An equation is a mathematical statement showing two algebraic or numeric expressions equal. A Linear equation is an equation with one or more variables which do not contain exponents. In other words, a linear equation is formed by adding or subtracting terms which can be
numerical multiples or quotients of variables. The name linear is derived from the word ' line' as as the graph of a
linear equation in two variable is a straight line in coordinate plane.

**Linear equation**, an algebraic equation in which each term is either a constant or the product of a constant and single variable.

- If a number is added to a variable or to an expression containing the variable, then subtract the same number on either side of the equation.
- If a number is subtracted to a variable or to an expression containing the variable, then add the same number on either side of the equation.
- If a variable or an expression containing a variable is multiplied by a number, then divide either side of the equation by the same number.
- If a variable or an expression containing a variable is divided by a number, then multiply either side of the equation by the same number.

Given equation 5x + 9 = 29

Step 1:

Subtract 9 from both side

=> 5x + 9 - 9 = 29 - 9

=> 5x = 20

Step 2:

Divide each side by 5,

=> $\frac{5x}{5}$ = $\frac{20}{5}$

=>**x = 4.**

Step 1:

Subtract 9 from both side

=> 5x + 9 - 9 = 29 - 9

=> 5x = 20

Step 2:

Divide each side by 5,

=> $\frac{5x}{5}$ = $\frac{20}{5}$

=>

Given equation 15x + 20a + 4 = 13

Step 1:

Subtract 4 from both side

=> 15x + 20a + 4 - 4 = 13 - 4

=> 15x + 20a = 9

Step 2:

Solve for x,

=> 15x = 9 - 20a

Divide each side by 15

=> $\frac{15x}{15}$ = $\frac{9 - 20a}{15}$

=> x = $ \frac{9 - 20a}{15}$

Step 3:

Put a = 1

=> x = $ \frac{9 - 20 * 1}{15}$

=> x = $ \frac{9 - 20}{15}$

=>**x = $ \frac{- 11}{15}$**

Step 1:

Subtract 4 from both side

=> 15x + 20a + 4 - 4 = 13 - 4

=> 15x + 20a = 9

Step 2:

Solve for x,

=> 15x = 9 - 20a

Divide each side by 15

=> $\frac{15x}{15}$ = $\frac{9 - 20a}{15}$

=> x = $ \frac{9 - 20a}{15}$

Step 3:

Put a = 1

=> x = $ \frac{9 - 20 * 1}{15}$

=> x = $ \frac{9 - 20}{15}$

=>

1.

Example, 3x - 5y + 7 = 0 is a general form of equation

where a = 3, b = -5 and C = 7

2.

Example : x - 2y = 10 is a standard form of equation

where a = 1, b = -2 and C = 10

3.

Example: y = 3x + 5.

where m = 3 and b = 5

A system of linear equations is classified into two parts based on their solutions:

**Consistent System**, A system has a unique or infinitely many solutions.**Inconsistent System**, An system does not have any solution.

Solved Example

6x + 2y = 9

3x + ky = - 7

Given system of equations

6x + 2y = 9

3x + ky = - 7

Step 1:

We know that

The system is inconsistent, has no solution

if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

By comparing the given equations with the general equations of the system we have

a_{1} = 6, b_{1 }= 2, c_{1} = 9

and b_{2} = 3, b_{2} = k, c_{2} = -7

Step 2:

Since system of equations have no solution

=> $\frac{6}{3} = \frac{2}{k} \neq \frac{9}{7}$

Step 3:

Find the value of k

=> $\frac{6}{3} = \frac{2}{k}$

=> 2 = $ \frac{2}{k}$

=> 2k = 2

=>**k = 1**

Hence for k = 1 the system of equations have no solution.

6x + 2y = 9

3x + ky = - 7

Step 1:

We know that

The system is inconsistent, has no solution

if $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

By comparing the given equations with the general equations of the system we have

a

and b

Step 2:

Since system of equations have no solution

=> $\frac{6}{3} = \frac{2}{k} \neq \frac{9}{7}$

Step 3:

Find the value of k

=> $\frac{6}{3} = \frac{2}{k}$

=> 2 = $ \frac{2}{k}$

=> 2k = 2

=>

Hence for k = 1 the system of equations have no solution.

- Substitution method
- Elimination method
- Graphical solution

x + y = 3

5x + 2y = 9

Given system of equations

x + y = 3 .....................(1)

5x + 2y = 9 .....................(2)

Step 1:

Apply substitution method to solve equations

Equation (1) => y = 3 - x

Substitute equation (1) in equation (2)

=> 5x + 2 (3 - x) = 9

=> 5x + 6 - 2x = 9

=> 3x = 9 - 6

=> 3x = 3

=> x = 1

Step 2:

Put x = 1 in equation (1)

=> 1 + y = 3

=> y = 3 - 1 = 2

=> y = 2

Hence solution of the system is**(x, y) = (1, 2).**

x + y = 3 .....................(1)

5x + 2y = 9 .....................(2)

Step 1:

Apply substitution method to solve equations

Equation (1) => y = 3 - x

Substitute equation (1) in equation (2)

=> 5x + 2 (3 - x) = 9

=> 5x + 6 - 2x = 9

=> 3x = 9 - 6

=> 3x = 3

=> x = 1

Step 2:

Put x = 1 in equation (1)

=> 1 + y = 3

=> y = 3 - 1 = 2

=> y = 2

Hence solution of the system is

3x + y = 2

5x + 2y = - 5

Given system of equations

3x + y = 2 ........................(1)

5x + 2y = - 5 ........................(2)

Step 1:

To eliminate y, multiply each side of equation (1) by 2

=> 2(3x + y) = 2 * 2

=> 6x + 2y = 4 ................(3)

Now the coefficients of y terms are same of both the equations.

Step 2:

Subtract the equation (2) from equation (3)

=> 6x + 2y - (5x + 2y) = 4 - (- 5)

=> 6x + 2y - 5x - 2y = 4 + 5

=> x = 9

Step 3:

Put x = 9 in equation (1)

=> 3 * 9 + y = 2

=> 27 + y = 2

=> y = 2 - 27 = - 25

=>**Solution of the system is (x, y) = (9, - 25).**

3x + y = 2 ........................(1)

5x + 2y = - 5 ........................(2)

Step 1:

To eliminate y, multiply each side of equation (1) by 2

=> 2(3x + y) = 2 * 2

=> 6x + 2y = 4 ................(3)

Now the coefficients of y terms are same of both the equations.

Step 2:

Subtract the equation (2) from equation (3)

=> 6x + 2y - (5x + 2y) = 4 - (- 5)

=> 6x + 2y - 5x - 2y = 4 + 5

=> x = 9

Step 3:

Put x = 9 in equation (1)

=> 3 * 9 + y = 2

=> 27 + y = 2

=> y = 2 - 27 = - 25

=>

Given linear equation, x + y = 6

The graph of x + y = 6 is shown here

The graph of x + y = 6 is shown here