basic algebra skills · basic algebra skills numeracy workshop [email protected]...
TRANSCRIPT
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Introduction
These slides are intended to give you a basic introduction to algebra.
Drop-in Study Sessions: Monday, Wednesday, Thursday, 10am-12pm, MeetingRoom 1.15, First floor, Guild Building, every week.
Ask a Maths Question: See the website.
Website: Slides, worksheet, solutions, online quiz.
www.studysmarter.uwa.edu.au → Numeracy → Online Resources
Email: [email protected]
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Introduction
In the Number Skills Workshop, we said that Addition is commutative. Thismeans that when you add two numbers together, the order does not matter.
3 + 17 = 17 + 3
Statements such as Addition is commutative, are mathematical in nature, and sowe wish to express these statements in the form of expressions and equations,
rather than English.
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Introduction
In the Number Skills Workshop, we said that Addition is commutative. Thismeans that when you add two numbers together, the order does not matter.
3 + 17 = 17 + 3
Statements such as Addition is commutative, are mathematical in nature, and sowe wish to express these statements in the form of expressions and equations,
rather than English.
[email protected] () Basic Algebra Skills 3 / 54
![Page 5: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/5.jpg)
Introduction
Instead of saying Addition is commutative, we can say that for any two numbers xand y , we have that
x + y = y + x
Using x and y to represent numbers, means we don’t have to be specific aboutwhich numbers we use. We can write our rules using pronumerals rather than
numerals.
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Introduction
Instead of saying Addition is commutative, we can say that for any two numbers xand y , we have that
x + y = y + x
Using x and y to represent numbers, means we don’t have to be specific aboutwhich numbers we use. We can write our rules using pronumerals rather than
numerals.
[email protected] () Basic Algebra Skills 4 / 54
![Page 7: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/7.jpg)
Multiplication
xy means “x times y”.
5x means “5 times x” or “5 lots of x”.
5xy2 means “5 times x times y squared”.
4y means “4 divided by y”.
2xy means “2 times x , divided by y”.
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Multiplication
xy means “x times y”.
5x means “5 times x” or “5 lots of x”.
5xy2 means “5 times x times y squared”.
4y means “4 divided by y”.
2xy means “2 times x , divided by y”.
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Multiplication
xy means “x times y”.
5x means “5 times x” or “5 lots of x”.
5xy2 means “5 times x times y squared”.
4y means “4 divided by y”.
2xy means “2 times x , divided by y”.
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Multiplication
xy means “x times y”.
5x means “5 times x” or “5 lots of x”.
5xy2 means “5 times x times y squared”.
4y means “4 divided by y”.
2xy means “2 times x , divided by y”.
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Multiplication
xy means “x times y”.
5x means “5 times x” or “5 lots of x”.
5xy2 means “5 times x times y squared”.
4y means “4 divided by y”.
2xy means “2 times x , divided by y”.
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Variables
In the previous slides, x and y are called variables.
This is because we haven’t identified them with specific numbers, and so theyrepresent a whole range of different numbers. In other words, they vary.
Variables are place-holders for numbers
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Variables
In the previous slides, x and y are called variables.
This is because we haven’t identified them with specific numbers, and so theyrepresent a whole range of different numbers. In other words, they vary.
Variables are place-holders for numbers
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Variables
In the previous slides, x and y are called variables.
This is because we haven’t identified them with specific numbers, and so theyrepresent a whole range of different numbers. In other words, they vary.
Variables are place-holders for numbers
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Expressions
An expression is a combination of numbers and variables.
Some examples are:
5x
4
3x + 5
2xy + 4x2y + 2x + 4x
7x2 −√x + 1
x
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Expressions
An expression is a combination of numbers and variables.
Some examples are:
5x
4
3x + 5
2xy + 4x2y + 2x + 4x
7x2 −√x + 1
x
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Terms
Expressions are made up of terms.
4x2 + 2xy − 4x
y+ 7xy − 2x2
The terms are the pieces of the above equation which are being added or
subtracted. We will put a box around each term:
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Terms
Expressions are made up of terms.
4x2 + 2xy − 4x
y+ 7xy − 2x2
The terms are the pieces of the above equation which are being added or
subtracted. We will put a box around each term:
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Terms
Every term in the expression is connected to the sign which is directly tothe left of it.
Here is an expression:
4x2+2xy−4x
y+7xy−2x2
We see that 2xy is connected to the + sign, and 4xy is connected to the − sign.
We see that 4x2 has no sign to the left of it, but this absence of a sign meansthat it is positive, and so in a sense there is a hidden + sign.
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Terms
Every term in the expression is connected to the sign which is directly tothe left of it.
Here is an expression:
4x2+2xy−4x
y+7xy−2x2
We see that 2xy is connected to the + sign, and 4xy is connected to the − sign.
We see that 4x2 has no sign to the left of it, but this absence of a sign meansthat it is positive, and so in a sense there is a hidden + sign.
[email protected] () Basic Algebra Skills 9 / 54
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Terms
Every term in the expression is connected to the sign which is directly tothe left of it.
Here is an expression:
4x2+2xy−4x
y+7xy−2x2
We see that 2xy is connected to the + sign, and 4xy is connected to the − sign.
We see that 4x2 has no sign to the left of it, but this absence of a sign meansthat it is positive, and so in a sense there is a hidden + sign.
