# 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,

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.

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