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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.
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).
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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:
Practice Task 2
What value of m will make the following number sentence true?
4m + 10 = 70
Do you now understand the basics of manipulating equations and that you need to maintain the equality of the equation by doing the same operation to both sides of the equal sign.
Please continue on to Big Idea 2
Pronumerals are the letters used in algebra and they stand for numbers. Repeated occurrences of the same pronumeral in an expression represent the same value.
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".
Learning Activity 1
Watch this YouTube video on the commutative and associative properties of numbers and algebra.
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) |
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) |
Learning Activity 2
Watch the YouTube video below which discusses the distributive property of numbers:
The following YouTube video explains how to combine like terms in algebraic expressions by working through some examples.
1. Consider the equation: x + y + z = x + p + z
Is this equation true? Is it always/never/sometimes true? Explain
2. Write an equation to represent the following situation: There are 6 students for every professor.
ExampleWatch the following video which conceptualises the distributive property in terms of the area of a rectangle. You may find this video useful for completing the next practice example.
Use the distributive property to expand the following expression:
(x+y)2
= (x + y)(x + y)
= x2 + xy + xy + y2
= x2 + 2xy + y2
We can see this visually using the area model below.
Image: Andrew Wilson
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
This also means that talking about 2a + 3b as 2 apples plus 3 bananas is incorrect and unhelpful even if it leads to correct answers in simple cases. Another important consideration is how we interpret the meaning of pronumerals in mathematical formulae.
Consider an example involving measuring units, such as finding the area of a rectangle (i.e., A = L W). The L and the W stand for the number of units and not for the words (i.e., length and width), even though we say the formula as if the pronumerals stand for words.
The purpose of this module was to help you understand;
Does this make sense to you now?
If so, please continue on to Big Idea 3
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.
Use the following interactive tool from Khan Academy for revision and to test your skill on Equivalent forms of expressions
When writing algebraic expressions from words, it is useful to be familiar with terms such as:
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
Match each algebraic expression with its corresponding worded explanation.
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Image Card Set 2: Educational Designer
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.
The purpose of Big Idea 3 was to help you understand
Is this clear to you now?
Continue on to Big Idea 4
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Find out more about what dividing by zero is undefined means at
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You may find the following interactive activities from the 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.
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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 is 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!).
Solve the following equations and make a note of each inverse operation you used in each step for each equation
a.
b.
c.
The purpose of Big Idea 4 is to give you an understanding of inverse operations
Do you now see that
are inverse operations?
Continue to onto Big Idea 5
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
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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: that 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 that 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.
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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.
Example 1
Let us reconsider Practice Task 1 from Big Idea 2.
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.
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.
The purpose of this Big Idea was to further develop your understanding of pronumerals.
In Big Idea 2, we looked at:
We have now built upon this and learned about
If you have completed all the Big Ideas and feel you have a good understanding of each Big Idea, then test yourself with the module Quiz for the Relationships between Fractions, Decimals, Ratios and Percentages
Good Luck!!
Authorised by the Director, Centre for University Pathways and Partnership
2 May, 2018
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