INTERACTIVE LESSONS

A.M. Online membership includes world-class, interactive lessons, both for students that have an Applied Maths teacher and those that are learning the subject from home.

Schoolbooks take a “fill in the gaps” approach, whereby a teacher is expected to expand on the core concepts of the book.

As we are not limited by what can fit into a schoolbook, our lessons instead describe every concept within Applied Maths with as much detail as possible.

Throughout each lesson, we also highlight three important aspects:

  • Key Points – the most important parts of the lesson.
  • Causes of Confusion – common mistakes that students make.
  • Exam Tips – tips to give you that extra edge on the exam!

Key Point

Vectors have both a magnitude and direction.

Cause of Confusion

Magnitudes are always positive. Coefficients, on the other hand, can be either positive or negative.

Exam Tip

Even if you are not asked to draw a diagram, it is always a good idea to draw one in your actual exam whenever possible so that you can more easily visualise the question.

Practice Question

A car uniformly accelerates from rest to a speed of \(20\mbox{ m/s}\) in \(8\) seconds.

Immediately afterwards, it uniformly decelerates to rest, covering a distance of \(150\mbox{ m}\).

(a) What was the acceleration rate in the first stage?

(b) What was the deceleration rate in the second stage?

(c) At what times was the car’s speed equal to \(10\mbox{ m/s}\)?

(a) \(2.5\mbox{ m/s}^2\)

(b) \(\displaystyle\frac{4}{3}\mbox{ m/s}^2\)

(c) \(4\mbox{ s}\) and \(15.5\mbox{ s}\)

vt(s)(m/s)208

(a)

First Stage

\begin{align} u = 0 && v=20 && t = 8 && a=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

\[v=u+at\]

\begin{align}\downarrow\end{align}

\begin{align}20=0+a(8)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a&=\frac{20}{8}\\&=2.5 \mbox{ m/s}^2\end{align}

(b)

Second Stage

\begin{align} u = 20 && v=0 && s = 150 &&a=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

\[v^2=u^2+2as\]

\begin{align}\downarrow\end{align}

\begin{align}0^2=20^2+2a(150)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a&=-\frac{20^2}{(2)(150)}\\&=-\frac{4}{3} \mbox{ m/s}^2\end{align}

The deceleration rate is therefore \(\dfrac{4}{3}\mbox{ m/s}^2\).

(c) 

vt(s)(m/s)10??
First Stage

\begin{align} u = 0 && v=10 && a = 2.5 && t=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

\[v=u+at\]

\begin{align}\downarrow\end{align}

\begin{align}10=0+(2.5)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{10}{2.5}\\&=4 \mbox{ s}\end{align}

\[\,\]

Second Stage

\begin{align} u = 20 && v=10 && a = -\frac{4}{3} && t=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

\[v=u+at\]

\begin{align}\downarrow\end{align}

\begin{align}10=20+\left(-\frac{4}{3}\right)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{10-20}{-\frac{4}{3}}\\&=7.5 \mbox{ s}\end{align}

The time since the beginning of the trip is therefore \(7.5+8=15.5\) seconds.

\[\,\]

Hence, the two times at which the car’s speed is \(10\mbox{ m/s}\) are \(4\) seconds and \(15.5\) seconds into the journey.

Although we are not asked to draw a time-velocity graph in this question, it is still a good idea to draw a rough graph for our own benefit.

In the first stage, the car accelerates from rest to a speed of \(20\mbox{ m/s}\) in a period of \(8\) seconds.

In the second stage, the car decelerates to rest, covering a distance of \(150\mbox{ m}\).

vt(s)(m/s)208

(a) To find the acceleration rate of the first stage, we can use one of our SUVAT equations.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 0 && v=20 && t = 8 \end{align}

and we are looking for \(a\).

\[\,\]

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told \(s\) nor do we want to find \(s\), we should choose the SUVAT equation which does not have \(s\) in it, namely

\[v=u+at\]

\[\,\]

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}20=0+a(8)\end{align}

Rearranging for \(a\), this gives

\begin{align}a&=\frac{20}{8}\\&=2.5 \mbox{ m/s}^2\end{align}

(b) To find the deceleration rate of the second stage, we can again use one of our SUVAT equations.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 20 && v=0 && s = 150 \end{align}

and we are looking for \(a\).

\[\,\]

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

As we are interested in both \(s\) and \(a\), we cannot use the graph and must instead use a SUVAT equation.