[email protected] () Basic Algebra Skills 9 / 54
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Terms
Every term in the expression is connected to the sign which is directly tothe left of it.
Here is an expression:
4x2+2xy−4x
y+7xy−2x2
We see that 2xy is connected to the + sign, and 4xy is connected to the − sign.
We see that 4x2 has no sign to the left of it, but this absence of a sign meansthat it is positive, and so in a sense there is a hidden + sign.
[email protected] () Basic Algebra Skills 9 / 54
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Terms
We put a box around each term so that each term is coupled with its sign.
We can move these blocks around to change how the expression looks, whilekeeping it equivalent.
If we move the block containing 4x2 from the left, we need to give it a + sign.
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Terms
We put a box around each term so that each term is coupled with its sign.
We can move these blocks around to change how the expression looks, whilekeeping it equivalent.
If we move the block containing 4x2 from the left, we need to give it a + sign.
[email protected] () Basic Algebra Skills 10 / 54
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Terms
We put a box around each term so that each term is coupled with its sign.
We can move these blocks around to change how the expression looks, whilekeeping it equivalent.
If we move the block containing 4x2 from the left, we need to give it a + sign.
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Moving terms
We can change how an expression looks by moving the terms around. When wemove a term, we must keep it connected to the same sign, that is, the sign on the
left of the term stays with it wherever it goes.
These three expressions are equivalent:
4x2 + 2xy − 4x
y+ 7xy − 2x2
4x2 − 4x
y− 2x2 + 7xy + 2xy
2xy − 2x2 + 7xy + 4x2 − 4x
y
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Moving terms
We can change how an expression looks by moving the terms around. When wemove a term, we must keep it connected to the same sign, that is, the sign on the
left of the term stays with it wherever it goes.
These three expressions are equivalent:
4x2 + 2xy − 4x
y+ 7xy − 2x2
4x2 − 4x
y− 2x2 + 7xy + 2xy
2xy − 2x2 + 7xy + 4x2 − 4x
y
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Exercise
Out of the following four expressions, which two are the same?
2x2 + 4x − 7− 2xy
−2xy + 4x − 7− 2x2
4x − 2x2 − 7− 2xy
−7 + 4x + 2xy − 2x2
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Exercise
Out of the following four expressions, which two are the same?
2x2 + 4x − 7− 2xy
−2xy + 4x − 7− 2x2
4x − 2x2 − 7− 2xy
−7 + 4x + 2xy − 2x2
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Like Terms
Some terms are made up using the same combination of variables, withmatching variables having matching powers. These are called like terms.
The terms 4xy and 17xy are like terms, as both consist of the same variables.
The terms 4xy2 and 3xy2 are like terms. Both consist of the same variables, andin both cases the y is being squared.
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Like Terms
Some terms are made up using the same combination of variables, withmatching variables having matching powers. These are called like terms.
The terms 4xy and 17xy are like terms, as both consist of the same variables.
The terms 4xy2 and 3xy2 are like terms. Both consist of the same variables, andin both cases the y is being squared.
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Like Terms
Some terms are made up using the same combination of variables, withmatching variables having matching powers. These are called like terms.
The terms 4xy and 17xy are like terms, as both consist of the same variables.
The terms 4xy2 and 3xy2 are like terms. Both consist of the same variables, andin both cases the y is being squared.
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Like Terms
The terms −6w3xyz2 and 22w3xyz2 are like terms. Both consist of the samevariables, and in both cases the w is being cubed, and the z is being squared.
The terms 7 and 3 are like terms, as they are both simply numbers.
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Like Terms
The terms −6w3xyz2 and 22w3xyz2 are like terms. Both consist of the samevariables, and in both cases the w is being cubed, and the z is being squared.
The terms 7 and 3 are like terms, as they are both simply numbers.
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Like Terms
The terms −6xy and yx are like terms. Both consist of the same variables. Theorder in which they appear is irrelevant. This is because multiplication iscommutative.
The terms 52y2x3z and −15zy2x3 are like terms. Both consist of the samevariables. The order in which they appear is irrelevant. It is also important tonote that in both terms, y is being squared, x is being cubed, and z is single.
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Like Terms
The terms −6xy and yx are like terms. Both consist of the same variables. Theorder in which they appear is irrelevant. This is because multiplication iscommutative.
The terms 52y2x3z and −15zy2x3 are like terms. Both consist of the samevariables. The order in which they appear is irrelevant. It is also important tonote that in both terms, y is being squared, x is being cubed, and z is single.
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Unlike Terms: Examples
The terms 5 and 6x are unlike. One contains an x whereas the other does not.
The terms 4xy2 and −2xy are unlike. In the first term, y is being squared but inthe second term, y is not.
The terms 2x2y and 3y2x are unlike. In the first term, x is being squared, but inthe second term y is being squared.
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Unlike Terms: Examples
The terms 5 and 6x are unlike. One contains an x whereas the other does not.
The terms 4xy2 and −2xy are unlike. In the first term, y is being squared but inthe second term, y is not.
The terms 2x2y and 3y2x are unlike. In the first term, x is being squared, but inthe second term y is being squared.
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Unlike Terms: Examples
The terms 5 and 6x are unlike. One contains an x whereas the other does not.
The terms 4xy2 and −2xy are unlike. In the first term, y is being squared but inthe second term, y is not.
