An equation in the form of ax + by + c = 0 is called linear equation, where a,b and c are real number and both a and b are not equal to zero.
A linear equation is an equation for a straight line.
For example y = 5x + 3 is a linear equation. The different values of x and y are:
If x = 0, then y = 3
If x = 1, then y = 8
If x = 2, then y = 13
All these values of (x,y) as (0 , 3), ( 1 , 8),(2 , 13),(-1 , -2) etc., are the solutions of the given linear equation.

Linear Equation in One Variable :-

A linear equation in which number of unknown variables is one, is known as linear equation in one variable.
Linear equation in one variable represents a point in number line.
For example- 7x + 5 = 12, y + 2 = 13 etc.

Linear Equation in Two Variable :-

A linear equation in which number of unknown variables are two, is known as linear equation in two variables.
Linear equation in two variables represents a line in XY-plane (Cartesian plane).
For example- 3x + 7y = 25, x + 5y = 12 etc.

Linear Equation in Three variables :-

A linear equation in which number of unknown variables are three, is known as linear equation in three variables.
Linear equation in three variables represents a plane in XYZ- coordinate system.
For example- 2x + 3y + 5z = 15, x + 3y + z = 20 etc.

Methods of Solving Linear Equations :-

Substitution Method :-

In this method, first the value of one variable must be represented in the form of another variable and the variable is inserted into another equation for solving the equation for that variable. Thus, a value of one variable is obtained and subsequently this value is used to find the value of another variable.

Example- Solve following equations with substitution method.
2x + 3y = 29, x - y = -3
Solution:-
x - y = -3 …..(1)
2x + 3y = 29 …..(2)
From Eq. (1)
x + 3 = y
or, y = x + 3
On putting the value of y in Eq.(2), we get
2x + 3 ( x + 3 ) = 29
or, 2x + 3x + 9 = 29
or, 5x = 29 - 9
or, 5x = 20

∴ x =	20	= 4
	5

On putting the value of x in Eq. (1), we get
4 - y = -3
or, 4 + 3 = y
∴ y = 7
Hence, x = 4 and y = 7

Elimination Method :-

In this method, the coefficients of one of the variables of each equation become same by multiplying a proper multiple. Then by solving these equations, we get the value of another variable. Subsequently with the help of the obtained value of the variable, we find the value of another variable.

Example- Solve following equations with elimination method.
3x + 7y = 75 , 5x - 5y = 25
Solution:-
3x + 7y = 75 …..(1)
5x - 5y = 25 …..(2)
Now, Eq.(1) is multiplied by 5 and Eq.(2) is multiplied by 7 and then adding them, we get
15x + 35y + 35x - 35y = 375 - 175
or, 50x = 200

∴ x =	200	= 4
	50

On putting the value of x in Eq.(1), we get
3 × 4 + 7y = 75
or, 12 + 7y = 75
or, 7y = 75 - 12
or, 7y = 63

∴ y =	63	= 9
	7

Hence, x = 4 and y = 9

Cross Multiplication Method :-

Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are two equations.
a₁x + b₁y + c₁ = 0 ….(1)
a₂x + b₂y + c₂ = 0 ….(2)
Eq.(1) is multiplied by b₂ and Eq.(2) by b₁
b₂a₁x + b₂b₁y + b₂c₁ = 0 ….(3)
b₁a₂x + b₁b₂y + b₁c₂ = 0 ….(4)
On subtracting Eq.(4) from Eq.(3)
( b₂a₁ - b₁a₂ )x + ( b₂b₁ - b₁b₂ )y + ( b₂c₁ - b₁c₂ ) = 0
or, ( b₂a₁ - b₁a₂ )x + ( b₂c₁ - b₁c₂ ) = 0
or, ( b₂a₁ - b₁a₂ )x = -( b₂c₁ - b₁c₂ )
or, ( b₂a₁ - b₁a₂ )x = ( b₁c₂ - b₂c₁ )

∴ x =	( b1c2 - b2c1 )
	( b2a1 - b1a2 )

