11+ Maths

Algebra

11+ Maths

Algebra

Objective: How to use algebra to solve problems and work with unknowns and to use this to solve equations, analyse functions and apply these methods to real-world problems.

Why Algebra Matters: Algebra improves logical thinking and problem-solving. It helps you tackle real-world problems by using variables and patterns.

11+ Maths

Algebra

​Algebra uses symbols (like $x$ and $y$) to represent numbers and relationships.

It helps solve problems involving unknown values and patterns.

Variables, Constants and Expressions

Variables: Symbols like $x,y,z$ that can change value.

Variables, Constants and Expressions

Constants: Numbers with fixed values.

Variables, Constants and Expressions

Expressions: Combinations of variables and constants (e.g., 3x+73x + 73x+7).

Variables, Constants and Expressions

Quick check:

What is the correct entity for '?' in this table?

Variables, Constants and Expressions

x, y & z are variables: they can hold different values.

5 and 12 are constant. They are always the same.  We have also added 1 as a constant because just "z" means there is 1z.

What Are Algebraic Formulae?

Algebraic formulae are equations that show relationships between variables (like $x,y,z$) and constants (numbers). They help solve problems by replacing variables with known values.

What Are Algebraic Formulae?

Algebraic formulae are equations that show relationships between variables (like $x,y,z$) and constants (numbers). They help solve problems by replacing variables with known values.

What Are Algebraic Formulae?

Algebraic formulae are equations that show relationships between variables (like $x,y,z$) and constants (numbers). They help solve problems by replacing variables with known values.

We have replaced 5

with the variable "x"

What Are Algebraic Formulae?

Algebraic formulae are equations that show relationships between variables (like $x,y,z$) and constants (numbers). They help solve problems by replacing variables with known values.

An easier way of writing two times x is simply 2x

What Are Algebraic Formulae?

Algebraic formulae are equations that show relationships between variables (like $x,y,z$) and constants (numbers). They help solve problems by replacing variables with known values.

We have replaced the value 5 with x. We know that in this case x = 5, but if we didn't, we could use maths to find the value (to "solve for x").

To solve the equation, we can add, subtract multiply and divide. But we have to keep the equation balanced by doing the same on both sides of the = sign.

+5

-5

x 5

/ 5

+5

-5

x 5

/ 5

Let's have a look at some samples:

10 = 10

This is quite obviously a true statement.

Let's have a look at some samples:

10 + 1 = 10 + 1

And it remains true even if we add or subtract, provided we do it on both sides of the equal sign.

Let's have a look at some samples:

10 x 2 = 10 x 2

And it remains true even if we add or subtract, provided we do it on both sides of the equal sign.

Let's have a look at some samples:

x = 10

Let's imagine that x = 10.  If so, this statement is obviously true.

Let's have a look at some samples:

x + 1 = 10 + 1

And if x remains 10, then this statement is clearly true too.  10 + 1 = 10 + 1.

Let's have a look at some samples:

x/2 = 10/2

Just like this would also be true: Half of x would be 5.  The important thing is that we do the same on both sides.

So, let's use this technique to solve the equation below:

Let's solve this step-by-step together.

We want x to be alone on one side of the equal sign and thus we start by moving +15 to the other side.

Let's solve this step-by-step together.

We can add, subtract, divide or multiply as we want, as long as we do the same thing on both sides.  So, we can subtract 15 on both sides to "move" +15 to the other side.  Some people like to think of this as just reversing the operation when moving a number. Plus becomes minus, divide becomes multiply and so on.

Let's solve this step-by-step together.

We can add, subtract, divide or multiply as we want, as long as we do the same thing on both sides.  So, we can subtract 15 on both sides to "move" +15 to the other side.  Some people like to think of this as just reversing the operation when moving a number. Plus becomes minus, divide becomes multiply and so on.

Let's solve this step-by-step together.

