Section 2: Arithmetic for College: Precedence, formulas and substitution.
Formulas occur a lot in college courses. The key to successfully doing calculations that
involve formulas is to realise that Mathematics isn’t always done in the order it is written
or read from left to right. Basic operations will be done in an order based on how difficult
the operation is and the key to using formulas is understanding this idea called
PRECEDENCE of operations.
This precedence is based on the following:
Multiplication is repeated addition
Raising to a power is repeated multiplication.
Some problems you will be able to do at the end of this section:
y 2  x 2  10
when y  6, x  4, z  2
10z  y
1.
Evaluate
2.
The distance s metres travelled in time t seconds is given by the formula
1
s  ut  at 2 ,where u is the initial velocity in m/s and a is the acceleration in m/s2.
2
Find the distance s the body travels in t = 6 seconds, with an initial velocity of u =
10 m/s at an acceleration a=5m/s2.
3
Find the roots of the quadratic equation x 2  6x  6  0 given that the roots of a
quadratic equation Ax 2  Bx  C  0 can be got from the formula
B  B2  4 AC
and that here A  1, B  6, C  6 .
x
2A
y  y1
Use the formula slope  2
to find the slope of the line between
x 2  x1
(x1, y1) = (-2, 1) and (x2, y2) = (3, 3)
4
5
Find the length of the side a in the triangle given
b sin A
using the sine rule formula: a 
sin B
b = 12
A = 50 degrees
B = 20 degrees
6.
a=?
Find the accumulated amount A using the compound Interest formula A=P(1+i)n
on a loan for the principal amount P = €1,500 taken out for a time n = 7 years at
an annual rate of interest, i = 6%.
Before looking at this in detail let us first consider the idea of raising to a power:
Page 1 of 17
Powers
.
Multiplications often involve multiplying several copies of the same number together. For
example, 6  6  36 , 2  2  2  8 . We have a short way to write such calculations using
powers. The power is how many copies of the same number we multiply. So we say that
6  6 is “6 to the power of 2” and we write
6  6  62 .
Similarly 2  2  2 is called “2 to the power of 3” and we write
2  2  2  23 .
Multiplying several copies of the same number is often called “raising the number to a
power”. So if we raise 2 to the power of 3, we get 2  2  2  23  8 .
Powers which are Positive Whole Numbers
Letting the letter a stand in for any number and the letter t stand in for any
positive whole number, a to the power of t is written as a t and means t
copies of a multiplied together. We can write this rule as
a t  a  a   a .



t times
Notice that when we have t  1, then whichever number is represented by a appears once
and is not multiplied by anything. So we get a1  a . Note also that a 0  1 regardless of
what a is.
Important Special Cases
a1  a
a0  1
Examples:
(a) 31  3.
3 to the power of 1 equals 3.
(b) 1001  100.
100 to the power of 1 equals 100.
(c) 01  0 .
0 to the power of 1 equals 0.
(d) 52  5  5  25.
5 to the power of 2 equals 25.
(e) 35  3  3  3  3  3  243 .
3 to the power of 5 equals 243.
(f) 29  2  2  2  2  2  2  2  2  2  512 .
2 to the power of 9 equals 512.
Page 2 of 17
(g)  7  7  7  7  343.
3
(h)  5  5  5  5  5  625.
4
7 to the power of 3 equals 343.
5 to the power of 4 equals 625.
Because 52  5  5  25 gives us the area of a square all of whose sides are of length 5, we
often say “5 squared” instead of “5 to the power of 2”. When we raise a number to the
power of 2, we often say that we square the number.
Examples:
(a) 52  5  5  25.
5 squared is equal to 25.
5
5
Area  5  5  25.
(b) 12  12  144 .
12 squared is equal to 144.
Because 2  2  2  23  8 gives us the volume of a cube all of whose sides are of length 2,
we often say “2 cubed” instead of “2 to the power of 3”. When we raise a number to the
power of 3, we often say that we cube the number.
Square roots:
Page 3 of 17
Examples:
(a) 2 3  2  2  2  8 .
2 cubed is equal to 8.
2
Volume  2  2  2  8 .
2
(b) 103  10  10  10  1000 .
10 cubed is equal to 1000.
2
Examples:
3
(a) Evaluate the power  9 .
 93  9  9  9  729.
(b) Evaluate the power 7 5 .
75  7  7  7  7  7  16807.
4
(c) Evaluate the power  3 .
 34  3  3  3  3  81.
(d) Evaluate the product 43  32 . 43  32  4  4  4  3  3  64  9  576.
3
(e) Evaluate the product  2  5 2 .
 23  52 = 2  2  2  5  5  8  25  200 .
(f) Evaluate the quotient 44  23 . 4 4  2 3 
4  4  4  4 256

