A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can have one or more variables
Example
ax+b=0
a and b are constant
ax+by+c=0
a,b,c are constants
In Linear equation ,No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction.
So ax^{2} +b=0 is not a linear equation
Similary (a/x)+b=0 is not a linear equation
Linear equations are straight lines when plotted on Cartesian plane
S.no 
Type of equation 
Mathematical representation 
Solutions 
1 
Linear equation in one Variable 
ax+b=0 ,a≠0 a and b are real number 
One solution 
2 
Linear equation in two Variable 
ax+by+c=0 , a≠0 and b≠0 a, b and c are real number

Infinite solution possible 
3 
Linear equation in three Variable 
ax+by+cz+d=0 , a≠0 ,b≠0 and c≠0 a, b, c, d are real number

Infinite solution possible 
Linear equation in two variables is represented by straight line the Cartesian plane.
Every point on the line is the solution of the equation.
Infact Linear equation in one variable can also be represented on Cartesian plane, it will be a straight line either parallel to x –axis or y –axis
x2=0 (straight line parallel to y axis). It means ( 2,<any value on y axis ) will satisfy this line
y2=0 ( straight line parallel to x axis ). It means ( <any value on xaxis ),2 ) will satisfy this line
1) Suppose the equation given is
ax+by+c=0 , a≠0 and b≠0
2) Find the value of y for x=0
y=c/b
This point will lie on Y –axis. And the coordinates will be (0,c/b)
3) Find the value of x for y=0
x=c/a
This point will lie on X –axis. And the coordinates will be (c/a, 0)
4) Now we can draw the line joining these two points
A pair of Linear equation in two variables
a_{1}x+b_{1}y+c_{1}=0
a_{2}x +b_{2}y+c_{2}=0
Graphically it is represented by two straight lines on Cartesian plane.
Simultaneous pair of Linear equation 
Condition 
Graphical representation 
Algebraic interpretation 
a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 Example x4y+14=0 3x+2y14=0 
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ 
Intersecting lines. The intersecting point coordinate is the only solution 
One unique solution only. 
a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0
Example 2x+4y=16 3x+6y=24

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ 
Coincident lines. The any coordinate on the line is the solution. 
Infinite solution. 
a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 Example 2x+4y=6 4x+8y=18

$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ 
Parallel Lines

No solution 
The graphical solution can be obtained by drawing the lines on the Cartesian plane.
S.no 
Type of method 
Working of method 
1 
Method of elimination by substitution 
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 2) Find the value of variable of either x or y in other variable term in first equation 3) Substitute the value of that variable in second equation 4) Now this is a linear equation in one variable. Find the value of the variable 5) Substitute this value in first equation and get the second variable 
2 
Method of elimination by equating the coefficients 
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0 2) Find the LCM of a_{1} and a_{2} .Let it k. 3) Multiple the first equation by the value k/a_{1} 4) Multiple the first equation by the value k/a_{2} 4) Subtract the equation obtained. This way one variable will be eliminated and we can solve to get the value of variable y 5) Substitute this value in first equation and get the second variable 
3 
Cross Multiplication method 
1) Suppose the equation are a_{1}x+b_{1}y+c_{1}=0 a_{2}x +b_{2}y+c_{2}=0
2) This can be written as
3) This can be written as 4) Value of x and y can be find using the x => first and last expression y=> second and last expression 
Three Linear equation in three 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
a_{3}x + b_{3}y+c_{3}z+d_{3}=0
1) Find the value of variable z in term of x and y in First equation
2) Substitute the value of z in Second and third equation.
3) Now the equation obtained from 2 and 3 are linear equation in two variables. Solve them with any algebraic method
4) Substitute the value x and y in equation first and get the value of variable z
Solved Examples
1) Solve the linear equation
3(x+3)=2(x+1)
Solution
3x+9=2x+2
or x+7=0
or x=7
2) Solve the Simultanous linear pair of equations
x+y=6
2x+y=12
Solution
We will go with elimination method
Step 1 ) Choose one equation
x+y=6
x=6y
Step 2) Substitute this value of x in second equation
2x+y=12
2(6y)+y=12
122y+y=12
or y=0
Step 3) Substitute of value of y in any of these equation to find the value of x
x+y=6
x=6
So x=6 and y=0 Satisfy both the equations
3) Solve the equations
x+2y =10 , 4x+8y=40
Solution
As per algebric condition
Condition 
Algebraic interpretation 
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ 
One unique solution only. 
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ 
Infinite solution. 
$\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ 
No solution 