Module 5: Algebra - Mathematics Pathways

Pre-Test for Module 5: Algebra

Click on the link below to take the Pre-Test for Module56.  The Pre-Test is optional but we recommend taking it to test your knowledge of Algebra. There are only 5 questions and it will only take about 10 minutes to complete. 

If you receive 80% or greater on the Pre-Test, you have a good knowledge of basic Algebra and can move on to the next module or review the materials in module 5.

If you receive less than an 80%, work your way through the module and then take the quiz at the end to test your knowledge.

Module 5: Algebra

Manipulations of Equations

 

Manipulation of equations relies on the fact that if two quantities are equal then “doing the same thing to each of them” will maintain the equality. In other words, if an equation is visualised as a scale, a balance must be kept at all times so that what you do to one side of the "=" you should also do to the other side!
  

 

Image: http://mathbitsnotebook.com/Algebra1/LinearEquations/scales.jpg 

 

Example

8 + 4 = 7 + 5 
8 + 4 – 4 = 7 + 5 – 4    (here we are subtracting 4 from both sides of the equation)
8 = 7 + 1          
(Note: that in the first line the left hand side and the right hand side of the equation sum to 12 and in the third line the two sides sum to 8. What is important is that equality is maintained at all times).

 

The following link to the Maths is fun Introduction to Algebra page deals with the idea of “keeping the balance” either side of the equal sign.
 

 

Common Misconceptions

Inappropriate use of the equal sign

1.  This misconception often arises when the equal sign is seen only as an instruction to “compute something”. For instance, consider the example  8 + 4 = + 5

Some people would incorrectly place 12 in the box because they think that they need to put “the answer” to what is on the left hand side. Whereas the correct response would be to place 7 in the box so that equality between the left hand side and the right hand side of the equation is maintained.
 
2.  A similar inappropriate use of the equal sign occurs when computations are completed as “run on” calculations.

Example, consider the following word problem

“A boy has five marbles and then a friend gave him ten more and then he lost two. How many marbles does the boy have now?”

Some people would incorrectly write 
5 + 10 = 15 - 2 = 13

Even though the final “answer” is correct the mathematical reasoning is not written correctly. This is because equality is not maintained.

A correct response to the problem could be:

5 + 10 = 15
15 – 2 = 13

Therefore the boy has 13 marbles

Practice Questions

Complete the following equations:

 

8 = 3 +

3 + 5 = 2 +

17 = + 17

 + 4 = 5 + 7

9 + 7 = + 9

 

Click here to check your answers

Pronumerals

Pronumerals are the letters used in algebra and they stand for a value (or an unknown amount).  Repeated occurrences of the same pronumeral in an expression represent the same value. For example, in the equation x + x + x = 12 we know that x is equal to 4.

We can think of Algebra as generalised arithmetic. That is, algebra allows us to make statements about the relationships between numbers that hold for all numbers – not just particular numbers that we deal with in arithmetic calculations. In Algebra, the laws of arithmetic (commutative, associative, and distributive) hold.  Note that the symbol ≠, means "does not equal" or "not equal to".

 

 

 

Commutative, Associative and Distributive Properties

Arithmetic	Generalised	Meaning
3 + 4 = 4 + 3	a + b = b + a	Any 2 numbers can be added in any order without affecting the result.
3 x 4 = 4 x 3	a x b=b x a (that is ab = ba)	Any 2 numbers can be multiplied in any order without affecting the result.
3 – 4 ≠ 4 − 3	a – b ≠ b − a
12 ÷ 4 ≠ 4 ÷ 12	a ÷ b ≠ b ÷ a

Associative Property

Arithmetic	Generalised
(5 + 6) + 8 = 5 + (6 + 8)	(a + b) + c = a + (b + c)
(7 – 3) – 2 ≠ 7 – (3 – 2)	a x b=b x a (that is ab = ba)
(5 x 2) x 4 = 5 x (2 x 4)	(a x b) x c = a x (b x c)
(20 ÷ 5) ÷ 2 ≠ 20 ÷ (5 ÷ 2)	(a ÷ b) ÷ c ≠ a ÷ (b ÷ c)

Watch the following video, it discusses the commutative, associative and distributive properties.

