In mathematics, linear expressions with equal sign that contains same set of variables can always be plotted as a straight line on xy graph. There are several methods to solve linear equations. Usually we consider substitution method, elimination method, matrix method or cross multiplication method while dealing with linear equations in two variables. In the section mostly we will deal with techniques used to solve linear equations in one and two variables only.

Solving Various Types of Linear Equations in One Variable

Solving Linear equations in one variable means to find the value of variable which satisfy both the equation. There are various techniques of balancing such expression on both sides of the equal sign. Arithmetic operations help in reducing a linear equation in one variable to simpler forms and find the value in one or two steps.

Solving a System of Linear Equations in Two Variables

A solution set of any system contains 2 variables is an ordered pair which satisfy both the equations or we can say that make equations true for resultant values. Before we start attempting such problems we should have an idea of some basic techniques of balancing the two linear equations on both the sides (RHS and LHS).

Popular Methods to Solve Linear equations in Two Variables:
1. Solving Linear Systems by Elimination Method
2. Solving Systems of Equations by Substitution Method
3. Solving Systems of Equations by Graphing Method
4. Solving Systems of Equations by Matrix Method

Examples

With the help of below illustrated examples will see how to use above listed methods and solve the problems.
Example 1: Solve 7x + 6y = 3800; 3x + 5y = 1750 using elimination method.
Solution:
Step 1: Name the equations
7x + 6y = 3800  ...(1)
3x + 5y = 1750  ...(2)

Step 2: Eliminate one of the variables
Use operation: {7 $\times$ equation (2)} - {3 $\times$ equation (1)}

Mulitiply equation (2) by number 7 and multiply equation (1) by number 3 and subtract from first one.

21x + 35y = 12250
21x + 18y = 11400
-   -      -
_____________________
0 + 17y = 850
_____________________

17y = 850
or y = 850/17 = 50

this implies y = 50

Step 3: Get the value of x
Substitute the value of y in equation (1)

7x + 6(50) = 3800
7x + 300 = 3800
7x = 3800-300 = 3500
x = 3500/7 = 500

value of x is 500
Answer: 
x = 500
y = 50

Example 2: Find the value of m: 2(m + 2) = 10 + m
Solution
2(m + 2) = 10 + m
2m + 4 = 10 + m
2m - m = 10 - 4
m = 6
Value of m is 6

Example 3: Solve : 5a + 10b = 9; -4a + b = 3 using substitution method.
Solution:
5a + 10b = 9   ...(1)
-4a + b = 3   ...(2)
Isolate equation (2) for b
b = 3 + 4a
Put value of b in equation (1)
5a + 10 ( 3 + 4a ) = 9
5a + 30 + 40a = 9
45a + 30 = 9
45a = 9 - 30
45a = -21
a = -21/45 = -7/15

Find the value of b.
Substitute the value of a in equation (2)
-4(-7/15) + b = 3
28/15 + b = 3
b = 3 - 28/17 = 17/15

Answer:
a = -7/15
b = 17/15

Practice Problem

Problem 1: Solve x + y = 2 and x + 5y = 10

Problem 2: Solve pair of linear equations: -5m + n = 2;  m - n = 3

Problem 3: Solve for x and y. 20x + 15y = 10 ; -x + 3y = -5 Problem 4: Solve for m, 12( m + 4) - 4/5 = 6m/10 - 15 Problem 5: Find the value of x: 5x + 6 = 3(x - 3)