 32.
222
8
Note: In examples (d), (e), (f), above, we calculate all powers first and then carry out the
indicated multiplications and divisions. We will say more about this in the next section.
Page 4 of 17
Calculating Expressions
Order of Precedence
Calculations must be performed in a specific order. This order is called the “rules of
precedence”
The rules of precedence are summarized by the following table:
Order of Precedence Table
Brackets
Powers
 and 
 and 
If there are brackets within brackets, you evaluate the inner most one first.
The procedure is to start at the top and work your way down.
Simple Examples:
Working in pairs figure out the story of the order in which these calculations have to
happen: (cover the answers below first!!)
(a) 3  4  5  7.
(b) 4  8  4  2
(c) (16  4)  2.
(d) 16  (4  2)
Answers to Simple Examples in full:
(a) 3  4  5  7  3  20  7  23  7  16.
(b) 4  8  4  2  4  8  2   4  2  2.
(c) (16  4)  2  4  2  2.
(d) 16  (4  2)  16  2  8.
COMMENT on 6(5) and its meaning as multiplication:
Page 5 of 17
Examples with brackets
Working in pairs figure out the story of the order in which these calculations have to
happen: ( cover the answers below first!!)
(a) 4  78  3
(c)  23  4  19  9
(b) 32  18  104
(d) 57  3  910  2
Answers to Examples with brackets in full.
(a) 4  78  3
 475
 4  35
 39.
(c)  23  4  19  9
 212  19  9
 2 7  9
 14  9
 23.
(b) 32  18  104
 32  84
 32  32
 64.
(d) 57  3  910  2
 54   98
 20  72
 92.
Page 6 of 17
Examples with nested brackets:
Working in pairs figure out the story of the order in which these calculations have to
happen: (cover the answers below first!!)
(a)
24  35  9
(c)
 34  572  5  3 6
(b) 5  3[2(3  4)  7(3  5)]
Answers to Examples with nested brackets in full.
(b) 5  3[2(3  4)  7(3  5)]
(a) 24  35  9
 5  327   7 2
 24  34
 5  314  14
 24   12
 24  12
 2 8
 16.
(c)
 5  328
 5  84
 79.
 34  572  5  3 6
 34  57 3  18
 34  5 21  18
 34  5 3
 34  15
 3 11
 33.
Page 7 of 17
REMEMBER:
In the rules of precedence, powers come after brackets, but before multiplication and
division.
Brackets,
Powers,
Multiplication, Division,
Addition, Subtraction.
Examples involving powers:
Working in pairs figure out the story of the order in which these calculations have to
happen: (cover the answers below first!!)
(a)
3  52
(b)
(d) 7  2  5  9
2
3
(c) 23  4
14  2 3  9
(e) 33  2 4
2
2
 4 2  52
2
(f)
6
2


