The document provides examples and step-by-step workings for evaluating algebraic expressions by substituting number values for variables. It demonstrates evaluating expressions including powers, integers, fractions, and evaluating expressions to find volumes. The examples show applying order of operations and substitution to simplify expressions down to a single numerical value.

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2. Evaluate the expression 4c2 + 3, for c = 2
2
4c + 3 (c = 2)
This is a typical Powers Substitution question.
The Algebra Variable letter “c” needs to be
replaced with the number value of “2”.
Working out is then performed to obtain a
number answer.

3. The steps for evaluating Powers are as follows:
 Put multiplication signs between the terms
 Replace the letters with the supplied numbers
 Expand out the Powers terms
 Work out the final number answer, using
Integer Rules for Operations where needed.

4. The little number “4” is called the
“Index” or “Exponent” and tells us
how many times to multiply out
the big number “3”
The big number “3” is called the “base”
and is what we multiply together
4
3 = 3 x 3 x 3 x 3 = 81
Multiply four of the Base Number

5. Here is a quick review of
the Integers Operations
Rules that are needed for
positive and negative
numbers.

6. The working out of answers after substituting in number values
needs to follow BODMAS / PEMDAS order of operations.

7. 2
Evaluate the expression 4c + 3, for c = 2
4c2 + 3 (c = 2)
2
4c means 4 x c x c
= (4) (2) (2) + 3
= 16 + 3
= 19

8. Evaluate 2a – 12 / b , for a = 3 and b = 2
3 2
2a3 – 12 / b2 (a=3 b=2)
= (2)(3)(3)(3) – 12
(2)(2)
= (2)(27) – 12 / 4
= 54 – 3 = 51

9. Evaluate 5a – 3a + b , for a = -2 b = -3
3 2 2
5a3 – 3a2 + b2 ( a = -2 b = -3 )
= (5)(-2)(-2)(-2) - (3) (-2)(-2) + (-3)(-3)
= (5) (4) (-2) - (3) (4) + 9
= -40 - 12 + 9 = -52 + 9 = -43

10. 2
Volume of any Cylinder = 3.14r h
Find the Volume when r = 2cm and h = 7cm
V = (3.14) (r) (r) (h) where r=2 h=7
V = (3.14) (2) (2) (7)
V = (3.14) (28)
3
V = 87.92 cm

11. Complete the following table using the
rule: y = x2 + 5
x y We need to use the expression
x2 + 5 to work out each y value.
0
1 Eg. For x = 0
2
(0)(0) + 5
3
= (0) + 5
= 5 (See next Slide)

12. Complete the following table using the
rule: y = x2 + 5
x Working Out Steps y
2
0 y = x + 5 = (0) (0) + 5 = 0+5 =5 5
1 y = x2 + 5 = (1) (1) + 5 = 1+5 =6 6
2 y = x2 + 5 = (2) (2) + 5 = 4+5 =9 9
3 y = x2 + 5 = (3) (3) + 5 = 9+5 = 14 14

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