By substituting value of x either in Eq. (1) or Eq. (2),we get

y =	( c1a2 - c2a1 )
	( a1b2 - a2b1 )

∴ By cross multiplication method,

⇒	x	=	y	=	1
	( b1c2 - b2c1 )		( c1a2 - c2a1 )		( a1b2 - a2b1 )

x =	( b1c2 - b2c1 )
	( a1b2 - a2b1 )

y =	( c1a2 - c2a1)
	( a1b2 - a2b1 )

Example- Solve the following equations with cross multiplication method.
2x + 3y - 17 = 0, 3x - 2y - 6 = 0
Solution:- 2x + 3y - 17 = 0 …..(1)
3x - 2y - 6 = 0 ……(2)
By cross multiplication method,

⇒	x	=	y	=	1
	( b1c2 - b2c1 )		( c1a2 - c2a1 )		( a1b2 - a2b1 )

⇒	x	=	y	=	1
	{3 × (-6) - (-2) × (-17)}		{(-17) × 3 - (-6) × 2}		{(2 × (-2) - 3 × 3}

⇒	x	=	y	=	1
	(-18 - 34)		(-51 + 12)		(-4 - 9)

⇒	x	=	y	=	1
	-52		-39		-13

⇒	x	=	1
	-52		-13

∴ x = 4

⇒	y	=	1
	-39		-13

∴ y = 3
Hence, x = 4 and y = 3

Consistency of the System of Linear Equations :-

A set of linear equations is said to be consistent, if there exists atleast one solution for these equations. A set of linear equation is said to be inconsistent, if there are no solutions for these equations.
Let a₁x + b₁y + c₁ = 0 …(1)
and a₂x + b₂y + c₂ = 0 …(2)
Here a₁, b₁, c₁ and a₂, b₂, c₂ are coefficient of y and real constants in equations.

Consistent System :-

The above system will be consistent, if

1. If	a1	≠	b1	, then system has unique solution and it represents a pair
	a2		b2

of intersecting lines.

2. If		a1	=	b1	=	c1	, then system has infinite solutions and it represent overlapping lines.
		a2		b2		c2

Inconsistent System :-

The above system will be inconsistent, If		a1	=	b1	≠	c1	and
		a2		b2		c2

in this case the equations do not have any solution. It represents a pair of parallel lines.

Example- Check whether the given system is consistent or not. If yes, then find the solution.
3x + 4y = 8
9x + 12y = 24
Solution:- 3x + 4y - 8 = 0 ….(1)
9x + 12y - 24 = 0 ….(2)

⇒		a1	=	3	=	1
		a2		9		3

⇒		b1	=	4	=	1
		b2		12		3

⇒		c1	=	-8	=	1
		c2		-24		3

∴		a1	=	b1	=	c1
		a2		b2		c2

Thus, system is consistent and has infinite solutions.

Example- Find the value of k for which the system of linear equations:
kx + 4y = k - 4, 16x + ky = k, has infinite solution.
Solution:-
kx + 4y - ( k - 4 ) = 0
16x + ky - k = 0
Here condition for infinite solutions is

∴		a1	=	b1	=	c1
		a2		b2		c2

⇒		k	=	4	=	k - 4
		16		k		k

⇒		k	=	4
		16		k

⇒ k² = 16 × 4
⇒ k = √( 16 × 4 )
∴ k = 8, -8

Also,		4	=	k - 4
		k		k

⇒ 4k = k ( k - 4 )
⇒ 4k = k² - 4k
⇒ k² - 4k - 4k = 0
⇒ k² - 8k = 0
⇒ k ( k - 8 ) = 0
⇒ k = 0 or k = 8
But k = 0 is not possible otherwise equation will be one variable.
∴ k = 8 is correct value for many solution.

Ex- For what value of k, the system of equations kx - 4y - 8 = 0 and 8x -6y - 12 = 0 has a unique solution?
Solution:- kx - 4y - 8 = 0
8x -6y - 12 = 0
For a unique solution, we have

		a1	≠	b1
		a2		b2

		k	≠	-4
		8		-6

⇒		k	≠	2
		8		3

∴ k ≠	16
	3