Solving single variable linear equations

Solving single variable linear equations - Transcript

Lecture 2 Solving single variable linear equations
An equation may contain any number of variables unknowns but we will consider how to solve equations which contain just one variable A linear equation is an equation in which the highest power of the unknown is one there are no squared cubic or higher terms eg 3x 2 5x 4
5x 4 x 7

The solution of an equation is the value or values of the unknown which fits the equation so that the left hand side LHS of the equation is equal to the right hand side RHS A solution of an equation may also be called a root In order to solve equations we manipulate them by performing simple operations on them addition subtraction multiplication division raising to a power squarin g square rooting etc The key thing to understand is that provided an operation is performed on the whole of both sides of the equation the equation remains in balance and the LHS remains equal to the RHS To solve single variable equations the unknown terms should be collected on one side of the equation and the constant terms on the other side Examples 1 x 4 11 x 4 4 11 4 x 7 x 3 2x 5 x 3 3 2x 5 3 x 2x 8 x 2x 2x 8 2x x 8 x 1 8 1 x 8 multiply both sides by 1 subtract 2x from both sides add 3 to both sides subtract 4 from both sides

2

Alternatively if you prefer to keep the unknown term positive x 3 2x 5 x 3 x 2x 5 x subtract x from both sides

1

3 x 5 3 5 x 5 5 8 x x 8 subtract 5 from both sides

3

5x 15

5x 5

15 5

divide both sides by 5

x 3

4

x 2
2 x 2

8

28

multiply both sides by 2

x 16 A common mistake when faced with x 4 which is quite wrong 5
3x 4 5 2

x 2

8 is to divide the 8 by 2 and obtain the answer of

4

3x 4

4

5 multiply both sides by 4 2

3x

20 2

3x 10
x 10 3

divide both sides by 3

x

10 or 3 1 3 3

2

6

x 2
2 x 2

x 3
x 3

2

22

multiply the whole of each side by 2

x

2x 3

4

multiply each term in the bracket by the 2 outside the bracket

3x

2x 3

34

multiply the whole of each side by 3

3x 2x 12 5x 12

x

12 5

x 2 4 Instead of multiplying by 2 and then by 3 these operations could have been combined into a single operation of multiplying by 6 since 6 2 3 xx 6 62 23 3x 2x 12 One operation is faster than two but if you feel more comfortable performing two separate operations rather than combining them into a single operation then do the operations separately

7

20 49 x x 20 x 4 9x x x x 20 4x 9x x
x 20 4x 9x 5x 20 9x 5x 20 5x 9x 5x 20 4x 3

x

multiply the whole of both sides by x notice how the 4 has been multiplied to become 4x

5 x x 5 3 x 4 2 x 3 10 3x 12 2x 6 10 x 18 10 x 18 18 10 18 x 8 multiply out the brackets being careful with the signs

8

9

3 x 2 2

2 2x 1 1

2

3 x 2 2 3 x 2 2

2 2x 1

12

2

2 2 2x 1 1 2

3 x 2 4 2x 1 2 3x 6 8x 4 2 5x 10 2 5x 12

5x 5
x 2 4

12 5

10

10 4 x 10 x x 4 x
10 4x

10 4

4x 4

2 5 x x 2 5 4

Checking your answers It is a good idea to perform a check on the value or values you have obtained for the variable This may be done by inputting the value into each side of the equation in turn and seeing if the results are equal Example 2 above x 3 2x 5 We calculated that for this relationship to be true x must equal 8 LHS 8 3 11 RHS 2 8 5 16 5 11 LHS RHS

20 49 x We calculated that for this relationship to be true x must equal 5
Example 7 above

x

LHS

5

20 5

4

25 5

45

49

RHS 9 LHS RHS

More about equations Be careful some apparently linear equations are not what they appear to be 1 eg x 2 x 1 xx x 2 x x x2 2x x x2 1 2x There is a squared term so this equation is not linear

All the equations met so far have had a unique solution root Some equations have infinitely many solutions though single variable linear equations cannot eg x y 10 There are an infinite number of solutions such as x 5 y 5 x 2 y 12 x 1 6 y 8 4 Where two variables are involved two independent equations are require d to obtain a unique solution However we can find x in terms of y make x the subject of the equation x y 10 x y y 10 y x 10 y

5

An equation may have no solution eg 0 x 9 The product of any number and 0 is 0 Therefore there is no value of x which can be multiplied by 0 to give 9 and the equation has no solution

Solving some nonlinear equations Not only is zero the product of any number and zero but it is also true that the only way for two or more numbers or terms to have zero as their product is for one or more of the numbers or terms to itself be zero If 5x 0 then x 0 If x was other than zero 5x would not be zero If then 3 x 5 0 x 5 0 x 5

We can make use of this to solve any nonlinear equations which can be written as the product of linear factors The RHS must equal zero 1 if x 2x 3 0 then either x 0 or 2x 3 0 x 0 or x
3 2

2x2 3x 0

2

if x 2 x 6 0 then either x 2 0 or x 6 0 x 2 or x 6

x2 4x 12 0

Some equations involving square roots and squares 1 Suppose we have the equation x4
x x x If we square the To obtain x on the LHS we must square x since x LHS of the equation we must square the RHS to keep the equation in balance x
2 2

42

xx x 16

44

6

2

Solve

x 2 2 52 x 2 x 2 5 5 x 2 5 x 7

3

Be careful about signs when square roots are involved Solve x 4 2 36 x 4 x 4 36 x 4 6 or 6 the square root of 36 is 6 or 6 x 6 4 or 6 4 x 2 or 10

4

Solve

5 L

5

L

5 L

L

5

multiply both sides by

L

5
5 5

L
L

5
5 5

divide both sides by

5

5 5
5 5
5

L

5

L

because 5

5

5

L
2

5

L

2

squaring both sides

5 L L 5

You could substitute x for L if that makes the calculation seem easier to you 5 Let x L 5 x
x 5 x x5

7

5
5 5

x5
x 5 5

5 5

x
2

5 5
5

x

x
L

x L so

5

5 L L 5

In general

Find L in terms of a and b for

a L

b

L

a L

Lb

a
a b

Lb
Lb b

a b

L
2

a b

L
2

squaring both sides

L

a b

a2 b2

The original equation was that a 5 and b 5

5 L

5 Comparing with the general equation

a L

b we see

8

L

a b

2

a2 b
2

52 5
2

25 5

5

Throughout this lecture I have tried to put down every step of the process when solving an equation so that you can see exactly what I have done In practice and as you gain confidence you do not need to show each step explicitly Do make sure you write down sufficient to avoid mistakes though if in doubt write it out

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