## Class 9 Mathematics Note

# LINEAR EQUATION

**Simultaneous Linear equation**

**Equations **

**Linear equations in two variables**

A linear system of two equations with two variables can be written in the form.

ax +by =c

cx + dy = q

Where any of the constants can be zero with the exception that each equation must have at least one variable in it

The x and y –coordinates gives the solutions of pair of simultaneous equations in two variables.

**Solving simultaneous linear equations in two variables **

**Substitution method **

The equations are:

x + y= 6……………….1

4x – 3y =-4…………………2

Equation 1 can be written as

y = 6 - x……………….3

4x – 3y = - 4

4x – 3*(6-x) = -4

4x - 18 + 3x = - 4

7x = 14

X= 2

Y= 4

**Elimination method **

The equations are:

x + y = 6……………….1

4x – 3y = -4…………………2

Multiplying equation 1 by 4, we get

**[x** + y= 6 ]*4

4x +4y= 24 ……………3

Subtracting 1 from 3

4x +4y= 24

4x – 3y = - 4

**-++**

7y = 28

Y= 4

X= 2

**Examples: 1**

**Solve each pair of simultaneous equations by elimination method**

**x + 2y = 8**

**x + y= 6**

Given,

x + 2y = 8…(i)

x + y = 6…(ii)

By elimination method,

Subtracting (i) and (ii)

x + 2y = 8

x + y = 6

or, y = 2

So, y = 2.

Then, x + 2 = 6

So, x = 4.

**2x + y = 13**

**x – y= 2**

Given,

2x + y = 13…(i)

x – y = 2…(ii)

By elimination method,

Adding (i) and (ii),

2x + y = 13

X – y = 2

Or, 3x = 15

So, x = 5.

So, 5 – y = 2

So, y = 3.

**Examples: 2**

**Solve each pair of simultaneous equations by substitution method**

**y = 2x**

**x + y = 6**

Given,

or, y = 2x..(i)

or, x + y = 6…(ii)

putting y = 2x in (ii),

or, x + 2x = 0

or, 3x = 6

So, x = 2

Putting x = 2 in (i),

Or, y = 2*2

= 4

So, y = 4.

**y = 3x**

**x +y = 20**

Given,

or, y = 3x….(i)

or, x +y = 20…(ii)

Putting y = 3x in (ii),

Or, x + 3x = 20

Or, 4x = 20

So, x = 5.

Now, putting x= 5 in (i),

Or, y = 3 * 5 = 15.

**Examples: 3**

**The total cost of a bicycle and a radio is Rs: 7000. If the cost of the bicycle Rs 1000 more than two times the cost of the radio, find the cost of each of these two items.**

Let the cost of bicycle and radio be x and y respectively.

Or, x + y = 7000.

Or, x = 7000 – y…(i)

Or, 2x = y – 1000….(ii)

Or, 2(7000 – y) = y – 1000

Or, 14000 – 2y – y = - 1000

Or, 15000 = 3y

So, y = 5,000

Now, putting y in (i),

or, x = 7000 – y

= 2,000.

The cost of a bicycle is Rs. 5,000 and radio is Rs. 2,000.

**Examples: 4**

**Three years hence a father will be four times as old as his son will be.Before two years he was seven times as old as his son was. Find their present ages.**

Let the age of father be x and his son be y. Then,

Or, (x + 3) = 4(y + 3)

Or, x + 3 = 4y + 12

Or, x = 9 + 4y…(i)

Or, (x – 2) = 7(y – 2)

Or, x – 2 = 7y – 14

Or, x – 7y = - 14 + 2

Or, x – 7y = - 12

Or, 7y – x = 12…(ii)

Putting value of x in (ii),

Or, 7y – (9 + 4y) = 12.

Or, 7y – 9 – 4y = 12.

Or, 3y = 21

So, y = 7.

Now, putting y = 7 in (i),

Or, x = 9 + 4 * 7

= 9 + 28

= 37

So, the age of the father is 37 yrs. And his son is 7 years.