UTAS Home › › Mathematics Pathways › Pathways to Health Science › Module 5: Algebra
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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.
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.
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
Complete the following equations:
8 = 3 +
3 + 5 = 2 +
17 = + 17
+ 4 = 5 + 7
9 + 7 = + 9
Click here to check your answers
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".
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) 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) You may remember calling this "expanding brackets" |
48 ÷ 4 = (40 ÷ 4) + (8 ÷ 4) Which is equivalent to: |
(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
Example 1
Consider the equation: x + y + z = x + p + z
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.
Solution:
Repeated occurrences of the same pronumeral represent the same value. In the case of this equation, the x on the left hand side is equal to the x on the right hand side and the z on the left hand side of the equation is equal to the z on the right hand side of the equation. We can therefore reason that the only time this equation can be true is when y = p.
Example 2
Write an equation to represent the following situation: There are 6 students for every professor.
Solution:
Let the number of students be S and let the number of Professors be P. The total number of students is equal to six times the number of professors. So we have the equation:
S = 6 x P
Which we can rewrite as S = 6 P
Example 3
Use the distributive property to expand the following expression:
(x+y)2
We can see this visually using the area model below
Image: http://ajwilson.co.uk/teaching/AF0/lecture_notes/Images/x+ysquared.GIF
As you can see in the rectangle on the left, the area of the small white square inside the rectangle is x2 and the area of each of the yellow rectangles inside the main rectangle is xy so the two together make 2xy and then the area of the larger square inside the main rectangle is y2. So altogether we have x2 + 2xy + y2
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.
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:
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.
You may find the following interactive activity from the Khan Academy useful revision
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
When writing algebraic expressions from words, it is useful to be familiar with terms such as:
3x + 8
x(x – 3)
(x must not be equal to 5. Why?)
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
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.
www.washoe.k12.nv.us/ecollab/washoemath/dictionary/vmd/images/i/erationsoppositeoperations.gif
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.
All Khan Academy content is available for free at www.khanacademy.org/
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
www.ndt-ed.org/EducationResources/Math/mathimages/algeq.jpg
x2 – 81 = 0
x2 = 81
x= +9 or x = -9
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
The following link to the Khan Academy is an interactive task that will provide you with the opportunity to construct and solve equations.
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.
Click here to check your answers
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.
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2 May, 2018
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