The terms 2x2y and 3y2x are unlike. In the first term, x is being squared, but inthe second term y is being squared.
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Adding and Subtracting Like Terms
We can add and subtract like terms!
7xy + 4xy = 11xy
9x2y + 5yx2 = 14x2y
14xyz3 − 5z3yx + 2yz3x = 11z3yx
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Adding and Subtracting Like Terms
We can add and subtract like terms!
7xy + 4xy = 11xy
9x2y + 5yx2 = 14x2y
14xyz3 − 5z3yx + 2yz3x = 11z3yx
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Adding and Subtracting Like Terms
We can add and subtract like terms!
7xy + 4xy = 11xy
9x2y + 5yx2 = 14x2y
14xyz3 − 5z3yx + 2yz3x = 11z3yx
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Adding and Subtracting Like Terms
Note that when we see expressions without a number at the front, such as x , xy2
and wxz , these really mean 1x , 1xy2 and 1wxz . We usually just don’t write inthe 1.
2xy + xy = 3xy
14x2y − yx2 = 13x2y
xyz + yzx + yxz = 3xyz
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Adding and Subtracting Like Terms
Note that when we see expressions without a number at the front, such as x , xy2
and wxz , these really mean 1x , 1xy2 and 1wxz . We usually just don’t write inthe 1.
2xy + xy = 3xy
14x2y − yx2 = 13x2y
xyz + yzx + yxz = 3xyz
[email protected] () Basic Algebra Skills 19 / 54
![Page 45: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/45.jpg)
Adding and Subtracting Like Terms
Note that when we see expressions without a number at the front, such as x , xy2
and wxz , these really mean 1x , 1xy2 and 1wxz . We usually just don’t write inthe 1.
2xy + xy = 3xy
14x2y − yx2 = 13x2y
xyz + yzx + yxz = 3xyz
[email protected] () Basic Algebra Skills 19 / 54
![Page 46: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/46.jpg)
Adding and Subtracting Like Terms
Note that when we see expressions without a number at the front, such as x , xy2
and wxz , these really mean 1x , 1xy2 and 1wxz . We usually just don’t write inthe 1.
2xy + xy = 3xy
14x2y − yx2 = 13x2y
xyz + yzx + yxz = 3xyz
[email protected] () Basic Algebra Skills 19 / 54
![Page 47: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/47.jpg)
Simplifying Expressions
We can use our knowledge of like terms, and shifting around terms, to simplifybig scary expressions!
Example: Simplify the following expression:
4xy2 + 3xy + 2xy2 − 2yx
How many distinct terms are there? We see that 4xy2 and 2xy2 are like terms,and also 3xy and −2yx are like terms.
Remembering to keep the signs the same, we shift the expression around to putlike terms next to each other:
4xy2 + 2xy2 + 3xy − 2yx
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Simplifying Expressions
We can use our knowledge of like terms, and shifting around terms, to simplifybig scary expressions!
Example: Simplify the following expression:
4xy2 + 3xy + 2xy2 − 2yx
How many distinct terms are there? We see that 4xy2 and 2xy2 are like terms,and also 3xy and −2yx are like terms.
Remembering to keep the signs the same, we shift the expression around to putlike terms next to each other:
4xy2 + 2xy2 + 3xy − 2yx
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Simplifying Expressions
We can use our knowledge of like terms, and shifting around terms, to simplifybig scary expressions!
Example: Simplify the following expression:
4xy2 + 3xy + 2xy2 − 2yx
How many distinct terms are there? We see that 4xy2 and 2xy2 are like terms,and also 3xy and −2yx are like terms.
Remembering to keep the signs the same, we shift the expression around to putlike terms next to each other:
4xy2 + 2xy2 + 3xy − 2yx
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Simplifying Expressions
4xy2 + 2xy2 + 3xy − 2yx
Now we can use our knowledge of adding and subtracting like terms to make itsimpler. We know that 4xy2 + 2xy2 = 6xy2, and also that 3xy − 2yx = xy . Doingthis we get:
6xy2 + xy
Our expression has been simplified into something which looks slightly less scary.
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Simplifying Expressions
4xy2 + 2xy2 + 3xy − 2yx
Now we can use our knowledge of adding and subtracting like terms to make itsimpler. We know that 4xy2 + 2xy2 = 6xy2, and also that 3xy − 2yx = xy . Doingthis we get:
6xy2 + xy
Our expression has been simplified into something which looks slightly less scary.
[email protected] () Basic Algebra Skills 21 / 54
![Page 52: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/52.jpg)
Simplifying Expressions
4xy2 + 2xy2 + 3xy − 2yx
Now we can use our knowledge of adding and subtracting like terms to make itsimpler. We know that 4xy2 + 2xy2 = 6xy2, and also that 3xy − 2yx = xy . Doingthis we get:
6xy2 + xy
Our expression has been simplified into something which looks slightly less scary.
