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I can use a letter to represent a generalised number and form a linear equation from a context.

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Licence

This content is © Oak National Academy Limited (2023), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

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Key learning points

Quantities that change can be assigned a variable.
If there is a relationship between these quantities, we can write this algebraically.
With two quantities, the algebraic statement can form an equation.

Common misconception

When forming expressions brackets can be missed out or incorrectly used.

Review priority of operations. Testing algebraic manipulations with numerical values can be useful when pupils are unsure if things are equivalent. E.g. the difference between 2(x + 4) and 2x + 4 can be shown with 2(5 + 4) and 2(5) + 4

Keywords

Equation - An equation is used to show 2 expressions that are equal to each other.
Interior angle - An interior angle is an angle formed inside a polygon by two of its edges.
Vertically opposite angles - Vertically opposite angles are pairs of opposite angles formed when two lines intersect at a point. They are equal.

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Hi, I'm Mrs. Wheelhouse, and welcome today's lesson on checking and securing understanding of forming linear equations.

This lesson is part of our unit on algebraic manipulation.

Now, algebra can be incredibly useful, so let's look at what skills we're learning with algebra today.

By the end of today's lesson, you'll be able to use a letter to represent a generalised number and form a linear equation from a context.

Some keywords we're gonna be using our lesson today: equation, interior angle, and vertically opposite angles.

Now, if these words are not familiar to you, I suggest you pause right now and have a read through their descriptions.

Now! you're ready.

Let's get started.

Our lesson today has two parts to it, and we're gonna begin by forming expressions.

Now, letters can be used to represent unknowns and variables in context.

Izzy is helping her dad to build a new shed.

The width of the shed has to be 2 metres, but the length can vary.

How could she form an expression for the area of the floor of the shed? Well, she knows that the width has to be 2, but that the length can vary.

Izzy has chosen the letter a to represent the length of the shed, so therefore, the area can be written as 2a.

Now, how could she write an expression for the perimeter of the floor? 'Member, perimeter is where we sum the edge lengths, so we have 2 + 2 + a + a, which is equivalent to 2a + 4, but you could have also written this in factorised form of 2 lots of a + 2.

It is important to consider the priority of operations and use brackets when manipulating variables, if necessary.

How can we write an expression for the area of this rectangle? Well, it would be 5 lots of a + 2.

If we want to write this in expanded form, we need to make sure you multiply the whole expression by 5, so we have 5a + 10.

How can we write an expression for the perimeter of the rectangle? That's right: we need to sum the edge lengths, so we have a + 2 + 5 + a + 2 + 5, but we can write this in other ways.

We could collect the like terms, which means I have 2a + 14.

I could, of course, treat the expression a + 2 like a term and add the opposite sides, so I could say I have 2 lots of a + 2, and I have 2 lots of 5.

Because both expressions are being multiplied by 2, I could write it like this: 2 lots of a + 7.

All of these expressions are equivalent.

They're all expressions for the perimeter of this rectangle.

We're going to cut the rectangle in half.

Can we write an expression for the area of the new shape? Well, I think we can because if I'm halving it and I know the expression for the total area, well, it's just half of that, right? So it's half of 5 lots of a + 2.

I could write it like this, of course: 5 lots of a + 2, all halved.

The order in which these three values are multiplied, of course, does not matter, so I could also write an expression for the area like any of the following.

Maybe you have one of these expressions, which is the one you prefer.

Since they're all equivalent, it doesn't matter which one you pick.

What about writing expression for the perimeter of the new shape? Can I do that? Huh.

No, I can't.

Well done if you didn't fall for that.

It depends on how I cut the shape, so what do I mean by that? So if the shape was cut in half horizontally, what would an expression for the perimeter look like? Well, I now know that one of the lengths will be 2.

5, but the other is unchanged; it stays as a + 2, so I could write it out like this, or I could gather the like terms. I could, of course, write it in the factorised form, and there are lots of different ways I could do that.

Which of these could be an expression for the area of this shape? So 'member, it could be more than one.

Pause the video while you make your choice.

Welcome back.