As we are not told \(t\) nor do we want to find \(t\), we should choose the SUVAT equation which does not have \(t\) in it, namely

\[v^2=u^2+2as\]

\[\,\]

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}0^2=20^2+2a(150)\end{align}

Rearranging for \(a\), this gives

\begin{align}a&=-\frac{20^2}{(2)(150)}\\&=-\frac{4}{3} \mbox{ m/s}^2\end{align}

Therefore, the deceleration rate of the second stage is \(\displaystyle\frac{4}{3}\mbox{ m/s}^2\).

(c) The maximum speed of the car is \(20\mbox{ m/s}\). Therefore, there will be two separate times that the car’s speed will be \(10\mbox{ m/s}\) – as the car accelerates towards the maximum speed and as it decelerates away from the maximum speed.

vt(s)(m/s)10??

To find this time in the first stage, we can again use a SUVAT equation.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 0 && v=10 && a = 2.5 \end{align}

and we are looking for \(t\).

\[\,\]

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told \(s\) nor do we want to find \(s\), we should choose the SUVAT equation which does not have \(s\) in it, namely

\[v=u+at\]

\[\,\]

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}10=0+(2.5)t\end{align}

Rearranging for \(t\), this gives

\begin{align}t&=\frac{10}{2.5}\\&=4 \mbox{ s}\end{align}

\[\,\]

To find this time in the second stage, we can again use a SUVAT equation.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} v = 10 && u=20 && a = -\frac{4}{3} \end{align}

and we are looking for \(t\).

\[\,\]

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told \(s\) nor do we want to find \(s\), we should choose the SUVAT equation which does not have \(s\) in it, namely

\[v=u+at\]

\[\,\]

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}10=20+\left(-\frac{4}{3}\right)t\end{align}

Rearranging for \(t\), this gives

\begin{align}t&=\frac{10-20}{-\frac{4}{3}}\\&=7.5 \mbox{ s}\end{align}

Note that this is the time only from the start of the second stage, and therefore the time since the beginning of the trip is \(7.5+8=15.5\) seconds.

\[\,\]

Hence, the two times at which the car’s speed is \(10\mbox{ m/s}\) are \(4\) seconds and \(15.5\) seconds into the journey.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

FULL SOLUTIONS

Schoolbooks typically only provide the final answers to practice questions in the back pages rather than the full details of how those answers are obtained.

Wouldn’t it again be awesome if your book instead provided those full details?

As you might have guessed, A.M. Online has got that covered!

We provide detailed, written solutions to EVERY exercise question, and we have just as many questions to practice with as you’d find in a schoolbook!

Can you imagine having a detailed written solution for every single exercise question in your schoolbook?! Wow!

In fact, we go even further! We provide a total of four different methods for students to check their understanding of every question:

  • First, check to see if you got the right Answer. If you didn’t…
  • Read the Solution, i.e. what you should write down when answering the question. If you don’t understand the solution…
  • Read the Walkthrough, i.e. what a teacher would say if they were explaining the solution to you. If you don’t understand the walkthrough…
  • Ask Mr. Kenny for further guidance on Teacher Chat!

These “stepping stones” greatly reduce the likelihood of a student “giving up” if they don’t understand the solution to a question.

Note: The equations on this page may load slowly (or not at all) on certain mobile/tablet devices. Don’t worry! This page is very resource-heavy compared to a typical lesson page that you’ll find on this website. As long as everything loads correctly on e.g. this preview lesson page, that device will work fine!

ANIMATIONS

Applied Maths is an incredibly dynamic subject with a significant portion of the curriculum dealing with the motion of objects. Animations, therefore, make for a much more helpful and beneficial learning experience compared to the static images of a schoolbook.