We can add, subtract, divide or multiply as we want, as long as we do the same thing on both sides.  So, we can subtract 15 on both sides to "move" +15 to the other side.  Some people like to think of this as just reversing the operation when moving a number. Plus becomes minus, divide becomes multiply and so on.

Let's solve this step-by-step together.

We can add, subtract, divide or multiply as we want, as long as we do the same thing on both sides.  So, we can subtract 15 on both sides to "move" +15 to the other side.  Some people like to think of this as just reversing the operation when moving a number. Plus becomes minus, divide becomes multiply and so on.

Let's solve this step-by-step together.

Now we have 2x on one side.  This represent two times x or simply x + x.  We want to have just one x.  We can do that by dividing by two on both sides, because half of two x is x.

Let's solve this step-by-step together.

Now we have 2x on one side.  This represent two times x or simply x + x.  We want to have just one x.  We can do that by dividing by two on both sides, because half of two x is x.

We now have a value for x.

We can test this answer by replacing the variable x with the value 5:

We now have a value for x.

We can test this answer by replacing the variable x with the value 5:

We now have a value for x.

We can test this answer by replacing the variable x with the value 5:

It's easy to see that x = 5 makes sense.

We have therefore successfully solved the equation and verified our answer.

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

5x take away 1x is 4x

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

40 minus 12 is 28

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Now we have the value of 4x.  To fully solve the equation, however, we need to find the value of x. If 4x is 28, then 1x will be a quarter of 28.

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Now we have the value of 4x.  To fully solve the equation, however, we need to find the value of x. If 4x is 28, then 1x will be a quarter of 28.

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Now we have the value of 4x.  To fully solve the equation, however, we need to find the value of x. If 4x is 28, then 1x will be a quarter of 28.

Let's do a few more algebraic equations:

We see that there are variables (x) and constants (12 and 40) on both sides.  Before we can solve for x, we need to move these so that x is on its own

Now we have the value of 4x.  To fully solve the equation, however, we need to find the value of x. If 4x is 28, then 1x will be a quarter of 28.

Equations with more than 1 unknown

An equation can have more than one variable (more than one letter representing an unknown value). Here is an example:

Equations with more than 1 unknown

An equation can have more than one variable (more than one letter representing an unknown value). Here is an example:

Here we have two unknown variables: x and y.

x = 1, y = 18

x = 2, y = 16

x = 3, y = 14

x = 4, y = 12

x = 5, y = 10

We don't have enough information to say what specific values x and y represents, but we can identify some possibilities:

Equations with more than 1 unknown

An equation can have more than one variable (more than one letter representing an unknown value). Here is an example:

If we are given more information, however, we may be able to solve.  Let's say we're ALSO told that

Now we can use different strategies to solve the equation and find the exact value for both x and y.

One way of solving these is to find a value for either x or y and then insert that into one of the equations.  Let's try this:

One way of solving these is to find a value for either x or y and then insert that into one of the equations.  Let's try this:

One way of solving these is to find a value for either x or y and then insert that into one of the equations.  Let's try this:

Now we have a value for x: 15 minus y.  Since we don't yet know what y is, we can't immediately find a constant value, but we can use this to move forward.

Now we have a constant value y = 10.  With this we can easily solve for x.

Now we have a constant value y = 10.  With this we can easily solve for x.

Now we have a constant value y = 10.  With this we can easily solve for x.

Now we have a constant value y = 10.  With this we can easily solve for x.

Now we have a constant value y = 10.  With this we can easily solve for x.

Let's insert the value for y, which we now know is 10, into the other equation.

Let's insert the value for y, which we now know is 10, into the other equation.

Let's insert the value for y, which we now know is 10, into the other equation.

Let's insert the value for y, which we now know is 10, into the other equation.

Finally, we have a value also for x.

Using two equations we solved for two unknowns:

Finally, we have a value also for x.

Using two equations we solved for two unknowns:

Finally, we have a value also for x.