Which ones did you go for? Well, you should have picked b, c, d, and f.

All of these could be expressions for the area of the triangle.

'Member, the area of the triangle is the base times the perpendicular height and then halved.

All the expressions you can see here that have been ticked are equivalent.

Which of these could be an expression for the perimeter of this rectangle? Pause the video and make your choice now.

Remember: could be more than one.

Welcome back.

Which ones did you go for? Well, c is absolutely one because I have 2 lots of 2a + 6, and I got that by adding the two lengths you can see, so 2a + 3 + 3 makes 2a + 6, and then I doubled it.

Now, forming expressions can also help us see how certain number tricks work.

What about this one here? So what I'd like you to do is pause the video, go through the steps, and see what you get.

Do that now.

Welcome back.

What value did you get? Well, you should always get the number 3 if you follow the steps correctly, and we can use algebra to show why this is the case, so I think of a number.

Now, remember, that number can be anything, so I'm going to use letter x to represent it, and I add 2, so the expression will be x + 2.

I then multiply by 3, so I have 3 lots of x + 2.

Now, 'member, I need to put the brackets here because the whole of the expression will be multiplied.

I now subtract the original number, so now I've got 3 lots of x + 2 - x, and then divide by 2.

Fractional notation groups the numerator, and this is showing that the whole expression is divided by 2.

I could, of course, have written the three lots of x + 2 - x inside a pair of brackets and then written divide by 2 outside, but I like using fractional notation.

I now need to subtract my original number again.

Whoo, that expression does look quite complicated.

Let's see what we can do with this, so I'm gonna expand the brackets, first of all, so I have 3x + 6 - x as the numerator.

Well, I can gather the like terms there.

That gives me 2x + 6 on the numerator.

That looks a lot easy to divide by 2.

What's half of 2x + 6? Well, that's x + 3, and now I can gather like terms again, so I'm just left with 3, and that is why it doesn't matter what number you think of: you'll always get 3 when you follow this number puzzle 3.

Time for a quick check.

Match the sentence with the expression.

Pause and do this now.

Welcome back.

a matches to a - 5, all divided by 2.

b matches to a - 5.

c matches to 2a - 11.

d matches to a divided by 2 and then - 5.

e matches to 2 lots of a - 5 and then takeaway 6, and then, f matches to 2a - 5 all divided by 2.

Well done if you've got these right.

It's now time for your first task.

For question 1, write an expression for the area of each shape in two different ways.

Pause and do this now.

Question 2.

We have a rectangular piece of paper with a width of 12 and a height of 2x + 4.

How could you write an expression for the perimeter? Can you think of more than one way? In part b, the paper is cut in half horizontally and the pieces fitted together as shown.

Write an expression for the perimeter of the new shape, and then c, imagine you went back to that first piece of paper, but now you cut it in half vertically and you fit the pieces together as shown.

Write an expression the perimeter of this shape.

Pause the video while you work on this.

Question 3, match the two expressions on the right to two of the number puzzles and then write an expression for the third number puzzle, and question 4, use your expressions to show the value you will always finish with.

Pause the video and do this now.

Welcome back.

Let's go through our answers, so question 1.

You can see here I've written equivalent expressions for the area of each shape.

Feel free to pause the video at this point so that you can see if your expressions are on the screen, and then you can check if they're right.

Don't worry if the final shape, so the trapezium, you've got different answers to what I've got.

You might still be right if they're equivalent.

There were so many possible answers, I couldn't put them all on the screen.

Question 2.

You had to write an expression for the perimeter of the rectangle, and what I've done here is given you some of the possible things you may have written.

In part b, we cut the paper in half horizontally.

Now, the important bit was that you got the dimensions of the new shape correct.

What I've then done is given some examples of how we could write an expression for the perimeter of the new shape.

Now, remember, you may have different answers to me, and that's okay as long as they're equivalent, and here was c.

Again, making sure you've got the dimensions of the new rectangle correct should mean that you got an expression for the perimeter correct, but again, if you don't match to one of the answers I have on screen, do feel free to check to see if yours are equivalent.