We therefore provide rich, custom-made animations, together with thousands of illustrations, throughout the lessons in order to provide students with what we believe are the best learning resources available for Applied Maths.

Note: Animations, equations and indeed many other advanced features of this website may not render correctly on browser versions released way back in 2017 or earlier. In the unlikely event that you are still using such an outdated version, updating the browser will get everything working smoothly!

Sample Paper Question

(a) Kruskal’s algorithm and Prim’s algorithm are both used to find the minimum spanning tree of a network.

(i) State one difference between these two algorithms.

(ii) State whether either algorithm is greedy or not and explain why.

(b) Consider the following network:

ABCDEF141792111201182

(i) State the total degree of this network.

(ii) Draw any two spanning trees of this network. 

(iii) Using either Kruskal’s algorithm or Prim’s algorithm, draw the minimum spanning tree of this network.

The weights of this network represent the cost (in millions of euro) to connect six Irish towns with roads.

(iv) If our goal is only to ensure that it’s possible to drive between any of the towns, state the minimum cost to achieve that goal.

(a)

(i) The first step of Kruskal’s algorithm is to choose the edge with the smallest weight. The first step of Prim’s Algorithm is to instead choose any random node.

\[\,\]

(ii) Both algorithms are greedy as both only consider local information when finding the optimal solution.

(b)

(i) \(18\)

\[\,\]

(ii)

ABCDEF17921118
ABCDEF142111112

(iii)

Network

ABCDEF141792111201182

MST

ABCDEF1491182

(iv) \(44\) million euro

(a)

(i) The first step of Kruskal’s algorithm is to choose the edge with the smallest weight. The first step of Prim’s Algorithm is to instead choose any random node.

\[\,\]

(ii) Both algorithms are greedy as both only consider local information when finding the optimal solution.

(b)

(i) Handshaking lemma: \(2\times 9=18\).

\[\,\]

(ii)

ABCDEF17921118
ABCDEF142111112

(iii)

Network

ABCDEF141792111201182

MST

ABCDEF1491182

(iv) \(14+9+2+8+11=44\) million euro

(a)

(i) There are many significant differences between both algorithms, one of which is how each algorithm begins.

The first step of Kruskal’s algorithm is to choose the edge with the smallest weight. The first step of Prim’s Algorithm is to instead choose any random node.

\[\,\]

(ii) Both algorithms are greedy as both only consider local information when finding the optimal solution.

(b)

(i) The total degree of the network can be found by summing the degrees of each node.

However, it can more quickly be found by using the handshaking lemma.

According to this lemma, the total degree of a network is simply equal to twice the number of edges of that network.

As this network has \(9\) edges, the total degree of this network is \(2\times 9=18\).

\[\,\]

(ii) Recall that a spanning tree of a graph is any subgraph that both contains all of the nodes of that graph and is also a tree.

Two of the many spanning trees of this network are therefore shown below.

ABCDEF141792111201182
ABCDEF141792111201182

(iii) We are told that we can use either algorithm. As there is no obvious benefit to choosing one over the other, we shall randomly choose to use Prim’s Algorithm and shall therefore follow the usual steps.

Step 1: Choose a random starting node. The edge with the smallest weight that is incident with that node is the first edge of the MST. Cross it off the network and add it to the MST.

We shall randomly choose node \(C\) as our starting node.  The edge with the smallest weight incident with this node is \(CD\). We therefore cross it off our network and add it to our MST.

Network

ABCDEF141792111201182

MST

CD2

Step 2: The edge with the next smallest weight that is incident with a node in the current MST and which does not create a cycle is the next edge of the MST. Cross it, and any edges that created cycles, off the network and add it to the MST.

Our MST currently has nodes \(C\) and \(D\) and our choices are therefore \(BC\), \(CE\), \(AD\), \(BD\), \(DE\) and \(DF\), none of which would create a cycle if added to the MST. The edge with the smallest weight of these options is \(DF\). We therefore cross it off our network and add it to our MST.

