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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
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
where a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2} are real numbers,
such that a_{1}^{2 }+ b_{1}^{2} ≠ 0, and a_{2}^{2 }+ b_{2}^{2} ≠ 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 1^{st} 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 2^{nd} 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
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 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.
- 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 1^{st} Equation
7(3 – 2y) – 15y =2
– 29y = –19
y = \(\frac{19}{29}\)
Substituting this value of y in Equation 2^{nd} equation
x = 3 – 2\(\frac{19}{29}\) = \(\frac{49}{29}\)
Thus, x = \(\frac{49}{29}\) and y= \(\frac{19}{29}\)
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