Question 3.

You had to match the two expressions on the right to two of the number puzzles and then write an expression for the third.

What you see here are the correct matchings and the expression for the third number puzzle.

Feel free to pause while you check this through.

Question 4.

What you can see here for each expression is how I've manipulated it to show what you always finish with, so for the first one, you'll always finish with the number 5.

The second one always ends with the number 6, and the final one, what the final value you end up with has to be whatever value you chose at the start.

Well done if you got these right.

It's now time for the second part of today's lesson on forming equations.

When we know two expressions are equal, we can form an equation with them.

For example, here.

What do you know about those two marked angles? That's right: They are vertically opposite angles, which means they must be equal, and because we have a statement of equality, we can form an equation.

In other words, 5a + 30 = 4a + 48.

What could we use to form an equation for a this time? So what facts do you know about angles and triangles? That's right: The interior angles in a triangle sum to 180 degrees, so because we have a statement of equality, we can form an equation.

We can say that 2a + 15 + 4a + a + 25 = 180.

Now, of course, we could simplify this.

Collecting like terms lets us reach an equation of the form 7a + 40 = 180.

Ah, Izzy, Lucas, and Laura are considering this shape.

We know it's a rectangle because we're told it is, and the pupils want to form an equation.

Well, Lucas says, \"We can use the fact \"that the interior angles of any quadrilateral \"sum to 360 degrees,\" and Izzy says, \"Well, \"a rectangle is a type of parallelogram, \"so the opposite angles will be equal,\" and Laura says, \"Well, hang on a second.

\"All the angles in a rectangle are right angles, \"so I'm gonna use that.\" What I'd like you to do is pause the video and use each pupil's idea to form a different equation for a.

Pause and have a go now.

Welcome back.

Let's see how you got on.

Well, we're gonna start with Lucas, who wants to use the fact that the interior angles of any quadrilateral sum to 360 degrees, so we write out our four expressions, and we sum them.

Now, of course, we can leave it like this, but it might be simpler to collect the like terms, which means that 6a + 228 = 360.

Let's consider Izzy, who said that because rectangles are a type of parallelogram, the opposite angles must be equal, so we can pick two opposite angles, for example, a + 68, and say that it's equal to the opposite, which is 112 - a.

We could also have used the other two opposite angles and said that 4a + 2 = 2a + 46.

Laura wanted to use the fact that all the angles in a rectangle are right angles, so she can make four possible equations: a + 68 = 90 or any of the other angles = 90.

Now, whose method did you like the most? Personally, if I wanted to know what a was and calculate all the angles, I think using one of Laura's was really simple.

Like a + 68 = 90 is really easy to work out the value of a, so I think I prefer Laura's, but what did you prefer? This shape is parallelogram.

Which of these could be equations for a? Remember, could be more than one.

Pause the video and make your choice now.

Welcome back.

Which ones did you pick? Well, you could have picked a, c, or d.

a was using the fact that the interior angles of any quadrilateral sum to 360 degrees, c was using the fact that the opposite angles in a parallelogram are equal to each other, and d is an equivalent form of a: it's just half of it.

Three consecutive integers sum to 66.

How can we form an equation for this information? Well, let's consider our number line.

If the first number is x, what would you call the next number? That's right: it'd be x + 1, and then, the next 1 would be x + 2.

Well, how can we form an equation now? Well, we take our three consecutive numbers, and remember, they sum to 66.

We could then, of course, collect the like terms, so we can form an equation that looks like 3x + 3 = 66.

Alex's uncle is currently three times his age.

In five years' time, their ages will sum to 70.

Hmm.

How can we form an equation from this information? Well, let's consider the current age of Alex and his uncle.

Well, if Alex's age is currently x, then because his uncle is three times Alex's age, so Alex's uncle's age can be expressed as 3x.

Now, let's consider their age in five years' time.

Well, Alex is currently x years old, so in five years' time, he'll be x + 5, and Alex's uncle's age in five years' time will be 3x + 5.