[email protected] () Basic Algebra Skills 21 / 54
![Page 53: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/53.jpg)
Simplifying Expressions: Example
Simplify the following expression:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 1: Identify like terms:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 2: Move terms so that like terms are next to each other:
xy2 + 2y2x − 3xyz + 5yzx − 4yxz
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Simplifying Expressions: Example
Simplify the following expression:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 1: Identify like terms:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 2: Move terms so that like terms are next to each other:
xy2 + 2y2x − 3xyz + 5yzx − 4yxz
[email protected] () Basic Algebra Skills 22 / 54
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Simplifying Expressions: Example
Simplify the following expression:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 1: Identify like terms:
xy2 − 3xyz + 2y2x + 5yzx − 4yxz
Step 2: Move terms so that like terms are next to each other:
xy2 + 2y2x − 3xyz + 5yzx − 4yxz
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Simplifying Expressions: Example
xy2 + 2y2x − 3xyz + 5yzx − 4yxz
Step 3: Now add the like terms together:
3xy2 − 2yxz
We have simplified our original expression!
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Simplifying Expressions: Example
xy2 + 2y2x − 3xyz + 5yzx − 4yxz
Step 3: Now add the like terms together:
3xy2 − 2yxz
We have simplified our original expression!
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Multiplying Terms
We have seen that if terms are like, then we can add or subtract them.
We would also like to multiply terms together.
It turns out, that two terms do not have to be like to be multiplied together.
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Multiplying Terms
We have seen that if terms are like, then we can add or subtract them.
We would also like to multiply terms together.
It turns out, that two terms do not have to be like to be multiplied together.
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![Page 60: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/60.jpg)
Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
We multiply the numbers together, we multiply the x ’s together, and multiply they ’s together.
Numbers are easy, we know that 3× 4 = 12.
Now we need to multiply x2 by x5. Remember that x2 really means xx (x timesx) and x5 really means xxxxx . Multiplying these we get:
x2x5 = xxxxxxx = x7
We just add the powers together!
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Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
We multiply the numbers together, we multiply the x ’s together, and multiply they ’s together.
Numbers are easy, we know that 3× 4 = 12.
Now we need to multiply x2 by x5. Remember that x2 really means xx (x timesx) and x5 really means xxxxx . Multiplying these we get:
x2x5 = xxxxxxx = x7
We just add the powers together!
[email protected] () Basic Algebra Skills 25 / 54
![Page 62: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/62.jpg)
Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
We multiply the numbers together, we multiply the x ’s together, and multiply they ’s together.
Numbers are easy, we know that 3× 4 = 12.
Now we need to multiply x2 by x5. Remember that x2 really means xx (x timesx) and x5 really means xxxxx . Multiplying these we get:
x2x5 = xxxxxxx = x7
We just add the powers together!
[email protected] () Basic Algebra Skills 25 / 54
![Page 63: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/63.jpg)
Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
We multiply the numbers together, we multiply the x ’s together, and multiply they ’s together.
Numbers are easy, we know that 3× 4 = 12.
Now we need to multiply x2 by x5. Remember that x2 really means xx (x timesx) and x5 really means xxxxx . Multiplying these we get:
x2x5 = xxxxxxx = x7
We just add the powers together!
[email protected] () Basic Algebra Skills 25 / 54
![Page 64: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/64.jpg)
Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
Now to multiply the y ’s together. We now know that we just add the powers, sowe have:
y3y2 = y5
Putting this all together we now know how to multiply the terms together:
(3x2y3)(4x5y2) = 12x7y5
[email protected] () Basic Algebra Skills 26 / 54
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Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
Now to multiply the y ’s together. We now know that we just add the powers, sowe have:
y3y2 = y5
Putting this all together we now know how to multiply the terms together:
(3x2y3)(4x5y2) = 12x7y5
[email protected] () Basic Algebra Skills 26 / 54
![Page 66: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/66.jpg)
Multiplying Terms: Example
Calculate the following multiplication of terms: (3x2y3)(4x5y2).
Now to multiply the y ’s together. We now know that we just add the powers, sowe have:
y3y2 = y5
Putting this all together we now know how to multiply the terms together:
(3x2y3)(4x5y2) = 12x7y5
[email protected] () Basic Algebra Skills 26 / 54
![Page 67: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/67.jpg)
Multiplying Terms: Examples
(2x2y)(3xy4) = 6x3y5
(−x3y3)(xy5) = −x4y8
(−5xy)(−2x2y3) = 10x3y4
(−2x2y)(−3x4y)(−4y) = −24x6y3
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Multiplying Terms: Examples
(2x2y)(3xy4) = 6x3y5
(−x3y3)(xy5) = −x4y8
(−5xy)(−2x2y3) = 10x3y4
(−2x2y)(−3x4y)(−4y) = −24x6y3
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![Page 69: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/69.jpg)
Multiplying Terms: Examples
(2x2y)(3xy4) = 6x3y5
(−x3y3)(xy5) = −x4y8
(−5xy)(−2x2y3) = 10x3y4
(−2x2y)(−3x4y)(−4y) = −24x6y3
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![Page 70: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/70.jpg)
Multiplying Terms: Examples
(2x2y)(3xy4) = 6x3y5
(−x3y3)(xy5) = −x4y8
(−5xy)(−2x2y3) = 10x3y4
(−2x2y)(−3x4y)(−4y) = −24x6y3
[email protected] () Basic Algebra Skills 27 / 54
![Page 71: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/71.jpg)
Single Bracket Expansion
We have just seen how to simplify expressions which look like:
term × term
Sometimes, we have to work with expressions of the form
term ×( term + term )
With numbers we use BIMDAS, and do what’s in the brackets first. However, thetwo terms in the brackets can only be added together if they are like terms.