Using two equations we solved for two unknowns:

Finally, we have a value also for x.

Using two equations we solved for two unknowns:

We can verify that these values make sense by inserting them in one of the original equations.

There is another method you can sometimes use when solving problems with two unknowns. This method works particularly well when adding the two equations together makes something ‘cancel out’.

You don't need to use this method, but if you feel you understand the principle, it may make equations easier to solve.

Let's rewind to the last question.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Looking at this, you may notice that equation 1 has +y, while equation 2 has a -y.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Looking at this, you may notice that equation 1 has +y, while equation 2 has a -y.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Looking at this, you may notice that equation 1 has +y, while equation 2 has a -y.

In algebra, we are allowed to add equations together.  If we could just rearrange the first statement, when we add them together, the +y and the -y should cancel each other out, just like a +1 and -1 would give a sum of 0.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Let's try to switch the first statement around.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Let's try to switch the first statement around.

Now if we were to add them together, it would look like this:

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Let's try to switch the first statement around.

Now if we were to add them together, it would look like this:

This is simply all the values of the two equations added together into one single equation.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Let's try to switch the first statement around.

Now if we were to add them together, it would look like this:

This is simply all the values of the two equations added together into one single equation.

Let's rewind to the last question.

As you may recall, we were presented with two statements:

Let's try to switch the first statement around.

Now if we were to add them together, it would look like this:

This is simply all the values of the two equations added together into one single equation.

Let's simplify this equation.

Let's simplify this equation.

First, y - y = 0.  We can remove this.

Let's simplify this equation.

Second, we can move +20 from left to right.  It will offset the +20 there.

Let's simplify this equation.

Next, we can move x to the other side, making it -x

Let's simplify this equation.

Finally, we can do 2x - x = x.  Remember x simply means 1x.  Think of this as bananas or apples.  2 apples, take away 1 apple = 1 apple

We have solved for x simply by adding the two equations together and simplifying.

From here we can solve for y as before by inserting the value we found for x into one of the equations and solving for y. 

We have solved for x simply by adding the two equations together and simplifying.

From here we can solve for y as before by inserting the value we found for x into one of the equations and solving for y. 

This is just an alternative method for solving equations with more than one unknown. If it feels confusing, don’t worry – you can stick to the first method. But feel free to experiment with equations like this on your own!

Solving worded problems with algrebra

We can use algebra to solve worded problems in a simple and logical way.

Solving worded problems with algrebra

To do this, we need to be able to use the information in the problem to write an equation that matches.

You can think of this like translating from English to Maths.

Clara has a number of marbles, but she won't tell you how many.  Instead she thinks for a while and tells you this:

If I double the number of marbles, then I add 8 marbles, I will have 40 marbles.

How many marbles did I start with?

We can write this sequence as an equation:

Let's start by calling the original number of marbles "x".  This represent the number we don't know, but we are trying to find.

We can write this sequence as an equation:

We were told that she first doubled the number of marbles.  Let's add that in.

We can write this sequence as an equation:

Then she added 8 marbles, so let's continue with +8.

We can write this sequence as an equation:

And finally, we were told that she ended up with 40 marbles.  So, let's finish the series of events with = 40.

We can write this sequence as an equation:

Now we have translated the story into "Maths".

Let's solve the equation and find the original number of marbles.

We can write this sequence as an equation:

Now we have translated the story into "Maths".

Let's solve the equation and find the original number of marbles.

We can write this sequence as an equation:

Now we have translated the story into "Maths".

Let's solve the equation and find the original number of marbles.

We can write this sequence as an equation:

Now we have translated the story into "Maths".

Let's solve the equation and find the original number of marbles.

We can write this sequence as an equation:

That's it.  Using algebra, we have found that the original number of marbles was 16!

Well done! You should now understand algebra, how to solve equations with one or two unknowns, and how to translate word problems into mathematical formulas.

With this information you should be one step closer to nailing the 11+ assessment.  Good luck!