 52 3   3 2 5  33
2

2
Answers to Examples involving powers in full:
Working in pairs figure out the story of the order in which these calculations have to
happen: (cover the answers below first!!)
(a)
3  52
(b)
(c) 23  4
14  2 3  9
 3  25
 14  8  9
 75.
 22  9
2
 27 
 249 
 98.
 13.
(d) 7  2  5  9
2
3
(e) 33  2 4
2
 4 2  52
2
2
(f)
6
2

2
 27  16  16  25
 25   64 
 112  9
 25  64
 121  81
 4  3  25
 89.
 40.
 16  75
2
2
2
2
2
2
 36  40  35
2
2
 59.
Page 8 of 17

 36  58  332  27
 5 2  4 
3

 52 3   3 2 5  33
2
Exercises
1. Working in pairs figure out the order in which these calculations have to happen:
2. Working on your own use your calculator to do the calculations:
(a) 2  5  3
(b) 1  8  2
(c) 2  4  1
(d) 3  5  4
(e) 6  4  1
(f) 5  8  2
(g) 2  4  3  2
(h) 5  7  9  3
(i) 23  1
(j) 46  3
(k) 7  32
(l) 4  52
(m) 2  42
(n) 72  32
(o) 1  2  3

(q) 2  3  2  32
2



(s) 2  2 2  1  2  4  1
2
(p) 6  1  5
2

(r) 1  5  2  12
2


2

(t) 3  32  1  3  3  2
Page 9 of 17
2
Introduction to Algebra
.
Algebra is a body of mathematical knowledge which has been developed to manipulate
symbols.
Some symbols take fixed and
unchanged values and these are
known as constants. e.g. let b
represent boiling point of water.
Some symbols represent quantities which
can vary and these are called variables.
e.g. Velocity of a car represented by the
symbol v which can vary.
A useful way to think of variables is to think of them as the address of a storage location
in which anything can be stored.
x
y
So if reading the expression
3x + 5y
it is useful to think of it as
3 times the content of x plus 5 times the content of y
The BOX / Empty Brackets skeleton approach to expressions:
Page 10 of 17
Evaluation of Algebraic Expressions (substitution of values)
In evaluating expressions, some things must be done before others. The order things are
done (the precedence) makes sure that an expression has a unique meaning. We have
already seen precedence in the Arithmetic Revision section.
Example
Evaluate 3xy  2 x 2  xz when x  4, y  3, z  5
Solution
3 xy  2 x 2  xz  3(4)( 3)  2(4)2  (4)(5)
 36  2.(16)  20
 36  32  20
 24
Example
Evaluate
t 2  20t  17 when (i) t  2
(ii) t  3
Solution
When t = 2
t 2  20t  17  (2) 2  20(2)  17
 4  40  17
 19
(i)
When t = -3
 t 2  20t  17  (3) 2  20(3)  17
(ii)
 9  60  17
 86
Example
Evaluate
xy  z 2
when x  4, y  5, z  3
2 xy
Solution
Example
Evaluate 2tw  t 2  4t  8w  4 when t = 4, w = -3
Solution
Page 11 of 17
We can now return to the problems we had at the start of the section:
Some examples we will be able to do at the end of this section:
y 2  x 2  10
when y  6, x  4, z  2
10z  y
1.
Evaluate
2.
The distance s metres travelled in time t seconds is given by the formula
1
s  ut  at 2 ,where u is the initial velocity in m/s and a is the acceleration in m/s2.
2
Find the distance s the body travels in t = 6 seconds, with an initial velocity of u =
10 m/s at an acceleration a=5m/s2.
3
Find the roots of the quadratic equation x 2  6x  6  0 given that the roots of a
quadratic equation Ax 2  Bx  C  0 can be got from the formula
B  B2  4 AC
and that here A  1, B  6, C  6 .
x
2A
y  y1
Use the formula slope  2
to find the slope of the line between
x 2  x1
(x1, y1) = (-2, 1) and (x2, y2) = (3, 3)
4
5
Find the length of the side a in the triangle given
b sin A
using the sine rule formula: a 
sin B
b = 12
A = 50 degrees
B = 20 degrees
6.
a=?
Find the accumulated amount A using the compound Interest formula A=P(1+i)n
on a loan for the principal amount P = €1,500 taken out for a time n = 7 years at
an annual rate of interest, i = 6%.
Q1
Evaluate
y 2  x 2  10
when y  6, x  4, z  2
10z  y
Solution
Page 12 of 17
Q2
The distance s metres travelled in time t seconds is given by the formula
1
s  ut  at 2 ,
2
where u is the initial velocity in m/s and a is the acceleration in m/s2.
Find the distance s the body travels in t = 6 seconds, with an initial velocity of u = 10 m/s
at an acceleration a=5m/s2.
Solution
Q3
Find the roots of the quadratic equation x 2  6x  6  0 given that the roots of a quadratic
B  B2  4 AC
equation Ax 2  Bx  C  0 can be got from the formula x 
and that here
2A
A  1, B  6, C  6 . (Hint first work out B 2  4 AC )
Solution
Page 13 of 17
Q4
Use the formula slope 
y 2  y1
to find the slope of the line between (x1, y1) = (-2, 1) and
x 2  x1
(x2, y2) = (3, 3)
Solution ( Answer: slope =0.4)
Further Exercises
Calculate the slope of the lines passing through the following points
1. (6,5) and (3,4)
2.
(2,4) and (6,8)
Page 14 of 17
Q5
b = 12
Find the length of the side a in the triangle given
b sin A
using the sine rule formula: a 
sin B
A = 50 degrees
B = 20 degrees
Further Exercise:
Find the length of the side a in the triangle
given using the cosine rule formula:
c 2  a 2  b 2  2ab cos(C )
a=?
C = 34˚ b = 6.1
a =11.2
c =?
Q6
Find the accumulated amount A using the compound Interest formula A=P(1+i)n on a
loan for the principal amount P = €1,500 taken out for a time n = 7 years at an annual
rate of interest, i = 6%.
Solution
Page 15 of 17
Further Exercise( challenge!!)
By picking out two points on the line
shown, find the slope of the line:
Hint look at the footnote first)
Solution
line 1
-2
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
-1
0
y
1
2
3
4
5
6
Footnote on Slope of a line.
The slope m of a straight line through the points
A ( x1, y1 ) and B ( x2 , y 2 ) is given by
y
B (x2, y2)
y2- y1
A (x1, y1)
slope 
difference in x - values y 2  y1