If the terms are unlike, then we need to cheat BIMDAS using expansion.
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Single Bracket Expansion
We have just seen how to simplify expressions which look like:
term × term
Sometimes, we have to work with expressions of the form
term ×( term + term )
With numbers we use BIMDAS, and do what’s in the brackets first. However, thetwo terms in the brackets can only be added together if they are like terms.
If the terms are unlike, then we need to cheat BIMDAS using expansion.
[email protected] () Basic Algebra Skills 28 / 54
![Page 73: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/73.jpg)
Single Bracket Expansion
We have just seen how to simplify expressions which look like:
term × term
Sometimes, we have to work with expressions of the form
term ×( term + term )
With numbers we use BIMDAS, and do what’s in the brackets first. However, thetwo terms in the brackets can only be added together if they are like terms.
If the terms are unlike, then we need to cheat BIMDAS using expansion.
[email protected] () Basic Algebra Skills 28 / 54
![Page 74: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/74.jpg)
Single Bracket Expansion: Example
Expand the following: 4(3 + x)
This looks like term ×( term + term ), where two of the terms are just numbers.
We will use the distributive law, which tells us how to get rid of the brackets,without being able to add 3 and x with each other, as they are unlike.
Distributive Law: a(b+c) = ab + ac
where a, b and c all represent terms.
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Single Bracket Expansion: Example
Expand the following: 4(3 + x)
This looks like term ×( term + term ), where two of the terms are just numbers.
We will use the distributive law, which tells us how to get rid of the brackets,without being able to add 3 and x with each other, as they are unlike.
Distributive Law: a(b+c) = ab + ac
where a, b and c all represent terms.
[email protected] () Basic Algebra Skills 29 / 54
![Page 76: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/76.jpg)
Single Bracket Expansion: Example
Expand the following: 4(3 + x)
This looks like term ×( term + term ), where two of the terms are just numbers.
We will use the distributive law, which tells us how to get rid of the brackets,without being able to add 3 and x with each other, as they are unlike.
Distributive Law: a(b+c) = ab + ac
where a, b and c all represent terms.
[email protected] () Basic Algebra Skills 29 / 54
![Page 77: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/77.jpg)
Single Bracket Expansion: Example
Expand the following: 4(3 + x)
Basically, each term inside the brackets gets multiplied by the term out thefront. As all of our terms are positive, multiplying them will keep them positive.
So we will multiply 4 by 3 to get 12, and we will multiply 4 by x to get 4x .
Expanding the brackets gives:
4(3 + x) = 12 + 4x
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Single Bracket Expansion: Example
Expand the following: 4(3 + x)
Basically, each term inside the brackets gets multiplied by the term out thefront. As all of our terms are positive, multiplying them will keep them positive.
So we will multiply 4 by 3 to get 12, and we will multiply 4 by x to get 4x .
Expanding the brackets gives:
4(3 + x) = 12 + 4x
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![Page 79: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/79.jpg)
Single Bracket Expansion: Example
Expand the following: 4(3 + x)
Basically, each term inside the brackets gets multiplied by the term out thefront. As all of our terms are positive, multiplying them will keep them positive.
So we will multiply 4 by 3 to get 12, and we will multiply 4 by x to get 4x .
Expanding the brackets gives:
4(3 + x) = 12 + 4x
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![Page 80: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/80.jpg)
Single Bracket Expansion: Examples
4(2x + 3) = 8x + 12
3(4− 2x) = 12− 6x
4x(7 + 5x) = 28x + 20x2
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Single Bracket Expansion: Examples
4(2x + 3) = 8x + 12
3(4− 2x) = 12− 6x
4x(7 + 5x) = 28x + 20x2
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Single Bracket Expansion: Examples
4(2x + 3) = 8x + 12
3(4− 2x) = 12− 6x
4x(7 + 5x) = 28x + 20x2
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Single Bracket Expansion: More examples
−2xy(3x − 2y) = −6x2y + 4xy2
−4x2y3(6xy + 3x3y3) = −24x3y4 − 12x5y6
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Single Bracket Expansion: More examples
−2xy(3x − 2y) = −6x2y + 4xy2
−4x2y3(6xy + 3x3y3) = −24x3y4 − 12x5y6
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Single Bracket Expansion: More examples
We can even expand expressions of the form:
term ×( term + term + term )
Once again, we multiply each term in the bracket by the term out the front.
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Single Bracket Expansion: More examples
2x(3 + x + 5x4) = 6x + 2x2 + 10x5
−4xy(2xy − 4x + 3x) = −8x2y2 + 16x2y − 12x2y
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Single Bracket Expansion: More examples
2x(3 + x + 5x4) = 6x + 2x2 + 10x5
−4xy(2xy − 4x + 3x) = −8x2y2 + 16x2y − 12x2y
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Double Bracket Expansion
Now we look at double bracket expansion, which involves expanding expressionsof the form:
( term + term )× ( term + term )
Each term in the first set of brackets gets coupled up with each term in thesecond set of brackets.
There are two terms to choose from in each, so there are 4 possiblecombinations in total.
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Double Bracket Expansion
Now we look at double bracket expansion, which involves expanding expressionsof the form:
( term + term )× ( term + term )
Each term in the first set of brackets gets coupled up with each term in thesecond set of brackets.
There are two terms to choose from in each, so there are 4 possiblecombinations in total.
