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Pair of Linear Equations in Two Variables: Notes, Formulae, Solutions

linear equation in two variable

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Linear equation in two variables:

When a polynomial equation is expressed in terms of ax + by + c = 0, where a, b, c are real numbers and x, y are variables also, a & b both are nonzero is known to be linear equation in two variables.

Solution of Pair of linear equations:

Each solution (x, y) of a linear equation in two variables, i.e. ax + by + c = 0,      it always corresponds to a sharp point on the line representing the equation, and vice versa.

Pair of Linear Equation in Two Variables

When two linear equations having same variables in both the equation is said to be pair of linear equations in two variables.

The most general form for representation of a pair of linear equations is
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
where a1, a2, b1, b2, c1, c2 are real numbers,
such that a12 + b12 ≠ 0, and a22 + b22 ≠ 0

We can represent and solve pair of linear equation in two variables by these two way:

  • Graphical method
  • Algebraic method

Graphical Method of solution of linear equation in two variables:

Two lines are used to represent pair linear equation in two variable on the graph:

Nature of Lines

Number of Solution

Type of equation

Intersecting lines Unique Solution Consistent
Coincide Infinite Solution Consistent (dependent)
Parallel No Solution Inconsistent

Algebraic Methods of solution of linear equation in two variables:

We have the following methods for finding the solution(s) of a pair of linear equations algebraically :

  • Substitution Method
  • Elimination Method
  • Cross-multiplication Method

Substitution method: Easy way to solve

Equation:
2x – y = 1
4x – 3y = 2

Step 1. From the two variable in the equation find the value of one variable “y” in terms of the other variable “x” from any one of the two equation whichever is easy to find.
as from 1st  y = 2x – 1

Step 2. Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved easily.
In some cases after substitution you will find that no variable is left if it satisfies the equality it will have infinite solution and if not then it will have no solution and thus inconsistent.
putting the value of y in 2nd equation
4x – 3(2x – 1)= 2
-2x= -1
x = ½

Step 3. Now we will substitute the value of x obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable y accordingly.

If a pair of linear equation in two variables is expressed as

a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
The following conditions obtained:

Comparision of ratios Graphical Representation Algebraic interpretation
\(\frac{{a}_{1}}{{a}_{2}}\) ≠ \(\frac{{b}_{1}}{{b}_{2}}\) Intersecting Lines Unique solution (Consistent)
\(\frac{{a}_{1}}{{a}_{2}}\) = \(\frac{{b}_{1}}{{b}_{2}}\) = \(\frac{{c}_{1}}{{c}_{2}}\) Coincidence Lines Infinite many solutions (Consistent)
\(\frac{{a}_{1}}{{a}_{2}}\) = \(\frac{{b}_{1}}{{b}_{2}}\) ≠ \(\frac{{c}_{1}}{{c}_{2}}\) Parallel Lines No Solution (Inconsistent)

 

Pair of Linear Equation Related Examples:

1. Two lines are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.

Solution:

First let’s take two solutions of each of the equations
x + 2y – 4 = 0
2x + 4y – 12 = 0

x 0 4
y = \(\frac{4-x}{2}\) 2 0

 

x 0 6
y = \(\frac{12 – 2x}{4}\) 3 0

Thus we plot the graph accordingly taking the coordinates.
We observe that the lines are parallel i.e., they do not intersect anywhere.

  1. Solve the following pair of equations by substitution method:
    7x – 15y = 2
    x + 2y = 3

Solution:

We take any of equations
x + 2y =3
x = 3 – 2y
Substitute the value of x in 1st Equation
7(3 – 2y) – 15y =2
– 29y = –19
y = \(\frac{19}{29}\)
Substituting this value of y in Equation 2nd equation
x = 3 – 2\(\(frac{19}{29})\) = \(\frac{49}{29}\)

Thus, x = \(\frac{49}{29}\) and y= \(\frac{19}{29}\)

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