Network

ABCDEF141792111201182

MST

CDF82

Step 3: Repeat Step 2 until the MST contains all of the nodes of the network.

Our MST currently has nodes \(C\), \(D\) and \(F\) and our choices are therefore \(BC\), \(CE\), \(AD\), \(BD\), \(DE\) and \(EF\), none of which would create a cycle if added to the MST. The edge with the smallest weight of these options is \(BC\). We therefore cross it off our network and add it to our MST.

Network

ABCDEF141792111201182

MST

BCDF982

Our MST currently has nodes \(B\), \(C\), \(D\) and \(F\) and our choices are therefore \(AB\), \(BD\), \(CE\), \(AD\), \(DE\) and \(EF\), of which \(BD\) would create a cycle if added to the MST. The edge with the smallest weight of the remaining options is \(EF\). We therefore cross both \(BD\) and \(EF\) off our network and add only \(EF\) to our MST.

Network

ABCDEF141792111201182

MST

BCDEF91182

Our MST currently has nodes \(B\), \(C\), \(D\), \(E\) and \(F\) and our choices are therefore \(AB\), \(CE\), \(AD\) and \(DE\), of which both \(CE\) and \(DE\) would create a cycle if added to the MST. The edge with the smallest weight of the remaining options is \(AB\). We therefore cross \(CE\), \(DE\) and \(AB\) off our network and add only \(AB\) to our MST.

Network

ABCDEF141792111201182

MST

ABCDEF1491182

Our MST now contains all of the nodes of our network and is therefore complete.

(iv) As our goal is to only ensure that it’s possible to drive between any of the towns, we are therefore looking for the total weight of the minimum spanning tree.

According to part (iii), the total weight of the minimum spanning tree is

\begin{align}14+9+2+8+11=44\end{align}

As we are told that each weight is in millions of euro, the minimum cost to connect the six towns together is \(44\) million euro. 

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

SAMPLE PAPERS

As there is a new curriculum for this subject for the 2023 Leaving Certificate onwards, the past papers for this subject have become, for the most part, unusable.

We have therefore included TEN sample examination papers so that both Higher and Ordinary Level students don’t have to worry about not having any exam questions to practice with!

These papers can only be found on A.M. Online and, just as with our practice questions, we also provide our four “stepping stones” for every sample paper question!

We organise these sample paper questions both 1) by year so that students can practice with full papers and 2) by topic so that students can improve their understanding in specific areas.

KNOWLEDGE CHECKS

Before jumping directly into answering full exercise questions, we first test each student’s basic understanding of the lesson content with quick-fire questions in the form of Knowledge Checks.

The questions within our 50+ Knowledge Checks questions give each student immediate feedback as to whether a lesson “clicked” or not, thereby indicating whether they are ready to answer the corresponding exercise set or whether they need to do some quick revision first.

/20

\[\,\]Knowledge Check 8

\[\,\]A quick check on your understanding of:

Displacement

Velocity

Acceleration

SUVAT equations

\[\,\]

The rate of change of a vector is itself a vector.

1 / 20

The rate of change of velocity with respect to time is the definition of

The rate of change of a vector is itself a vector.

2 / 20

The rate of change of displacement with respect to time is the definition of

What quantity has that unit?

3 / 20

"A car brakes from \(\mathbf{15\mbox{ m/s}}\) east to \(0\mbox{ m/s}\) east at a rate of \(5\mbox{ m/s}^2\) within \(3\) seconds". The quantity in bold refers to:

What quantity has this unit?

4 / 20

"A car brakes from \(15\mbox{ m/s}\) east to \(\mathbf{0}\mbox{ m/s}\) east at a rate of \(5\mbox{ m/s}^2\) within \(3\) seconds". The quantity in bold refers to:

Displacement is a vector quantity.

5 / 20

A pedestrian walks \(25\) metres west from \(o\) followed by \(10\) metres east. Their total displacement from \(o\) is

Distance is a scalar quantity.