Distributive Property

 

Arithmetic	Generalised
35 x 6 = (30 x 6) + (5 x 6) Let us look at how this works: 6 x (30 + 5) can you see that we have split the 35 up into 30 + 5. Another piece of shorthand in algebra is to omit the 'x' sign so it becomes 6(30 + 5)and so: 6(30 + 5)= (6 x 30) + (6 x 5)	a x (b + c) = a (b + c) = (a x b) + (a x c) = ab + ac You may remember calling this "expanding brackets"
48 ÷ 4 = (40 ÷ 4) + (8 ÷ 4) Let us look at how this works: We rewrite 48 as (40 + 8) so now we can write: (40 + 8) ÷ 4 We can rewrite this as: Which is equivalent to: =( 40 ÷ 4) + (8 ÷ 4) = 10 + 2 = 12 It is not necessary to include all these steps, they have been provided here to show you how it works.	(a + b) ÷ c =  =
(5 x 2) x 4 = 5 x (2 x 4)	(a x b) x c = a x (b x c)
(20 ÷ 5) ÷ 2 ≠ 20 ÷ (5 ÷ 2)	(a ÷ b) ÷ c ≠ a ÷ (b ÷ c)

Watch the following video, it explains how to combine like terms in algebraic expressions by working through some examples

Is this equation true? Is it always/never/sometimes true? Explain
= (x + y)(x + y)
=  x² +  xy +  xy + y²
= x² +  2xy + y²
We can see this visually using the area model below.

Common Misconceptions

Juxtaposition 

The use of juxtaposition as a short-hand for multiplication is common in algebra.

For example 6y is used to denote 6 multiplied by y.

Examples

Simplify the following expressions without using the multiplication symbol.

1. 5 x y = 5y
2. t x 8 = 8t
3. x x y (it would be easy to confuse the multiplication sign with the pronumeral x, so it is more convenient and less confusing to rewrite this expression as xy).
4. M x 4 x N= 4MN  (order is not important because multiplication is commutative but by convention we put the pronumerals in alphabetical order after the number).

Vinculum

The vinculum (horizontal fraction bar) acts as a set of brackets for the expressions, in the numerator and in the denominator so is equal to (6x+4)÷2 

In the above expression, firstly x was multiplied by 6, then 4 was added to the result, then then the new result was divided by 2.

An equivalent way of representing the expression (please refer to Big Idea for more detail on what we mean by an algebraic expression) is:

• (6x + 4) because dividing by 2 is the same as multiplying by . This expression can be simplified further. One approach is to recognise that the expression could also be written as:

• now we can simplify this to 3x +2.

We can explain an algebraic expression in terms of a series of actions on an unknown number. For example, 3 (x + 4 ) + 2 can be explained as a series of actions on x in a flow chart.

 

Practice Task 1

1.  Write an equivalent way of expressing the following algebraic expression without using the fraction bar (vinculum)

 

2.  Explain how the expression 3x + 2 is different to the expression 3(x + 2). 

3.  Simplify the expression 

+4

     

Please click here to check your answers 

 3x + 8

 x(x – 3)

(x must not be equal to 5. Why?)

Practice Task 2

Match each algebraic expression with its corresponding worded explanation.

Image: http://www.educationaldesigner.org/ed/volume1/issue1/article3/images/figure_3a_large.png

 

Image: http://www.educationaldesigner.org/ed /volume1/issue1/article3/images/figure_3b_large.png 

 

Click here to check your answers

Practice Task 3

Simplify the expression

 

 

Click here to check your answer 

Common Misconceptions

• Students may only “cancel” one part of an expression (e.g., might be simplified to 3x+4 because the student may think that once the 2 has been cancelled in the 6 and the 2 (denominator) it is no longer there to be cancelled as a factor of 4). 

