Exercises:   [4]   [5]   [6]

UNIT 1B - ALGEBRA

ELEMENTARY ALGEBRA

Why are numbers often represented in mathematics by symbols? One reason may be that a symbol is used because the actual number is unknown. For example, in a bag you have some oranges and you want to give six of them to a friend, without leaving yourself short. Obviously, you cannot say you have 10 – 6 or 20 – 6 left, as you just do not know. For the initial number in the bag you select some arbitrary symbol, say q (for quantity) then the number you have left will be q – 6. You still do not know how many oranges you have but at least you have a way of expressing it.

Now someone tells you that there were originally 20 oranges, now many are left? To answer the question you substitute the number 20 for the unknown value q ,

Number of oranges left = q – 6 = 20 – 6 = 14.

Another reason for the use of symbol is that you may wish it to represent a variety of possible numbers. For example, if there were originally 20 oranges, then

number left = 20 – 6 = 14;

If there were 13 to start with, then

number left = 13 – 6 = 7,

and so on. Expressions involving symbols are more general than those involving numbers.

The symbols used to represent numbers are completely arbitrary. However, it is common practice to use small letters of the alphabet.. Often those from the beginning of the alphabet a, b, c,… are chosen to represent constant values and those from the end …x, y, z to represent variable values such as measurements. For example the equation

y = a ´ x (often written y = ax)

can be used to convert from miles to kilometres; if a has the value and x is the number of miles (measured in some way) then y is the equivalent number of kilometres,

e.g. if x = 30, y = ´ 30 = 48,

if x = 145, y = ´ 145 = 232.

 

SUBSTITUTION

Substitution is a major source of error. To avoid mistakes, do not take short cuts.

 

Exercises 1.4 Solutions

Evaluate the following expressions:

1. 5d + 6 for d = 2 and d = 4.

2. a ´ b + 2b for a = 1, b = 2 and a = 6, b = 1.

3. q 2 + 2q + 5 for q = 3. (q is the Greek letter ‘theta’; see the list of Greek letters at the back of the book.)

4. ax + for a = 1, b = 5, x = 2.

5. ax2 + bx + c for x = 2.

6. 2x + 3 for x = ½ and x = a.

 

 

ALGEBRAIC OPERATIONS

Symbols which represent real numbers are manipulated in exactly the same way as real numbers. While you should now be thoroughly familiar with applying arithmetic operations to real numbers, you may not be consciously aware of the algebraic equivalents, so some are listed here.

  1. For any pair of real numbers a and b,

    a + b = b + a (commutative law of addition).

    e.g. 2 + 3 = 3 + 2 = 5.

    e.g. 2.01 + p = p + 2.01 = 5.1515926… .

    But be careful with subtraction: 2 – 3 is not equal to 3 – 2 (written as 2 – 3 ¹ 3 – 2) because

    2 – 3 = 2 + (-3) = (-3) + 2

    whereas 3 –2 = 3 + (-2) = (-2) + 3.

  2. For any two real number a and b,

    ab = ba (commutative law of multiplication).

    e.g. and .

  3. For any three real numbers a, b and c,

    (a + b) + c = a + (b + c) (associative law of addition).

    e.g. (1 +4) + 2 = 5 + 2 = 7 and

    1 + (4 + 2) = 1 + 6 = 7.

  4. For any three real numbers a, b and c,

    (ab)c = a(bc) (associative law of addition).

    e.g. ´ 6 = ´ 6 = 3p and

    ´ (p ´ 6) = ´ 6p = 3p .

  5. For any three real numbers a, b and c

ab + ac = a(b + c) (distributive law).

Check this using a =1, b =2 and c= 3

 

Exercises 1.5 Solutions

1. If a = 2 , b = 3, and c = 4 evaluate:

(a) a + bc (b) (a + b)c

(c) (a + b)/c (d) a + b/c

2. Simplify the following expressions;

(a) (b)

(c) x(y + z) = 2{xy + y(z – x)}
  1. State which, if any, of these expressions are equal:

(1 + y)/100, 1/y + 100, y/1 + 100, y + 1/100, 1 + y/100.

 

SOLVING A LINEAR EQUATION

Returning again to the oranges, remember you had a bagful and gave 6 away to a friend, when you get home you remember that you forgot to pay the greengrocer. You want to give him the money but do not know how many you bought. All you do know is the equation

Number left = q -6

Where q is the number that were in the bag to start with. You look in the bag and find there are 8 left so

8 = q –6 .

At his stage, common sense probably tells you that you must have had 14 oranges in the first place. However, lets go through the process of finding q systematically. You have

8 = q –6

And you want q by itself so add 6 to both sides of the equation (if one side of the equation is altered, the other side must be altered in exactly the same way),

8 + 6 = q – 6 +6 ;

but 8 + 6 = 14 and –6 + 6 = 0, so

q = 14 .

Any equation of the form

ax + b = c

where a, b and c are known, constant values and x(or q in the orange example) is an unknown variable, is called a linear equation is one variable x. To solve such an equation try to isolate the unknown variable on one side of the equation as shown in the following example.

Example 1f. 2x + 3 = 7, find x.

Subtract 3 from both sides 2x = 7 – 3 = 4,

Divide both sides by 2

so the answer is x = 2.

Whenever you solve an equation check your answer by substituting the value you obtained into the original equation and making sure that the two sides of the equation are equal.

Put x = 2 into the left-hand side of the equation:

2x + 3 = 2 ´ 2 + 3 = 4 + 3 = 7

which does equal the right-hand side, so all is well.

Example 1g. 5t – 3 = 3t + 9, find t.

Subtract 3t from both sides 5t – 3 – 3t = 9,

Simplify 2t – 3 = 9,

Add 3 to both sides 2t = 12,

Divide both sides by 2 t = 6.

To check this answer put t = 6 into the left-hand side 5 ´ 6 3 = 30 3 = 27, put t = 6 into the right-hand side 3 ´ 6 + 9 = 18 + 9 = 27, the two sides are equal so t = 6 is a solution of the equation.

Example 1h. 1/(2 +y) = 4, find y.

Multiply both sides by (2 + y) 1 = 4(2 + y),

Simplify 1 = 8 + 4y

Subtract 8 from both sides -7 = 4y

Divide both sides by 4

Check the answer by substitution.

 

Exercises 1.6 Solutions

Solve the following equations:

1. 4x + 3 = 5

2. 5x + 8 = 7x – 3

3. t +1 = (4t – 2) + 2t

4. 3(2-4q) = 6

5. 2/z + 1/z = 3