6 / 20

A pedestrian walks \(25\) metres west from \(o\) followed by \(10\) metres east. Their total distance travelled is

Speed is a scalar quantity.

7 / 20

A cyclist travels \(60\) metres east in \(2\) seconds at a constant acceleration. The cyclist's speed is

Velocity is a vector quantity.

8 / 20

A cyclist travels \(60\) metres east in \(2\) seconds at a constant acceleration. The cyclist's velocity is

Speed is equivalent to the magnitude of velocity.

9 / 20

Speed and velocity have the same units.

Acceleration is the rate of change of velocity.

10 / 20

A cyclist increases her velocity from \(10\mbox{ m/s}\) east to \(20\mbox{ m/s}\) east before reducing it again to \(10\mbox{ m/s}\) east. As her starting and final velocities are the same, she did not accelerate during this motion.

A vector has both a magnitude and a direction.

11 / 20

When a car uses a roundabout, it must accelerate.

Acceleration is the rate of change of velocity.

12 / 20

If an object is moving, it must be accelerating.

If an object is not moving, its velocity is zero.

13 / 20

In order for a moving car to come to a stop, it must decelerate.

Displacement is a vector quantity.

14 / 20

In the below image, the car is parked and is not moving. What is its displacement from \(o\) after 10 seconds?

\[\,\]

\[\,\]

If there is no acceleration, then there is no change in velocity.

15 / 20

In the SUVAT vector equations, if \(\vec{a}=0\), then \(\vec{u}=\vec{v}\).

What SUVAT variable is not included in this triangle?

16 / 20

"Dad's Silly Triangle" can be used:

What quantities do we know and what quantity are we looking for?

17 / 20

A cyclist increases his velocity from \(15\mbox{ m/s}\) east to \(25\mbox{ m/s}\) east over a period of \(5\) seconds. Which of the following SUVAT equations should we use to find the cyclist's acceleration in that time?

What quantities do we know and what quantity are we looking for?

18 / 20

A cyclist increases his velocity from \(15\mbox{ m/s}\) east to \(25\mbox{ m/s}\) east over a period of \(5\) seconds. Which of the following SUVAT equations should we use to find the cyclist's displacement in that time?

What happens when we multiply a scalar by a vector?

19 / 20

\(5t^3\vec{a}\) is a vector

What assumption was made when deriving the SUVAT equations?

20 / 20

If the acceleration of an object is gradually changing over time, we cannot use the SUVAT equations.

Your score is

0%

Key Point

A linear equation is any equation of the form

\begin{align}ax+b=0\end{align}

where \(a\) and \(b\) are constants.

All of the following are therefore considered linear equations.

\begin{align}2x+3=0 && x-7=0\end{align}

\begin{align}\frac{x}{3}+2=0 && 1.7x-\pi=0\end{align}

In fact, even the following equation

\begin{align}5x=x+3\end{align}

can be considered a linear equation since, as we shall see shortly, it can be arranged as

\begin{align}4x-3=0\end{align}

In general, as long as the highest power of the unknown variable is \(1\), the equation is considered linear.

To solve such equations, we perform the following steps

  • Move all of the terms containing \(x\) to one side of the equation and all of the terms not containing \(x\) to the other side. When switching any term from one side to the other, we must flip the sign of that term!
  • Add/subtract all of the terms on each side.
  • Divide both sides by any coefficient in front of the final term containing \(x\).

Guided Example

Solve the following equation:

\begin{align}6x-7 = 2x+5\end{align}

To begin, we place all of the terms containing \(x\), i.e. \(6x\) and \(2x\), on one side, e.g. the left side, and all of the terms not containing \(x\), i.e. \(-7\) and \(+5\), on the other side.

To do that, we need to move the \(2x\) term from the right side to the left side, causing its sign to switch to \(-2x\). We also need to move the \(-7\) term from the left side to the right side, causing its sign to instead switch to \(+7\).

\begin{align}6x-7 = 2x+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x-2x = 5+7\end{align}

We now add/subtract the term on both sides

\begin{align}6x-2x = 5+7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x = 12\end{align}

Finally, as our goal is to find \(x\), i.e. to write \(x=\mbox{‘something’}\), we need to remove the coefficient in front of the \(x\) by dividing both sides by \(4\).

\begin{align}4x = 12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4x}{4}=\frac{12}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=3\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

Exam Tip

If you’re ever unsure as to whether you solved an equation correctly, you can quickly check by reinserting the value that you obtained for the unknown variable back into the original equation.