The reason why this is wrong is because the student has only divided 6x by 2 and not divided 4 by 2. Remember the vinculum acts as a set of brackets around the 6x + 4 and, as division is distributive, we must divide both  6x and 4 by 2.

• Students may not distinguish between expressions such as 3x+2 and 3(x+2). 

One and 2-step Equations

You may find the following interactive activities from Khan Academy useful. The first link focuses on one step equations and the second link deals with 2 step equations.

Think about how you are using inverse operations to solve the equations.

One Step Equations

 2-Step Equations  

  

All Khan Academy content is available for free at www.khanacademy.org/

Practice Task 1

Solve the following equations and make a note of each inverse operation you used in each step for each equation.

1.  14x = 98

2.  5y + 10 = 3y

3.

4.  For what value of x are the following expressions undefined:

a) 

b)  

c)  

 

Click here to check your answers 

Common Misconceptions

• Students may fail to “do the same thing to BOTH sides” of an equation. The example shown below is correct. You will notice if you follow through the reasoning line by line, every operation is performed on both sides of the equation to maintain equality at all times. Think about the inverse operations that are performed in order to finally get to the solution that x = 4.

www.ndt-ed.org/EducationResources/Math/mathimages/algeq.jpg

• It’s easy to forget to consider both square roots when required, or may fail to eliminate a square root that has no meaning in the physical circumstances. 

x² – 81 = 0

x² = 81

x= +9 or x = -9

• In the worked example above both the negative and the positive solutions to this equation must be considered because x can take on two values, either -9 or +9 to make this equation true. On the other hand consider the following scenario: A child asks you to work how old her sister by telling you “If you square my sister’s age and then subtract 81 you get 0, how old is my sister?”. This scenario is explained by the equation above. This time, however, we can say that the answer is 9 because we can eliminate -9 as a solution because it has no physical meaning in this particular context (clearly the sister could not be -9 years old!). Often only one solution is physically reasonable.

• Students may either divide by zero or divide by expressions that may be equal to zero. Dividing by zero is not mathematically allowable. Dividing by zero is undefined. As suggested earlier, more about what dividing by zero is undefined means at Maths is Fun Dividing by Zero

 

Pronumeral Representing an Unknown Number

There are some situations where the requirement is to “solve for x” (i.e., find the particular value of x) or some other pronumeral as in the following examples:

Example 1

Find the length of the hypotenuse (h) of the right angle triangle below.

Solve 

 

Image: http://www.cimt.plymouth.ac.uk/projects/mepres/book8/bk8i3/bk8_3i2.htm

 

In the triangle above, we want to find the length of the hypotenuse (h).

You may recall Pythagoras' Theorem, which tells us: The square of the hypotenuse is equal to the sum of the squares of the two shorter sides.

In this case, this means that h² is equal to 5² + 12².

h²=5² + 12² (work out the squares of the two shorter sides)
h²=25 + 144 (add the squares of the two shorter sides together)
h²=169 (square root both sides to find the value of h)
h=13

Note: we are only interested in positive 13 metres for our answer. Recall that the square root of 169 is either be positive 13 or negative 13. We can omit the negative answer because -13 metres makes no practical sense. 

Example 2

Solve the following equation for x.

9x + 3 = 21

9x + 3 – 3 = 21 – 3

(Note: 3 has been subtracted from both sides of the equation to maintain equality as discussed in Big Idea 1).

9x = 18

(both sides of the equation have been divided by 9)

x = 2 

Example 3

Solve the following equation for x.

 

Image: http://tkendall.edublogs.org/files/2010/03/Notation2_350.jpg

Construct and Solve Equations

The following link to the Khan Academy is an interactive task that will provide you with the opportunity to construct and solve equations.

