A set of linear equations with two or more variables having degree one. Before we go in details it is also recommended to have a look on other form of linear equations like linear equations with one variable and with two variables and so on. System of linear equation is one the most prominent topics in algebra. In this section will learn more about system of linear equations with two variables and different methods to solve them such as substitution method and elimination methods etc.


What is the System of Linear Equations

System of linear equations is a set of two or more linear equations working together and involving the same set of variables. A general system can contains m linear equations with n unknowns. 

For Example:
System of linear equations with 2 variables is:

$a_1 x + b_1 y + c_1 = 0$ and
$a_2 x + b_2 y + c_2 = 0$.

System of linear equations with 3 variables:
$a_1 x + b_1 y + c_1 z + d_1= 0$
 $a_2 x + b_2 y + c_2 z + d_2 = 0$ and
$a_3 x + b_3 y + c_3 z + d_3 = 0$

General Form

For $x_1,x_2, x_3,.....x_n$ are the unknowns and $b_1, b_2,.... b_m $ are the constant terms and $a_{11}, a_{12},.......,a_{mn}$ are the coefficients of the system. A general system of m linear equations with n unknowns can be written as:

$a_{11}x_1+a_{12}x_2+....+a_{1n}x_n=b_1$
$a_{21}x_1+a_{22}x_2+....+a_{2n}x_n=b_2$
.
.
.
$a_{m1}x_1+a_{m2}x_2+....+a_{mn}x_n=b_m$

Solving System of Linear Equations in Two Variables

Let us consider, $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$. 

There can be different ways/methods to solve these homogeneous system of linear equations. Some are
1. Elimination Method
2. Substitution Method.
3. Matrix Method
4. Cross Multiplication Method

The solutions for the system of equations can be consistent or inconsistent. All based upon the ratios of the coefficients. For the above system of equations:

 System of Equations  Condition  Solution Type
Consistent    $\frac{a_{1}}{a_{2}}\: \neq \: \frac{b_{1}}{b_{2}}$ Unique solution 
 Consistent  $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$  Infinite solution
 Inconsistent    $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}}\: \neq \:  \frac{c_{1}}{c_{2}}$  Solution does not exists

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 three-dimensional co-ordinate 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,
$\frac{\mathrm{d^n}y }{\mathrm{d} x^n}+P_{1}\frac{d^{n-1}y }{dx^{n-1}}+P_{2}\frac{d^{n-2}y }{dx^{n-2}}+...................P_{n}y=X$, 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 non-homogeneous 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 = $e^{-\int Pdx}\int Qe^{\int Pdx}dx$

Examples

Example 1: Solve : 2x + 3y = 25  and  3x + 2y =25
Solution:
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)

Solve equation 3 and equation 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 (5, 5).

Example 2: Solve for x and y. 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.
Since $\frac{a_{1}}{a_{2}}\: = \: \frac{b_{1}}{b_{2}}\: \neq \:  \frac{c_{1}}{c_{2}}$ = 1/3
So solution does not exist for this particular example.

Word Problems

Follow below steps to solve word problems.
Step 1: Read and understand the problem carefully.
Step 2: Identify the unknown quantities. Represent with the variable.
Step 3: Formulated the equations.
Step 4: Simplify the equations, using any of the methods.
Step 5: Write your answer.

Example: 4 chairs and 3 tables cost Rs. 1400 and 5 chairs and 2 tables cost  Rs. 1400. Find the cost of a chair and a table.
Solution:
Let the cost of a chair be Rs. x , and the cost of a table be Rs. y.
   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 Rs. 200 and cost of a Table is Rs. 200.