For example, if we reinsert \(x=3\) back into our original equation above, we obtain

\begin{align}6x-7 = 2x+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6(3)-7 = 2(3)+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}11 = 11\end{align}

As this is of course true, our solution must indeed be correct!

Consider the following statement:

“The sum of two numbers equals five. What are those numbers?”

In this case, as we have two distinct unknown numbers, we shall represent them using two variables \(x\) and \(y\).

Answering this question is then equivalent to solving the following equation:

\begin{align}x+y=5\end{align}

So, what is the solution to this equation?

Well, we can easily see that \(x=2\) and \(y=3\) satisfies this equation.

\begin{align}2+3=5\end{align}

However, \(x=1\) and \(y=4\) also satisfies this equation

\begin{align}1+4=5\end{align}

as does \(x=1.7\) and \(y=3.3\)

\begin{align}1.7+3.3=5\end{align}

and so on. This equation in fact has an infinite number of solutions!

Let us therefore instead consider the following different statement.

“The sum of two numbers equals five. The difference between those two numbers is three. What are those numbers?”

This information is instead represented by the following two equations:

\begin{align}x+y&=5\\ x-y&=3\end{align}

To answer this question, we need to solve these two equations at the same time, i.e. simultaneously.

This time, rather than obtaining an infinite number of solutions, we will instead only obtain one solution. This fact is indeed true more generally if we have a larger number of unknown variables.

Key Point

If we wish to find \(n\) unknown variables, we need to solve \(n\) simultaneous equations.

For now, we shall consider only solving two simultaneous equations for two unknowns.

So, how do we solve two simultaneous equations? More, specifically how do we solve such equations if they are both linear equations?

\begin{align}x+y&=5\\ x-y&=3\end{align}

Well, there are in fact two methods that we can use.

1) Substitution

This method is summarised as follows:

  • Rewrite one of the equations as either \(x=\mbox{‘stuff’}\) or \(y=\mbox{‘stuff’}\).
  • Substitute the \(\mbox{‘stuff’}\) into the other equation.

Guided Example

Solve the following simultaneous equations using substitution:

\begin{align}2x+3y&=7\\ 4x-5y&=3\end{align}

To begin, we shall rewrite one of the equations as either \(x=\mbox{‘stuff’}\) or \(y=\mbox{‘stuff’}\). As it seems equally as easy to do that for either equation, we shall randomly choose to do it for the first equation.

For that equation, it also seems equally as easy to rearrange it as either \(x=\mbox{‘stuff’}\) or \(y=\mbox{‘stuff’}\). We shall randomly choose to rearrange it as \(y=\mbox{‘stuff’}\).

\begin{align}2x+3y&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3y=7-2x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y=\frac{7-2x}{3}\end{align}

Using this, we can now replace \(y\) in our second equation.

\begin{align}4x-5y=3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x-5\left(\frac{7-2x}{3}\right)=3\end{align}

We now have an equation just for \(x\) which we can solve as usual!

Multiplying the above equation by \(3\), we obtain

\begin{align}12x-5(7-2x)=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x-35+10x=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}22x=44\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\end{align}

Now that we know what \(x\) is, we can find \(y\) by inserting this value into any of our equations that contain both \(x\) and \(y\), e.g.

\begin{align}y=\frac{7-2x}{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{7-2(2)}{3}\\&=\frac{3}{3}\\&=1\end{align}

Therefore, the solution to this set of simultaneous equations is \(x=2\) and \(y=1\).