Now, remember, in five years' time their ages sum to 70, so I take the expression for Alex's age in 5 years and Alex's uncle's age in five years, sum them, and write them equal to 70.

I can, of course, collect like terms, so 4x + 10 = 70.

Let's do a quick check.

Izzy has four cats.

Each cat is two years younger than the previous cat, and this year, their combined age is 32.

Which of these equations best shows this relationship? Pause the video and make your choice now.

So which equations best show this relationship? Well, you could have picked either d or e.

Now, you might be confused 'cause they look very different, so why are they both allowed? Well, for d, in the equation, a represents the age of the youngest cat, so my youngest cat is a years old.

The next cat is a + 2, and the next cat is a + 4, and the fourth cat is a + 6, but for e, I said that a is the age of the oldest cat, so it really depends where I'm starting my count from as to how I write my expressions for the other cats' ages.

It's now time for our final task.

In question 1, for each question, use your angle knowledge to form an equation.

Pause and do this now.

Question 2.

Houses on Oak Road are numbered so adjacent houses increase by one, so, in other words, number 56 is in between 55 and 57.

For a, three adjacent houses sum to 57, so please form an equation with that information, and in b, five adjacent houses sum to 45.

Form an equation for that.

In question 3, our houses are now numbered so adjacent houses increase by two, so, in other words, we would go 54, 56, 58.

What I'd now like you to do is to form an equation for four adjacent houses that sum to 40 and equation for six adjacent houses that sum to 174.

Pause the video and do this now.

Welcome back.

Question 4.

A large rectangle is made by joining four small rectangles as shown.

The perimeter of one small rectangle is 32 centimetres, so part a, what's the relationship between the length and the width of one small rectangle, and then, in b, label the smaller length of one small rectangle a and form an equation for the perimeter of one of the small rectangles.

Pause and do this now.

Welcome back.

Question 5.

This large square is made up of four identical rectangles surrounding a smaller square.

The length of one rectangle is 3 centimetres longer than its width.

The perimeter of the whole shape is 36 centimetres.

Form an equation from this information.

Pause and do this now.

Welcome back.

Time to go through our answers, so for question 1, we had to use our angle knowledge to form equations.

1a, we have vertically opposite angles, so 3x + 40 = x + 80.

In b, we know that the two base angles in isosceles triangle are equal, so 3x + 10 is equal to 5x - 20.

In c, we know that the interior angles of a triangle sum to 180 degrees, so I have 3x - 1 + 7x + 21 + 90 = 180, and then I could gather the like terms. Of course, what I could do is I could say, \"Well, I know that one of the angles in this triangle \"is equal to 90 degrees \"because I have a right angle triangle, \"and that means the other two angles \"must sum to 90 degrees,\" and then I collected the like terms. In d, I used the fact that I have a quadrilateral, and therefore, all interior angles sum to 360 degrees, and I collected the like terms to write it as simply as I could.

Well done if you got these right.

Question 2.

You had to form an equation.

Now, again, you might have a different equation.

It depended on where you started your count from.

For these, I started with the first house number, so the smallest house number, being x, so the next house along would be x + 1, and the next house along would be x + 2, so my three adjacent houses sum to 57, and then I collected the like terms. I did the same for part b, but remember: if you started your count from the house with the greatest number on it, then, obviously, yours will look different.

Again, for question 3, I started with the first house being called x, and then I increased in value.

Question 4.

We can see that the length is three times the width of a small rectangle, so the length of the small rectangle is three times the width of the small rectangle, and I can see that from the diagram because one small rectangle stood on its width, so stood on the smaller side, I can see that three small rectangles fit alongside that longer length, so three lots of the width is equal to one lot of the length.

In part b, I had to label the smaller length of one small rectangle a and then form an equation for the perimeter of one of the small rectangles.

Well, if I know that the width is a and the length is 3a, then I've got 3a + a + 3a + a, which is 8a, and that's equal to 32.

Question 5.

This square was made up of four identical rectangles surrounding a smaller square.

I asked you to form an equation, so what I've done is I've marked on the length and the width of one of the rectangles so that you can see how I formed the expression you can see here for the perimeter.

