lec-2-38.ppt

Lec-2 solving linear

Equations

A2.1 Solving simple linear equations

Contents

A2 Equations

A2.5 Non-linear equations A2.4 linear Equations with two variables

A2.2 Equations with the unknown on both sides

Equations

For example,

x

+ 7 = 13 is an equation

In an equation the unknown usually has a particular value. Finding the value of the unknown is called solving the equation.

x + 7 = 13



linear equation in one variable

a

x

+ b =

c

Using inverse operations to solve equations

We can use inverse operations to solve simple equations.

For example,

x + 5 = 13

x = 13 – 5

x = 8

Always check the solution to an equation by substituting the solution back into the original equation.

If we substitute x = 8 back into x + 5 = 13 we have

Using inverse operations to solve equations

Solve the following equations using inverse operations.

5x = 45

x = 45 ÷ 5

x = 9 Check:

5 × 9 = 45

17 – x = 6

17 = 6 + x

17 – 6 = x

Check:

11 = x x = 11

Using inverse operations to solve equations

Solve the following equations using inverse operations.

x = 3 × 7

x = 21 Check:

3x – 4 = 14

3x = 14 + 4 3x = 18

Check:

3 × 6 – 4 = 14

x = 18 ÷ 3

What number am I thinking of …?

I’m thinking of a number.

When I multiply the number by 2 and add 5 the answer is 11.

What number am I thinking of …?

We can write this as an equation.

Instead of using ? for the number I am thinking of, let’s use the letter x.

Start with x

x

Multiply by 2

2x

and add 5

2x + 5

to give you 11.

What number am I thinking of …?

Start with x

x

Multiply by 2

2x

and add 5

2x + 5

to give you 11.

2x + 5 = 11

We can solve this equation using inverse operations in the opposite order.

Start with the equation:

2x + 5 = 11

Subtract 5: 2x = 11 – 5

Divide by 2:

2x = 6

x = 6 ÷ 2

A2.2 Equations with the unknown on both sides

Contents

A2 Equations

A2.4 A2.4 linear Equations with two variables A2.1 Solving simple linear

equations

• 3

n

– 11=

2

n

– 3

n

– 11 = –3

n = 8

A2.2 Equations with the unknown on both sides

3



8 – 11 = 2



8 – 3

• 4

n

2x + 5 = 65 – 2x

4x = 60

2a+4 = 3a – 2

A2.3 Solving more difficult equations

Contents

A2 Equations

A2.4 linear Equations with two variables A2.1 Solving simple linear

equations

Equations with brackets

Equations can contain brackets. For example:

2(3x – 5) = 4x

To solve this we can

Multiply out the brackets: 6x –10 = 4x

6x -4x = 10

Sometimes we can solve equations such as:

2(3x – 5) = 4x

by first dividing both sides by the number in front of the bracket:

Divide both sides by 2: 3x – 5 = 2x

3x -2x = 5

x = 5

Solving equations involving division

Linear equations with unknowns on both sides can also involve division.

For example, 3x + 2

4 = 11 – x

In this case we must start by multiplying both sides of the equation by 4.

3x + 2 = 4(11 – x) 3x + 2 = 44 – 4x

3x + 4x = 44 -2 7x = 42

Solving equations involving division

Sometimes the expressions on both sides of the equation are divided.

For example,

In this example, we can multiply both sides by (x + 3) and (3x – 5) in one step to give:

4(3x – 5) = 5(x + 3) 12x – 20 = 5x + 15

12x – 5x = 15 +20 7x = 35

x = 5 4

(x + 3)

5

A2.4 linear Equations with two variables

Contents

A2 Equations

A2.1 Solving simple linear equations

A2.2 Equations with the unknown on both sides

linear Equations with two variables

The most commonly used algebraic methods of

solving simultaneous linear equations in two

variables are:

1-

Method of elimination by substitution

2-

Method of elimination by equating the

coefficient



ELIMINATION BY SUBSTITUTION

•For example: we want to solve,

x + 2y = -1 --- (i)

2x – 3y = 12 ---(ii)

x = -2y -1 --- ----(iii)

Substituting the value of x in equation (ii), we get

2x – 3y = 12

2 ( -2y – 1) – 3y = 12

- 4y – 2 – 3y = 12

Putting the value of y in eq (iii), we get

x = - 2y -1

x = - 2 (-2) – 1

= 4 – 1

= 3

ELIMINATION METHOD

• For example: we want to solve,

•

• 2x + 3y = 4

Let 3x + 2y = 11 --- (i)

2x + 3y = 4 ---(ii)

Multiply 3 in equation (i) and 2 in equation (ii) and

subtracting eq. iv from iii, we get

• putting the value of y in equation (ii) we get,

2x + 3y = 4

2 (5 )+ 3y = 4

10 + 3y = 4

3y = 4 – 10

3y = - 6

y = - 2

No

. Answer

1 u = 4 , z = 3

2 u = 2 , y = 5

3 c= 3 , u = 6

4 u = 2 , v = 6

5 a = 5 , x = 4

6 a = 6 , v = 1