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

2) Subtraction

This method is instead summarised as follows:

  • rewrite one of the equations such that the coefficient of either the \(x\) term or the \(y\) term is the same in both equations
  • subtract one equation from the other

Guided Example

Solve the following simultaneous equations using subtraction:

\begin{align}2x+3y&=7\\ 4x-5y&=3\end{align}

Currently, both equations have different coefficients in front of their \(x\) and \(y\) terms (\(2\) and \(4\) for the \(x\) terms and \(3\) and \(-5\) for the \(y\) terms).

For either pair of coefficients to be the same, we need to multiply one (or both) equations by a constant. We shall choose to make the \(x\) coefficients the same as this only requires multiplying the first equation by \(2\)

\begin{align}4x+6y&=14\\ 4x-5y&=3\end{align}

Now, the coefficient of the \(x\) term is the same in both equations. Therefore, if we now subtract one equation from the other, the \(x\) terms will disappear, leaving us with an equation just for \(y\)!

\begin{align}4x+6y=14&\\\underline{-\,\,\,(4x-5y=3)}&\\11y=11&\end{align}

since \(4x-4x=0\), \(6y-(-5y)=11y\) and \(14-3=11\).

This equation can now be easily solved for \(y\) to give

\begin{align}y&=\frac{11}{11}\\&=1\end{align}

Again, we can now insert this value for \(y\) into any equation involving \(x\) and \(y\) to find \(x\), e.g.

\begin{align}2x+3y&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x+3(1)&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x=4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\end{align}

As expected, the solution is the same as that obtained using the first method, i.e. \(x=2\) and \(y=1\).

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

Key Point

A quadratic equation is any equation of the form

\begin{align}ax^2+bx+c=0\end{align}

where \(a\), \(b\) and \(c\) are constants.

All of the following are therefore considered quadratic equations.

\begin{align}x^2+5x+3=0 && 2x^2-6=0\end{align}

\begin{align}8x^2+17x=0 && \frac{3x^2}{5}-12x+1=0\end{align}

The following equation

\begin{align}12x=-5+4x^2\end{align}

is also a quadratic equation as it can be arranged as

\begin{align}4x^2-12x-5=0\end{align}

In general, as long as the highest power of the unknown variable is \(2\), i.e. the degree of the polynomial is \(2\), the equation is considered quadratic.

So, how do we solve equations of this type?

Well, first note that if \(a=0\), the equation is now longer quadratic but instead linear

\begin{align}bx+c=0\end{align}

and we already know how to solve such equations.

If instead \(b=0\), i.e. if the equation is of the form

\begin{align}ax^2+c=0\end{align} then the equation can easily be solved by rearranging.

Guided Example

Solve the following equation:

\begin{align}3x^2-48=0\end{align}

This equation can be rearranged as follows

\begin{align}3x^2-48=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2=48\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2=16\end{align}

Note that there are in fact two numbers that satisfy this equation: \(-4\) and \(4\) (since both \((-4)^2=16\) and \(4^2=16\)).

Therefore, this quadratic equation has two solutions: \(x=-4\) and \(x=4\). (This can also be written in one line as \(x=\pm 4\).)

In general, quadratic equations will have (at most) two solutions. The solution(s) to quadratic equations (and indeed equations of an even higher degree) are often referred to as the roots of the equation.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

In all other cases, we cannot simply rearrange the equation to find the solution. For example, consider the most general case of all three coefficients being non-zero, e.g.

\begin{align}x^2+3x+2=0\end{align}

There is no way to rearrange this equation to obtain \(x=\mbox{‘stuff’}\) (or even \(x^2=\mbox{‘stuff’}\)).

Of course, we can rearrange it as

\begin{align}x^2=-3x-2\end{align}

but that is of no help as we still have an \(x\) term on the right hand side!

So, how do we proceed?

Well, we can use factorisation! We can factorise the following quadratic expression

\begin{align}x^2+3x+2\end{align}

as

\begin{align}(x+1)(x+2)\end{align}

Therefore, we can rewrite this quadratic equation as

\begin{align}(x+1)(x+2)=0\end{align}

How does that help us solve this equation?