Constructing and Solving Equations 

Pronumerals to Represent a Variable

There are other situations where pronumerals are used to describe the relationship between quantities that vary (such as v = s/t, which represents the relationship between velocity (v), displacement (s), and time (t), or y = 3x+2, which describes a linear function relating y and x; each of the variables can vary but only in ways constrained by the relationship). In situations like this we refer to the unknowns as variables because they can and do vary with respect to the other variables in the relationship.  

These ideas underpin many of the calculations that are performed in scientific fields of endeavour, including the health sciences. 

The formula used calculate Body Mass Index (BMI), shown below,  illustrates the relationship between  several variables, Body Mass Index, height and weight.  That is, as weight increases the BMI increases, so the BMI is proportional to weight. On the other hand, as height increases BMI decreases, so there is an inversely proportional relationship between the variables height and BMI.

Body Mass Index = weight/(height)²

There is a linear relationship between BMI and weight (e.g., halve weight -> halve BMI).

There is a non-linear relationship between BMI and height (e.g., halve height-> 4 x BMI)

The equation above can also be rearranged  to make one of the other variables such as weight or height, the subject. 

w=BMI×h²

 

Example 1 

Let us reconsider example 1 from Big Idea 4. The question asked, is the following relationship true? Is it always/never/sometimes true? Explain

x + y + z = x + p + z 

This relationship is true when y = p 

Now we call y and p variables because they can vary, provided that the value of y is equal to the value of p. In this case y and p can take on any value, provided that y = p.

So we can say that the relationship is true if and only if y = p

Example 2

Consider the relationship

y = 2x + 3

The pronumerals in this relationship are variables because the value of y varies with the value of x in a particular way. For example, when x = 1, y = 5, when x = 2, y = 7 and so on. In other words the value of y is equal to two times the value of x plus 3.

Example 3

Archaeologists can estimate a person’s height from the length of the femur (thigh bone). The formula for doing so is:

H = 2.3 L + 61.4

Where all measurements are in centimetres and H represents height (of person) and L represents length (of femur). This formula suggests that a person’s height is equal to 2.3 times the length of his/her femur plus 61.4. 

Let us say that the archaeologist has found a femur of a human being, and the bone is 45cm long. How tall do we think that person must have been?

Solution:

The relationship between height and femur length is

H = 2.3 L + 61.4 we know the that femur length is 45cm
H = 2.3 x 45 + 61.4
H = 164.9

Therefore the archaeologist predicts that the person was about 165cm tall.

Note that H and L are variables because the value of H is dependent upon the value of L. 

 

Example 4

Suppose that the archaeologist found another femur of a human that measured 52 cm in length. How tall do we think that person must have been?

H = 2.3 L + 61.4
H = 2.3 x 52 + 61.4
H = 181

Therefore the archaeologist predicts the person must have been 181cm tall.

Practice Task 1

1. Solve the equation 3x – 2 = x + 4
2. Solve the equation 2(t + 3) = t + 7
3. A delivery company charges according to the equation C = 5W + 8 where ‘C’ is the charge in dollars and ‘W’ is the weight of the item in kg. What will to cost to have a 6kg package delivered?
4. Sheldon is twice as old as Alan, who is 5 years older than Claire. If their total age is 27 years, how old is Claire?
5. The same delivery company in question 3 above, charged $63 to deliver a particular package. What was the weight of the package. 

Common Misconceptions

Some people have difficulty distinguishing between situations in which the pronumeral  is an unknown and when it is a variable. It does not help that many resources, text books, and explanations tend to use the term variable to describe pronumerals even in situations where the pronumeral is representing a particular unknown number.  

Click on the link below to take the online self-assessed quiz.

There are 10 mathematics questions on the Quiz and they are about the information in this module.  

To pass this quiz, you will need to get a mark of 80%.  Feedback will be provided for both correct and incorrect answers at the end of the Quiz.  If you answer questions incorrectly, then it is strongly recommended that you review the sections of the modules to review those topics.  You will be able to re-take the quiz if needed.

Make sure to enter your name and email address in the quiz so your results can be mailed to you for your records.  You may need to show your results to your university.

Module 5: Algebra Quiz