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Double Bracket Expansion
Now we look at double bracket expansion, which involves expanding expressionsof the form:
( term + term )× ( term + term )
Each term in the first set of brackets gets coupled up with each term in thesecond set of brackets.
There are two terms to choose from in each, so there are 4 possiblecombinations in total.
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Double Bracket Expansion: Example
Expand the following:
(3 + 2x)(5 + 3x)
We need to calculate each of the following:
3 times 5
3 times 3x
2x times 5
2x times 3x
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Double Bracket Expansion: Example
Expand the following:
(3 + 2x)(5 + 3x)
We need to calculate each of the following:
3 times 5
3 times 3x
2x times 5
2x times 3x
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Double Bracket Expansion: Example
Expand the following:
(3 + 2x)(5 + 3x)
We need to calculate each of the following:
3 times 5
3 times 3x
2x times 5
2x times 3x
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Double Bracket Expansion: Example
Expand the following:
(3 + 2x)(5 + 3x)
We need to calculate each of the following:
3 times 5
3 times 3x
2x times 5
2x times 3x
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Double Bracket Expansion: Example
Expand the following:
(3 + 2x)(5 + 3x)
We need to calculate each of the following:
3 times 5
3 times 3x
2x times 5
2x times 3x
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Double Bracket Expansion: Example
We can do each of these:
3 times 5 equals 15
3 times 3x equals 9x
2x times 5 equals 10x
2x times 3x equals 6x2
So we can expand as follows:
(3 + 2x)(5 + 3x) = 15 + 9x + 10x + 6x2
Usually, once we expand we simplify like terms, so we get:
15 + 19x + 6x2
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Double Bracket Expansion: Example
We can do each of these:
3 times 5 equals 15
3 times 3x equals 9x
2x times 5 equals 10x
2x times 3x equals 6x2
So we can expand as follows:
(3 + 2x)(5 + 3x) = 15 + 9x + 10x + 6x2
Usually, once we expand we simplify like terms, so we get:
15 + 19x + 6x2
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Double Bracket Expansion: Example
We can do each of these:
3 times 5 equals 15
3 times 3x equals 9x
2x times 5 equals 10x
2x times 3x equals 6x2
So we can expand as follows:
(3 + 2x)(5 + 3x) = 15 + 9x + 10x + 6x2
Usually, once we expand we simplify like terms, so we get:
15 + 19x + 6x2
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Double Bracket Expansion: Example
Expand the following:
(4x − 5)(2 + 4x)
We need to calculate each of the following:
4x times 2
4x times 4x
−5 times 2
−5 times 4x
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Double Bracket Expansion: Example
Expand the following:
(4x − 5)(2 + 4x)
We need to calculate each of the following:
4x times 2
4x times 4x
−5 times 2
−5 times 4x
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Double Bracket Expansion: Example
Expand the following:
(4x − 5)(2 + 4x)
We need to calculate each of the following:
4x times 2
4x times 4x
−5 times 2
−5 times 4x
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Double Bracket Expansion: Example
Expand the following:
(4x − 5)(2 + 4x)
We need to calculate each of the following:
4x times 2
4x times 4x
−5 times 2
−5 times 4x
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Double Bracket Expansion: Example
Expand the following:
(4x − 5)(2 + 4x)
We need to calculate each of the following:
4x times 2
4x times 4x
−5 times 2
−5 times 4x
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Double Bracket Expansion: Example
We can do each of these:
4x times 2 equals 8x
4x times 4x equals 16x2
−5 times 2 equals − 10
−5 times 4x equals − 20x
So we can expand as follows:
(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x
Usually, once we expand we simplify like terms, so we get:
16x2 − 12x − 10
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Double Bracket Expansion: Example
We can do each of these:
4x times 2 equals 8x
4x times 4x equals 16x2
−5 times 2 equals − 10
−5 times 4x equals − 20x
So we can expand as follows:
(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x
Usually, once we expand we simplify like terms, so we get:
16x2 − 12x − 10
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Double Bracket Expansion: Example
We can do each of these:
4x times 2 equals 8x
4x times 4x equals 16x2
−5 times 2 equals − 10
−5 times 4x equals − 20x
So we can expand as follows:
(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x
Usually, once we expand we simplify like terms, so we get:
16x2 − 12x − 10
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Double Bracket Expansion: Examples
(3− 4x)(2x − 7) = 6x − 21− 8x2 + 28x
= −8x2 + 34x − 21
(6− y)(4 + x) = 24 + 6x − 4y − xy
(6x + 7xy)(3x2 − 5y3) = 18x3 − 30xy3 + 21x3y − 35xy4
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Double Bracket Expansion: Examples
(3− 4x)(2x − 7) = 6x − 21− 8x2 + 28x
= −8x2 + 34x − 21
(6− y)(4 + x) = 24 + 6x − 4y − xy
(6x + 7xy)(3x2 − 5y3) = 18x3 − 30xy3 + 21x3y − 35xy4
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Double Bracket Expansion: Examples
(3− 4x)(2x − 7) = 6x − 21− 8x2 + 28x
= −8x2 + 34x − 21
(6− y)(4 + x) = 24 + 6x − 4y − xy
(6x + 7xy)(3x2 − 5y3) = 18x3 − 30xy3 + 21x3y − 35xy4
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Equations
An equation is a mathematical object of the form:
expression = expression
For example:
3x2y + 6xy = 7xy2 − 4y
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Equations
An equation is a mathematical object of the form:
expression = expression
For example:
3x2y + 6xy = 7xy2 − 4y
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Equations
An equation which involves the variable x , tells us how x engages with othernumbers.
