Introduction:
We are already familiar with solving linear equations of one variable and two variables. The general form of linear equation of one variable is ax + b = 0 where a is not 0 and a and b are real numbers, and the general form of linear equations of two variables is
ax + by + c =0 where a, b and c are real numbers. In this section let us discuss with system of linear equations.
What is a system of linear Equations:
System of Linear Equations Definition:
Linear Equations in one variable:The general form of linear equation in one variable is ax + b = 0, where a and b are Real numbers and a is not equal to 0.
This system has unique solution, which is x =  b/a
For example : 2x + 6 = 0
=> 2x = 6
= > x =  6/2 =  3
The only solution to this equation is x =  3
This solution can be represented on a number line.
System of linear Equations in two variables:
The general form of linear equation in two variables is ax + by +c = 0, where a, b, c are real numbers and a and b both not equal to 0.
Let us consider the example x + y = 6
This equation will satisfy for infinitely many pairs of the form (x,y) satisfying this condition.
For example ( 1,6), 2,4), (3,3), (7, 1),.........etc
so that in all the cases x + y = 5, meaning the sum of the coordinates = 6
Hence these pairs of values satisfying the given equation are called the solution to the given equation.
System of linear Equations Solver:
1. For the linear equations of one variable, there will be unique solution which can be solved from the given equation.
2. For the linear equations of two variables, we need to have two equations to solve for the variables.
Let us assume that the given two equations are of the form
a
_{1} x + b
_{1} y + c
_{1} = 0 and a
_{2} x + b
_{2} y + c
_{2 }= 0.
It is possible to solve the system of linear equations for the unknown variables x and y.
The methods involved in solving these homogeneous system of linear equations are,
1. Substitution Method.
2. Elimination Method.
3. Cross Multiplication Method
4. Matrix Method etc.
To solve a system of linear equations we should know about the types of solution(s) that exists for the system.
Consistent system of linear equations:
The system of linear equations is said to be consistent if the solution exists.
For the above system of equations, if the following ratio satisfies we can say about the type of solutions accordingly.
If $\frac{a_{1}}{a_{2}}\: \neq \: \frac{b_{1}}{b_{2}}$, then the system has unique solution.
If $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$, then the system has infinite solution.
Inconsistent System of Equations:
The system of linear equations is said to be inconsistent if the solution does not exists.
For the above system of equations, if the following ratio satisfies then we can say that the system is inconsistent
If $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}}\: \neq \: \frac{c_{1}}{c_{2}}$
Dependent system of Linear Equations:
System of Linear Equations is said to be dependent if both the equations have infinitely common solutions.
So the coinciding lines are the dependent system of linear equations.
They satisfy the condition,
$\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}}\: = \: \frac{c_{1}}{c_{2}}$
Independent system of Linear Equations:
System of Linear Equations is said to be independent if they do not have infinite number of common solution.
So the intersecting lines and the parallel lines are the Independent system of Linear Equations.
Solving system of linear equations:
While solving a system of linear equations, we will come to know if they are consistent or inconsistent, dependent or independent.
System of Linear Equations Examples:
Solve system of linear equations 2x + 3y = 25 ; 3x + 2y =25
Solution:
Let us number the equations
2x + 3y = 25 (1)
3x + 2y = 25 (2)
Multiplying Equation (1) by 3, 3 ( 2x + 3y ) = 3 ( 25)
=> 6x + 9y = 75 (3)
Multiplying equation (2) by 2, 2 ( 3x + 2y ) = 2 ( 25)
=> 6x + 4y = 50 (4)
6x + 9y = 75 (3)
6x + 4y = 50 (4)
______________
Subtracting (4) from (3) 5y = 25
=> y = 25 /5 = 5
Substituting y=5 in (1) we get, 2x + 3(5) = 25
=> 2x + 15 = 25
=> 2x = 25  15 = 10
=> x = 10/2 = 5
Therefore, the two equations intersect at the point (2,2).
Hence the system of equations is consistent and independent.
If we verify the condition, $\frac{a_{1}}{a_{2}}\: \neq \: \frac{b_{1}}{b_{2}}\:$ , we see that,$ \frac{2}{3}\neq \frac{3}{2}$
Example 2: x + y = 4, 2x + 2y = 9
Solution:
Let x + y = 4 (1)
2x + 2y = 9  (2)
Multiplying (1) by 2, we get,
2 ( x+ y ) = 2(4)
=> 2x + 2y = 8 (3)
=> 2x + 2y = 9 (2)
Subtracting , (2) from (3), we get, 0 = 1, which is not true.
Hence the system of equations have no solution.
From the above relation, $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}}\: \neq\: \frac{c_{1}}{c_{2}}$
we have $\frac{1}{2}= \frac{1}{2}\neq \frac{4}{9}$
Hence the above pair of equations is inconsistent and independent.
Example 3: x + 3y = 8 ; 3x + 9y = 24
Solution:
Let x + 3y = 8 (1)
3x + 9y = 24  (2)
Substituting x = 8  3y in (2), 3 ( 8  3y ) + 9y = 24
=> 24  9y + 9y = 24
This condition is true for all values of y.
we get 24 = 24 which is true.
Therefore the system of equations has infinitely many solution.
Hence the system is consistent and dependent.
Hence by solving system of equations we can conclude if the system of linear equations is consistent or inconsistent and independent or dependent.
System of Linear Equations word problems:
We can follow the following steps while solving the word problems.
Step 1: Read the problem carefully and identify the unknown quantities. Give these quantities a variable name like x,y,u,v,w, etc.
Step 2: Identify the variables to be determined.
Step 3: Read the problem carefully and formulate the equations in terms of the variables to be determined.
Solve 4: Solve the equations obtained in step 3, using any one of the method you are comfortable with.
Example:4 chairs and 3 tables cost 1400 dollars and 5 chairs and 2 tables cost 1400 dollars. Find the cost of a chair and a table.
Solution:
Let the cost of a chair be x dollars, and the cost of a table be y dollars.

