Pair of Linear Equations in two variables
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 a2+ b2 ≠ 0)
For example, let us substitute x = 2 and y = 2 in the left hand side (LHS) of the equation 3x + 4y = 14
Then
LHS = 3(2) + 4(2)
= 6 + 8
= 14,
which is equal to the right hand side (RHS) of the equation.
∴ x = 2 and y = 2 is a solution of the equation 3x + 4y = 14
Now let us substitute x = 1 and y = 1 in the equation 3x + 4y = 14.
Then,
LHS = 3(1) + 4(1)
= 3 + 4
= 7
which is not equal to the RHS
∴ x = 1 and y = 1 is not a solution of the equation.
Geometrically, what does this mean? It means that the point (2, 2) lies on the line representing the equation 3x + 4y = 14, and the point (1, 1) 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.
The general form for a pair of linear equations in two variables x and y is
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,
where a1, b1, c1, a2, b2, c2 are all real numbers and
a12 + b12 ≠ 0,
a22 + b22 ≠ 0.
Example 5x + 34y – 7 = 0 and 5x – 22y + 9 = 0
To summarise the behaviour of lines representing a pair of linear equations in two variables and the existence of solutions
(i) x – y = 1 and x + y = 3 (The lines intersect) 
(ii) the lines may be parallel. In this case, the equations have no solution (inconsistent pair of equations).
x + y = 2 and x + y = 3 
(iii) the lines may be coincident. In this case, the equations have infinitely many solutions [dependent (consistent) pair of equations].
x + y = 2 and 2x + 2y = 4 
Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0,
(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
Substitution Method : We shall explain the method of substitution by taking some examples.
Solve the following pair of equations by substitution method:
x – y = 1 ⇒ (1)
x + y = 3 ⇒(2)
Solution :
Step 1 :
We pick either of the equations and write one variable in terms of the other.
Let us consider the Equation (2)
x + y = 3
and write it as x = 3 – y
Step 2 :
Substitute the value of x in Equation (1)
We get
(3 – y) – y = 1
3 – 2y = 1
2y = 2
∴ y = 1
Step 3 :
Substituting this value of y in Equation (1), we get
x = 2
Real Life application:
You have a specific area of plot where you are building your home. At one corner you plan to have experimental room. You don't know the dimension of it, but there is a relation where in area of a room gets reduced by 9 square units when length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the experimental room
Elimination Method
2x – y = 3 ⇒ (1)
x + y = 3 ⇒(2)
Step 1 :
Multiply Equation (1) by 1 and Equation (2) by 2 to make the coefficients of x equal. Then we get the equations:
2x – y = 3 ⇒(3)
2x + 2y = 6 ⇒(4)
Step 2 :
Subtract Equation (3) from Equation (4) to eliminate x, because the coefficients of y are the same. So, we get
3y = 3
y = 1
Step 3 :
Substituting this value of y in (1), we get
x = 2
Cross multiplication
Lines represented by
a1x + b1y + c1 = 0 and
a2x + b2y + c2 = 0
We can express it as
x/(b1c2 - b2c1)
= y/(c1a2 - c2a1)
= 1/(a1b2 - a2b1)
Real Life application:
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Let the speed of 1st car and 2nd car be u and v
Respective speed of both cars while they are travelling in same direction = (u - v) km/h
Respective speed of both cars while they are travelling towards each other = (u + v) km/h
Given: When the travel in same direction they meet after 5 hours
but require 1 hour when they travel towards each other.
Place is 100km apart
5(u - v) = 100
u - v = 20 km/h ⇒ (i)
1(u + v) = 100 ⇒ (ii)
Adding both the equations, we get
2u = 120
u = 60 km/h ⇒ (iii)
Substituting this value in equation (ii), we obtain
v = 40 km/h
∴ speed of one car = 60 km/h and speed of other car = 40 km/h