What I've then done is I've collected the like terms, and I've tried writing this in different ways, so you can see lots of the equivalent equations that you could have written.

Well done if you've got one of these, but remember, you might have a different one but still be correct.

It's now time to sum up what we've learned today.

Quantities that change can be assigned a variable.

If there is a relationship between quantities, we can write this algebraically.

When we have a statement of equality, we can form an equation.

We can use properties of shapes to form equations for shapes whose angles and/or dimensions are described algebraically.

Well done.

You've worked really well today, and I look forward to seeing you for more lessons on algebraic manipulation.

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" }, { "text": "

x(2x + 8)

" } ], "id": "0cf3ec21-7e23-fbea-e2c0-7046b1298109", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" } ], "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1", "meta": { "gradeStageId": "ks4", "subject": "Maths", "tier": "Foundation", "grade": "Year 10", "topic": null, "unit": "Algebraic manipulation", "lessonCount": "13", "age": [ 14, 15 ], "gradeId": "age_14_15", "gradeStage": "GCSE", "lesson": "Checking and securing understanding of forming linear equations" } }, { "id": "2c2fc389-e830-be45-6709-de2f87c9c592", "type": "Resource", "resourceType": "Quiz", "title": "Checking and securing understanding of forming linear equations", "quizType": "Exit quiz", "items": [ { "type": "Question", "question": { "text": "A rectangle has side lengths of

3x + 5

and

2x - 3

. Write an expression for the perimeter of the shape." }, "questionType": "Short answer", "correct": [ { "text": "10x + 4" }, { "text": "2(5x+2)" } ], "id": "c6e59ca1-2047-bd0e-bd43-e865b9df0e7a", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" }, { "type": "Question", "question": { "text": "A triangle has side lengths of

x + 2

2x + 3

, and

3x + 4

. Write an expression for the perimeter of the shape." }, "questionType": "Short answer", "correct": [ { "text": "6x + 9" }, { "text": "3(2x + 3)" } ], "id": "0d770208-fe15-5b01-9680-d94ee8b4724f", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" }, { "type": "Question", "question": { "text": "A regular quadrilateral (square) has side length

4y - 5

. Write an expression for the perimeter of the shape." }, "questionType": "Short answer", "correct": [ { "text": "16y-20" }, { "text": "4(4y-5)" } ], "id": "3489b8b1-0345-ace2-1a1a-b1cc967ae658", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" }, { "type": "Question", "question": { "text": "A rectangle has side lengths of

5a - 6

and

3a + 2

. Write an expression for the perimeter of the shape." }, "questionType": "Multiple choice", "correct": [ { "text": "

16a - 8

" } ], "choices": [ { "text": "

16a - 8

" }, { "text": "

16a + 8

" }, { "text": "

8a - 4

" }, { "text": "

8a + 4

" } ], "id": "1b2d29b1-9f4f-7edb-2695-e7c0db0849c3", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" }, { "type": "Question", "question": { "text": "A regular pentagon has side length

2z + 1

. Write an expression for the perimeter of the shape." }, "questionType": "Short answer", "correct": [ { "text": "10z + 5" }, { "text": "5(2z + 1)" } ], "id": "6aada667-1142-ee7a-c92e-772d0a9dc09b", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" }, { "type": "Question", "question": { "text": "A regular hexagon has side length

3n - 4

. Write an expression for the perimeter of the shape." }, "questionType": "Short answer", "correct": [ { "text": "18n - 24" }, { "text": "6(3n-4)" } ], "id": "464ad469-7adc-c2e2-aaac-dd1d232bf7f7", "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1" } ], "lessonId": "11d0fc30-1299-22c5-b27e-e66d6a7247f1", "meta": { "gradeStageId": "ks4", "subject": "Maths", "tier": "Foundation", "grade": "Year 10", "topic": null, "unit": "Algebraic manipulation", "lessonCount": "13", "age": [ 14, 15 ], "gradeId": "age_14_15", "gradeStage": "GCSE", "lesson": "Checking and securing understanding of forming linear equations" } } ] }