From this, we might be able to deduce the identity of x . This is called solving theequation for x .
Perhaps we are given the following equation:
x + 5 = 8
This equation tells us that when you add 5 to x , you end up with 8. There is onlyone known number which acts like this and that number is 3. So x must be equal
to 3. We write this as:
x = 3
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Equations
An equation which involves the variable x , tells us how x engages with othernumbers.
From this, we might be able to deduce the identity of x . This is called solving theequation for x .
Perhaps we are given the following equation:
x + 5 = 8
This equation tells us that when you add 5 to x , you end up with 8. There is onlyone known number which acts like this and that number is 3. So x must be equal
to 3. We write this as:
x = 3
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Equations
An equation which involves the variable x , tells us how x engages with othernumbers.
From this, we might be able to deduce the identity of x . This is called solving theequation for x .
Perhaps we are given the following equation:
x + 5 = 8
This equation tells us that when you add 5 to x , you end up with 8. There is onlyone known number which acts like this and that number is 3. So x must be equal
to 3. We write this as:
x = 3
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Equations
What if we were given this equation:
4 + 2x = 20
This says, when you add four with two times x , you end up with 20.
We know from experience, that adding four with sixteen gives us 20.
So this 2x has been put there in place of 16. We now know that two times x issixteen, and so x must be equal to 8. We write:
x = 8
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Equations
What if we were given this equation:
4 + 2x = 20
This says, when you add four with two times x , you end up with 20.
We know from experience, that adding four with sixteen gives us 20.
So this 2x has been put there in place of 16. We now know that two times x issixteen, and so x must be equal to 8. We write:
x = 8
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Equations
What if we were given this equation:
4 + 2x = 20
This says, when you add four with two times x , you end up with 20.
We know from experience, that adding four with sixteen gives us 20.
So this 2x has been put there in place of 16. We now know that two times x issixteen, and so x must be equal to 8. We write:
x = 8
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Equations
Sometimes equations can be solved as we did above, by thinking.
Other times they are slightly harder to work through.
We wish to have a process which will allow us to systematically solve linearequations, perhaps the most common type.
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Equations
Sometimes equations can be solved as we did above, by thinking.
Other times they are slightly harder to work through.
We wish to have a process which will allow us to systematically solve linearequations, perhaps the most common type.
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Equations
Sometimes equations can be solved as we did above, by thinking.
Other times they are slightly harder to work through.
We wish to have a process which will allow us to systematically solve linearequations, perhaps the most common type.
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![Page 121: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/121.jpg)
Linear Equations
Linear equations are equations which only contain number terms like 3 and 6 andsingle variables with no power such as 2x and −4y .
Here are some examples:
4 + 3x = 20
3− 5y = −2
17w = 3− w
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Linear Equations
Linear equations are equations which only contain number terms like 3 and 6 andsingle variables with no power such as 2x and −4y .
Here are some examples:
4 + 3x = 20
3− 5y = −2
17w = 3− w
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Solving Linear equations
An equation is a statement that two quantities are the same. For example:
2x + 3 = 15
The above equation says that 2x + 3 is equal to 15, and our task is to find thevalue of x that makes this so.
As both sides of the equation represent the exact same quantity, adding orsubtracting the same number to both sides will keep the equation true. We may
also multiply or divide both sides by the same number.
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Solving Linear equations
An equation is a statement that two quantities are the same. For example:
2x + 3 = 15
The above equation says that 2x + 3 is equal to 15, and our task is to find thevalue of x that makes this so.
As both sides of the equation represent the exact same quantity, adding orsubtracting the same number to both sides will keep the equation true. We may
also multiply or divide both sides by the same number.
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Solving Linear Equations
This idea of tweaking both sides by the same amount proves most useful, and isthe basis for solving all sorts of equations in algebra.
2x + 3 = 15
We are trying to change the above equation to a new equation, a simpler one ofthe form
x = number.
The way we get to this new equation is to move all the numbers away from x ,so that it is sitting on a side of the equals sign all by itself.
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Solving Linear Equations
This idea of tweaking both sides by the same amount proves most useful, and isthe basis for solving all sorts of equations in algebra.
2x + 3 = 15
We are trying to change the above equation to a new equation, a simpler one ofthe form
x = number.
The way we get to this new equation is to move all the numbers away from x ,so that it is sitting on a side of the equals sign all by itself.
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Solving Linear Equations
This idea of tweaking both sides by the same amount proves most useful, and isthe basis for solving all sorts of equations in algebra.
2x + 3 = 15
We are trying to change the above equation to a new equation, a simpler one ofthe form
x = number.
The way we get to this new equation is to move all the numbers away from x ,so that it is sitting on a side of the equals sign all by itself.
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Solving Linear Equations: Example
Solve the following equation for x :
2x + 3 = 15
The first thing to do is to move away the numbers which are not in the sameterm as x .
The 3 is being added to it is a different term.
The 2 is part of the same term as x .
In this sense, 2 and x are closer to each other (they are part of the same term),and so we should move the 3 away first.