.
difference in x - values x2  x1
Page 16 of 17
O
x2- x1
x
Selected answers to“Some examples we will be able to do at the end of this section”
Q1. When you are square rooting a large expression, think of the large expression as
having a bracket round it. Precedence will then tell you to calculate this expression before
you do the square root
 (6) 2  42  10 
y 2  x 2  10

 

10 z  y
 10(2)  6 
 36  16  10 
 

 20  6 
 42 
   3
 14 
Q2
1 2
at
2
1
s  [10](6)   5  (6) 2
2
s  60  90
s  ut 
s  150 metres
Q3
roots of the quadratic equation Ax 2  Bx  C  0 are given formula
 [6]  [6]2  4(1)  6 
B  B2  4 AC
. Here A  1, B  6, C  6 . => x 
x
2A
2(1)
6  36  24
6  12
=> x 
2
2
6  3.46410
6  3.46410
6  3.46410
=> x 
=> x 
or x 
2
2
2
=>x =4.73205 or x =1.26795
=> x 
Q4
y 2  y1
x 2  x1
(3)  (1)
2
=> slope 

(3)  (2) 5
(x1, y1) = (-2, 1) and (x2, y2) = (3, 3) => slope 
Q5 B = 20 degrees, A = 50 degrees side b =12
b sin A
(12) sin(50)
(12)0.766
=> a 
=> a 
a
sin B
0.342
sin(20)
Page 17 of 17
=> a  26.88