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 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.

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).

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

3. Solving Systems of Equations by Graphing Method

4. Solving Systems of Equations by Matrix Method

7x + 6y = 3800 ...(1)

3x + 5y = 1750 ...(2)

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

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

x = 500

y = 50

2(m + 2) = 10 + m

2m + 4 = 10 + m

2m - m = 10 - 4

m = 6

Value of m is 6

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)

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

28/15 + b = 3

b = 3 - 28/17 = 17/15

a = -7/15

b = 17/15

**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)