Chairs 
Tables 
Total Cost

Equation 
No. of Items

4 
3

1400


Cost 
4x 
3y 
4x + 3y 
4x + 3y = 1400 
No. of items 
5 
2 
1400 

Cost 
5x 
2y

5x + 2y 
5x + 2y = 1400

Let us solve linear system of equations we have framed here.
Hence we have the equations, 4x + 3y = 1400 (1)
5x + 2y = 1400 (2)
Multiplying (1) by 5, 5( 4x + 3y ) = 5( 1400)
=> 20 x + 15 y = 7000 (3)
Multiplying (2) by 4, we get 4 ( 5x + 2y ) = 4 ( 1400)
=> 20 x + 8 y = 5600 (4)
20 x + 15 y = 7000 (3)
______________________
Subtracting (3) from (4),  7 y = 1400
=> y = 1400/7 = 200
Substituting y= 200 in Equation (1), we get,
4x + 3 ( 200) = 1400
=> 4x + 600 = 1400
=> 4x = 1400  600 = 800
=> x = 800/4 = 200
Therefore
cost of a Chair is 200 dollars and cost of a Table is 200 dollars.
System of linear equations in three variables:
The general form of linear equation in three variables, x, y and z is
ax + by + cz +d =0, where a, b, c are real numbers and a, b, c not all equal to 0.
This represent the equation of a plane in threedimensional coordinate system, where a, b, c are the direction ratios of the normal to the plane.
To solve the equation in three variables, we need to have three conditions (equations) relating the variables x, y and z.
Elimination method is the most suitable method to solve the equations.
Linear System of Differential Equations:
A differential equation in which the dependent variable and its derivatives appear only in first degree is called a linear differential equation.
An ordinary linear differential equation of order n is of the form,
$\frac{\mathrm{d^n}y }{\mathrm{d} x^n}+P_{1}\frac{d^{n1}y }{dx^{n1}}+P_{2}\frac{d^{n2}y }{dx^{n2}}+...................P_{n}y=X, where P_{1}, P_{2},....................P_{n}$ and X are functions of x.
If X = 0, it is called homogeneous equation, otherwise it is a nonhomogeneous equation.
First order linear differential equation:
The general form of a linear equation of the first order is,
$\frac{\mathrm{d} y}{\mathrm{d} x}+Py = Q$, where P and Q are functions of x.
The solution to the above equation is given by,
y = $e^{\int Pdx}\int Qe^{\int Pdx}dx$
*****