = 16, & x – y = 4

Solution: The two solutions of each equation

x 2 4 6 8

y = 16 – x 14 12 10 8

x 2 4 6 8

y = x – 4 -2 0 2 4

From the graph, we find the common point of intersection, i.e., (10,6).

More Solved Examples for You

Question: A number consists of two digits such that the digit in the

ten’s place is less by 2 than the digit in unit’s place. Three times the

number added to 5/7 times the number obtained by reversing the digits

equal to 108. What is the sum of the digit of the number?

Solution: Let the two digits of the number be x & y in which x is in

the ten’s place & y is in the unit’s place. The number can be written as

10 × x + y (as 36 = 3 × 10 + 6).

Case 1: x = y – 2 …(i)

Case 2: The number obtained by reversing the digits = 10 × y + x (as

63 = 6 × 10 + 3).

A.T.Q., 3(10x + y) + 5 (10y + x) / 7 = 108

⇒ 30x + 3y + 50y / 7 + 5x / 7 = 108

⇒ 210x + 21y + 5x + 50y = 756

⇒ 215x + 71y = 756 …(ii)

Solving (i) and (iii), we get x ≈ 2 & y ≈ 4. The sum of the numbers = 2

+ 4 = 6.

Question: The sum of two numbers is 20. Five times one number is

equal to 4 times the other. Find the bigger of the two numbers.

Solution: Suppose the two numbers be x & y. We have,

x + y = 20 …(i)

5x = 4y …(ii)

Multiplying (i) by 5 & subtracting (ii) from it, we get

5x + 5y = 100

5x – 4y = 0

– + = –

—————

9y = 100

y = 100/9 or y ≈ 11 and substituting this value of x in any of the above

equations, we get x ≈ 9. Thus, the bigger of the two numbers is 11.

Consistency of Pair of Linear Equations in Two Variables

A pair of linear equations in two variables have the same set of

variables across both the equations. These equations are solved

simultaneously to arrive at a solution. In this article, we will look at

the various types of solutions of equations in two variables. 

Types of Linear Equations in Two Variables

Solutions of linear equations in two variables can be of three types

1. Single Solution

2. Infinite Solutions

3. No Solution

Understanding these types will help us in solving linear equations in

two variables effectively. We will look at each of them in details.

Type 1: A single solution of a pair of linear equations in two variables

Consider the following pair of linear equations in two variables,

● x – 2y = 0

● 3x + 4y = 20

The solution of this pair would be a pair (x, y). Let’s find the solution,

geometrically. The tables for these equations are:

x 0 2

y = (1/2)x 0 1

x 0 4

y = (20 – 3x)/4 5 2

Now, take a graph paper and plot the following points:

● A(0, 0)

● B(2, 1)

● P(0, 5)

● Q(4, 2)

Next, draw the lines AB and PQ as shown below.

From the figure above, you can see that the two lines intersect at the

point Q (4, 2). Therefore, point Q lies on the lines represented by both

the equations, x – 2y = 0 and 3x + 4y = 20. Hence, (4, 2) is the

solution of this pair of equations in two variables. Let’s verify it

algebraically:

● x – 2y = 4 – 2(2) = 4 – 4 = 0 = RHS

● 3x + 4y = 3(4) + 4(2) = 12 + 8 = 20 = RHS

Further, from the graph, you can see that point Q is the only common

point between the two lines. Hence, this pair of equations has a single

solution.

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Type 2: Infinite solutions of a pair of linear equations in two variables

Consider the following pair of linear equations in two variables,

● 2x + 3y = 9

● 4x + 6y = 18

The solution of this pair would be a pair (x, y). Let’s find the solution,

geometrically. The tables for these equations are:

x 0 4.5

y = (9 – 2x)/3 3 0

x 0 3

y = (18 – 4x)/6 3 1

Now, take a graph paper and plot the following points:

● A(0, 3)

● B(4.5, 0)

● P(3, 1)

Next, draw the lines AB and AP as shown below.

From the figure above, you can see that the two lines coincide.

Therefore, there is no point of intersection between these two lines.

Every point on the line represented by 2x + 3y = 9 is present on the

line represented by 4x + 6y = 18. Hence, this pair of equations has an

infinite number of solutions.

Type 3: No solution of a pair of linear equations in two variables

Consider the following pair of linear equations in two variables,

● x + 2y = 4

● 2x + 4y = 12

The solution of this pair would be a pair (x, y). Let’s find the solution,

geometrically. The tables for these equations are:

x 0 4

y = (4 – x)/2 2 0

x 0 6

y = (12 – 2x)/4 3 0

Now, take a graph paper and plot the following points:

● A(0, 2)

● B(4, 0)

● P(0, 3)

● Q(6, 0)

Next, draw the lines AB and PQ as shown below.

From the figure above, you can see that the two lines are parallel to

each other. Therefore, these lines don’t intersect at all. Hence, this pair

of equations has no solution.

Consistency of a Pair of Linear Equations in Two Variables

From the three examples above, we define the following terms

● Inconsistent pair of linear equations: A pair of linear equations

which has no solution. (Type 3 explained above)

● Consistent pair of linear equations: A pair of linear equations

which has a solution. (Type 1 and 2 explained above)

○ Independent pair of linear equations – If the pair of

equations has only one solution.

○ Dependent pair of linear equations – If the pair of

equations has infinite solutions.

To summarise, the lines represented by

● (x – 2y = 0) and (3x + 4y = 20) intersect each other.

● 2x + 3y = 9 and 4x + 6y = 18 coincide with each other.

● ‘x + 2y = 4’ and ‘2x + 4y = 12’ are parallel to each other.

In General Form

If a1, b1, c1, a2, b2 and c2 are the coefficients of the equations in

general form, then we can write the following table by comparing the

values of a1/a2, b1/b2, and c1/c2 in all three examples.

Pair of lines

a1/a2

b1/b2

c1/c2 Comparison

Graphical Representati

on

Algebraic Interpretati

on

x – 2y = 0

1/3 – 2/4 0/20 (a1/a2) ≠ (b1/b2) Intersecting

lines One

solution 3x + 4y –

20 = 0

2x + 3y – 9 = 0

2/4 3/6 9/18 (a1/a2) = (b1/b2) = (c1/c2)

Coincident lines

Infinite solutions

4x + 6y – 18 = 0

x + 2y – 4 = 0

1/2 2/4 4/12 (a1/a2) = (b1/b2) ≠ (c1/c2) Parallel lines No solution

2x + 4y – 12 = 0

From the table above, you can observe that if the lines represented by

the equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are,

i. Intersecting, then (a1/a2) ≠ (b1/b2)

ii. Coincident, then (a1/a2) = (b1/b2) = (c1/c2)

iii.Parallel, then (a1/a2) = (b1/b2) ≠ (c1/c2)

It is important to note that the converse is also true.

Solved Examples for You

Question: Check graphically whether the pair of equations ‘x + 3y =

6’ and ‘2x – 3y = 12’ is consistent. If so, solve them graphically.

Solution: Let us draw the graphs of both the equations. The tables for

these equations are:

x 0 6

y = (6 – x)/3 2 0

x 0 3

y = (2x – 12)/3 – 4 – 2

Now, take a graph paper and plot the following points:

● A (0, 2)

● B (6, 0)

● P (0, – 4)

● Q (3, – 2)

Next, draw the lines AB and PQ as shown below.

From the figure above, you can see that the two lines intersect at the

point B (6, 0). Therefore, x = 6 and y = 0 is the solution of this pair of

equations in two variables. Hence, it is Consistent.