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Solving Linear Equations: Example
Solve the following equation for x :
2x + 3 = 15
The first thing to do is to move away the numbers which are not in the sameterm as x .
The 3 is being added to it is a different term.
The 2 is part of the same term as x .
In this sense, 2 and x are closer to each other (they are part of the same term),and so we should move the 3 away first.
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Solving Linear Equations: Example
Solve the following equation for x :
2x + 3 = 15
The first thing to do is to move away the numbers which are not in the sameterm as x .
The 3 is being added to it is a different term.
The 2 is part of the same term as x .
In this sense, 2 and x are closer to each other (they are part of the same term),and so we should move the 3 away first.
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![Page 131: Basic Algebra Skills · Basic Algebra Skills Numeracy Workshop adrian.dudek@uwa.edu.au adrian.dudek@uwa.edu.au Basic Algebra Skills 1 / 54. Introduction These slides are intended](https://reader036.vdocuments.us/reader036/viewer/2022071020/5fd45ed30dde822df33ddbad/html5/thumbnails/131.jpg)
Solving Linear Equations: Example
Solve the following equation for x :
2x + 3 = 15
The first thing to do is to move away the numbers which are not in the sameterm as x .
The 3 is being added to it is a different term.
The 2 is part of the same term as x .
In this sense, 2 and x are closer to each other (they are part of the same term),and so we should move the 3 away first.
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Solving Linear Equations: Example
2x + 3 = 15
We basically want to subtract away 3 from the left hand side of the equation.
But we know, that we must do the same operation to both sides, to preserve theequation.
So we take away 3 from both sides:
2x + 3−3 = 15−3
Which gives us
2x = 12
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Solving Linear Equations: Example
2x + 3 = 15
We basically want to subtract away 3 from the left hand side of the equation.
But we know, that we must do the same operation to both sides, to preserve theequation.
So we take away 3 from both sides:
2x + 3−3 = 15−3
Which gives us
2x = 12
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Solving Linear Equations: Example
2x + 3 = 15
We basically want to subtract away 3 from the left hand side of the equation.
But we know, that we must do the same operation to both sides, to preserve theequation.
So we take away 3 from both sides:
2x + 3−3 = 15−3
Which gives us
2x = 12
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Solving Linear Equations: Example
2x + 3 = 15
We basically want to subtract away 3 from the left hand side of the equation.
But we know, that we must do the same operation to both sides, to preserve theequation.
So we take away 3 from both sides:
2x + 3−3 = 15−3
Which gives us
2x = 12
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Solving Linear Equations: Example
2x = 12
It is probably pretty clear now that the answer is 6. However, we still need tocomplete our method, so that we can complete harder problems later on.
We now need to get rid of the 2. We divide both sides by 2 here to get:
2x
2=
12
2
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Solving Linear Equations: Example
2x = 12
It is probably pretty clear now that the answer is 6. However, we still need tocomplete our method, so that we can complete harder problems later on.
We now need to get rid of the 2. We divide both sides by 2 here to get:
2x
2=
12
2
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Solving Linear Equations: Example
2x
2=
12
2
If you look at the left hand side of the above equation, we can see that x is beingmultiplied by 2, and then divided by 2. But doubling a number and then halving
it, gets you back to where you started. So the left hand side is really just x . Soour equation is:
x =12
2
We know that 12 divided by 2 is 6. So we end up with:
x = 6
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Solving Linear Equations: Example
2x
2=
12
2
If you look at the left hand side of the above equation, we can see that x is beingmultiplied by 2, and then divided by 2. But doubling a number and then halving
it, gets you back to where you started. So the left hand side is really just x . Soour equation is:
x =12
2
We know that 12 divided by 2 is 6. So we end up with:
x = 6
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Solving Linear Equations: Example
2x
2=
12
2
If you look at the left hand side of the above equation, we can see that x is beingmultiplied by 2, and then divided by 2. But doubling a number and then halving
it, gets you back to where you started. So the left hand side is really just x . Soour equation is:
x =12
2
We know that 12 divided by 2 is 6. So we end up with:
x = 6
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Solving Linear Equations: Example
Solve the following equation for x :
10− 2x = 18
First we get rid of the 10 on the LHS, by subtracting 10 from both sides:
10− 2x−10 = 18−10
Which we simplify to get:
−2x = 8
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Solving Linear Equations: Example
Solve the following equation for x :
10− 2x = 18
First we get rid of the 10 on the LHS, by subtracting 10 from both sides:
10− 2x−10 = 18−10
Which we simplify to get:
−2x = 8
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Solving Linear Equations: Example
Solve the following equation for x :
10− 2x = 18
First we get rid of the 10 on the LHS, by subtracting 10 from both sides:
10− 2x−10 = 18−10
Which we simplify to get:
−2x = 8
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Solving Linear Equations: Example
−2x = 8
Now we divide both sides by −2:
−2x
−2=
8
−2
Which becomes:
x =8
−2
So the solution of our equation is x = −4.
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Solving Linear Equations: Example
−2x = 8
Now we divide both sides by −2:
−2x
−2=
8
−2
Which becomes:
x =8
−2
So the solution of our equation is x = −4.
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Using STUDYSmarter Resources
This resource was developed for UWA students by the STUDYSmarter team forthe numeracy program. When using our resources, please retain them in their
original form with both the STUDYSmarter heading and the UWA crest.
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