Well, we now have ‘something’ multiplied ‘something else’ equalling zero. We know with regular numbers that this can only be the case if (at least) one of the two numbers is zero!

\begin{align}0\times 6=0 && 4 \times 0=0\end{align}

There is no way of multiplying two numbers together to get zero unless (at least) one of those numbers is zero!

This is also true with algebra – (at least) one of these factors must also be zero, i.e.

\begin{align}x+1=0 && \mbox{or}&& x+3=0\end{align}

This therefore gives us our two solutions (roots)

\begin{align}x=-1 && \mbox{or} && x=-3\end{align}

Guided Example

Solve the following equation:

\begin{align}2x^2-2x-12=0\end{align}

The quadratic expression on the left hand side can be factorised, giving

\begin{align}(2x+4)(x-3)=0\end{align}

and therefore

\begin{align}2x+4=0 && \mbox{or}&& x-3=0\end{align}

Hence, the two roots of this equation are \(x=-2\) and \(x=3\).

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

We can also use factorisation if \(c=0\), i.e. if the equation is of the form

\begin{align}ax^2+bx=0\end{align}

Guided Example

Solve the following equation:

\begin{align}3x^2-12x=0\end{align}

The quadratic expression on the left hand side can be factorised, giving

\begin{align}(2x+4)(x-3)=0\end{align}

and therefore

\begin{align}2x+4=0 && \mbox{or}&& x-3=0\end{align}

Therefore, the two roots of this equation are \(x=-2\) and \(x=3\).

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

Unfortunately, it is not possible to factorise all quadratic expressions!

For example, consider the following expression

\begin{align}2x^2-5x+2=0\end{align}

The quadratic expression on the left cannot be factorised using usual methods.

However, let us check if \(x=0.5\) is a possible root.

\begin{align}2(0.5)^2-5(0.5)+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5-2.5+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=0\end{align}

Therefore, \(x=0.5\) is indeed one of the roots!

Of course, it makes sense that this root cannot be found using usual factoring methods as such methods don’t consider the possibility that factor pairs may be decimals, fractions etc. And for good reason – we would then have an infinite number of factor pairs to check!

So, how do we determine that \(x=0.5\) is a root without guessing? And what is the other root of this equation?

Well, it turns out that there is in fact a formula that we can use!

Key Point

The roots of a quadratic equation of the form

\begin{align}ax^2+bx+c=0\end{align}

are given by

\begin{align}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{align}

where \(a\), \(b\) and \(c\) are constants.

Note the \(\pm\) sign in this equation: for one of the roots, we use a \(+\) sign, and for the other, we use a \(-\) sign.

Guided Example

Solve the following equation:

\begin{align}2x^2-5x+2=0\end{align}

This is a quadratic equation of the form

\begin{align}ax^2+bc+c=0\end{align}

with \(a=2\), \(b=-5\) and \(c=2\).

Inserting these values into our quadratic formula, we obtain

\begin{align}x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\&=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)}\\&=\frac{5\pm \sqrt{25-16}}{4}\\&=\frac{5\pm 3}{4}\end{align}

Therefore, our two roots are

\begin{align}x=\frac{5-3}{4} && \mbox{and} && x=\frac{5+3}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=0.5 && \mbox{and} && x=2\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

This quadratic formula can in fact always be used to find the roots of a quadratic equation, even when factorisation is possible!

Guided Example

Solve the following equation:

\begin{align}2x^2-2x-12=0\end{align}

using the quadratic formula.

Note that this is the same quadratic equation that was solved in the second Guided Example above using factorisation.

The coefficients in this case are \(a=2\), \(b=-2\) and \(c=-12\).

Inserting these values into our quadratic formula, we obtain

\begin{align}x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-12)}}{2(2)}\\&=\frac{2\pm \sqrt{100}}{4}\\&=\frac{2\pm 10}{4}\end{align}

Therefore, our two roots are

\begin{align}x=\frac{2-10}{4} && \mbox{and} && x=\frac{2+10}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=-2 && \mbox{and} && x=3\end{align}

which are the same roots that we obtained by using factorisation!

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

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