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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]

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.

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.

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.

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.

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”.

Multiplication

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

Variables

Expressions

An expression is a combination of numbers and variables.

Some examples are:

3x + 5

2xy + 4x2y + 2x + 4x

7x2 −√x + 1

Expressions

An expression is a combination of numbers and variables.

Some examples are:

3x + 5

2xy + 4x2y + 2x + 4x

7x2 −√x + 1

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:

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:

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.

4x2+2xy−4x

y+7xy−2x2

4x2+2xy−4x

y+7xy−2x2

4x2+2xy−4x

y+7xy−2x2

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.

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

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

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

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

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.

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.

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.

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.

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.

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.

Adding and Subtracting Like Terms

We can add and subtract like terms!

7xy + 4xy = 11xy

9x2y + 5yx2 = 14x2y

14xyz3 − 5z3yx + 2yz3x = 11z3yx

7xy + 4xy = 11xy

9x2y + 5yx2 = 14x2y

14xyz3 − 5z3yx + 2yz3x = 11z3yx

7xy + 4xy = 11xy

9x2y + 5yx2 = 14x2y

14xyz3 − 5z3yx + 2yz3x = 11z3yx

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

2xy + xy = 3xy

14x2y − yx2 = 13x2y

2xy + xy = 3xy

14x2y − yx2 = 13x2y

2xy + xy = 3xy

14x2y − yx2 = 13x2y

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

4xy2 + 3xy + 2xy2 − 2yx

4xy2 + 2xy2 + 3xy − 2yx

4xy2 + 3xy + 2xy2 − 2yx

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.

4xy2 + 2xy2 + 3xy − 2yx

6xy2 + xy

4xy2 + 2xy2 + 3xy − 2yx

6xy2 + xy

Simplifying Expressions: Example

Simplify the following expression:

xy2 − 3xyz + 2y2x + 5yzx − 4yxz

Step 1: Identify like terms:

Step 2: Move terms so that like terms are next to each other:

xy2 + 2y2x − 3xyz + 5yzx − 4yxz

Step 3: Now add the like terms together:

3xy2 − 2yxz

We have simplified our original expression!

Step 3: Now add the like terms together:

3xy2 − 2yxz

We have simplified our original expression!

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.

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.

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!

x2x5 = xxxxxxx = x7

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

y3y2 = y5

(3x2y3)(4x5y2) = 12x7y5

y3y2 = y5

(3x2y3)(4x5y2) = 12x7y5

Multiplying Terms: Examples

(2x2y)(3xy4) = 6x3y5

(−x3y3)(xy5) = −x4y8

(−5xy)(−2x2y3) = 10x3y4

(−2x2y)(−3x4y)(−4y) = −24x6y3

(2x2y)(3xy4) = 6x3y5

(−x3y3)(xy5) = −x4y8

(−5xy)(−2x2y3) = 10x3y4

(−2x2y)(−3x4y)(−4y) = −24x6y3

(2x2y)(3xy4) = 6x3y5

(−x3y3)(xy5) = −x4y8

(−5xy)(−2x2y3) = 10x3y4

(−2x2y)(−3x4y)(−4y) = −24x6y3

(2x2y)(3xy4) = 6x3y5

(−x3y3)(xy5) = −x4y8

(−5xy)(−2x2y3) = 10x3y4

(−2x2y)(−3x4y)(−4y) = −24x6y3

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.

term × term

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.

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

Single Bracket Expansion: Examples

4(2x + 3) = 8x + 12

3(4− 2x) = 12− 6x

4x(7 + 5x) = 28x + 20x2

4(2x + 3) = 8x + 12

3(4− 2x) = 12− 6x

4x(7 + 5x) = 28x + 20x2

4(2x + 3) = 8x + 12

3(4− 2x) = 12− 6x

4x(7 + 5x) = 28x + 20x2

Single Bracket Expansion: More examples

−2xy(3x − 2y) = −6x2y + 4xy2

−4x2y3(6xy + 3x3y3) = −24x3y4 − 12x5y6

−2xy(3x − 2y) = −6x2y + 4xy2

−4x2y3(6xy + 3x3y3) = −24x3y4 − 12x5y6

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.

2x(3 + x + 5x4) = 6x + 2x2 + 10x5

−4xy(2xy − 4x + 3x) = −8x2y2 + 16x2y − 12x2y

2x(3 + x + 5x4) = 6x + 2x2 + 10x5

−4xy(2xy − 4x + 3x) = −8x2y2 + 16x2y − 12x2y

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.

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

(3 + 2x)(5 + 3x)

3 times 5

3 times 3x

2x times 5

2x times 3x

(3 + 2x)(5 + 3x)

3 times 5

3 times 3x

2x times 5

2x times 3x

(3 + 2x)(5 + 3x)

3 times 5

3 times 3x

2x times 5

2x times 3x

(3 + 2x)(5 + 3x)

3 times 5

3 times 3x

2x times 5

2x times 3x

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

3 times 5 equals 15

(3 + 2x)(5 + 3x) = 15 + 9x + 10x + 6x2

15 + 19x + 6x2

3 times 5 equals 15

(3 + 2x)(5 + 3x) = 15 + 9x + 10x + 6x2

15 + 19x + 6x2

(4x − 5)(2 + 4x)

4x times 2

4x times 4x

−5 times 2

−5 times 4x

(4x − 5)(2 + 4x)

4x times 2

4x times 4x

−5 times 2

−5 times 4x

(4x − 5)(2 + 4x)

4x times 2

4x times 4x

−5 times 2

−5 times 4x

(4x − 5)(2 + 4x)

4x times 2

4x times 4x

−5 times 2

−5 times 4x

(4x − 5)(2 + 4x)

4x times 2

4x times 4x

−5 times 2

−5 times 4x

−5 times 2 equals − 10

−5 times 4x equals − 20x

(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x

16x2 − 12x − 10

(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x

16x2 − 12x − 10

(4x − 5)(2 + 4x) = 8x + 16x2 − 10− 20x

16x2 − 12x − 10

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

(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

(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

Equations

An equation is a mathematical object of the form:

expression = expression

For example:

3x2y + 6xy = 7xy2 − 4y

Equations

An equation is a mathematical object of the form:

expression = expression

For example:

3x2y + 6xy = 7xy2 − 4y

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:

Equations

x + 5 = 8

Equations

x + 5 = 8

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:

Equations

4 + 2x = 20

Equations

4 + 2x = 20

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.

Equations

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

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

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.

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.

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.

2x + 3 = 15

x = number.

2x + 3 = 15

x = number.

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.

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

2x + 3 = 15

2x + 3−3 = 15−3

Which gives us

2x = 12

2x + 3 = 15

2x + 3−3 = 15−3

Which gives us

2x = 12

2x + 3 = 15

2x + 3−3 = 15−3

Which gives us

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 = 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:

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:

We know that 12 divided by 2 is 6. So we end up with:

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

10− 2x = 18

10− 2x−10 = 18−10

−2x = 8

10− 2x = 18

10− 2x−10 = 18−10

−2x = 8

Now we divide both sides by −2:

Which becomes:

So the solution of our equation is x = −4.

−2x = 8

Now we divide both sides by −2:

Which becomes:

So the solution of our equation is x = −4.

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