If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

If it’s a formula you need, here are the two that are relevant this week. In the 2023 exams, you will be given these formulae.

Pythagoras’ Theorem

This week, we’re exploring:

1. Integer Solutions to Inequalities
2. Forming and Solving Equations
3. Reflection
4. Writing Ratios in the Form 1:𝑛
5. Pythagoras’ Theorem

Question 1: Integer Solutions to Inequalities

When finding the integer (whole number) solutions to an inequality, we need to pay attention to the inequality signs. Remember, < or > mean less than and greater than but don’t include the number itself, while ≤ or ≥ do include the number.

For example, 𝑥 > 3 would be all numbers larger than 3 and the first integer solution would be 4. However, 𝑥 ≥ 3 would be all numbers greater than or equal to 3 and the first integer solution would be 3.

Write down all the integer values of 𝑛 such that -2 ≤ 𝑛 + 2 < 7.

In this case, we need to start by solving the inequality. Just like when solving linear equations, we use inverse operations. In this case, we just need to subtract 2 from all three parts of the inequality.

-2 ≤ 𝑛 + 2 < 7

-4 ≤ 𝑛 < 5

Since -4 ≤ 𝑛, we include -4 in our solution. However, since 𝑛 < 5 we don’t include 5. So, the integer solutions are -4, -3, -2, -1, 0, 1, 2, 3 and 4.

Question 2: Forming and Solving Equations

In exam questions, you are often asked to apply algebra to a particular topic – for example, the perimeter of a shape or the ages of different people.

Arjun is five years older than Bobby. Bobby is twice as old as Cooper. Given that the sum of their ages is 40, determine their ages.

The key to this question is identifying the person whose age doesn’t depend on anyone else’s age.

Arjun is 5 years older than Bobby, so Arjun’s age depends on Bobby’s age. Bobby is twice as old as Cooper, so Bobby’s age depends on Cooper’s. Cooper’s age doesn’t depend on anyone else’s age, so we assign a variable (letter) to his age. Which letter we use doesn’t matter.

Let Cooper be 𝑎 years old.

Bobby is twice as old as Cooper, so Bobby will be 2 × 𝑎 or 2𝑎 years old.

Arjun is 5 years older than Bobby, so we add 5 to Bobby’s age. Arjun is (2𝑎 + 5) years old.

Next, we find the sum of their ages. In other words, we add them up and simplify.

𝑎 + 2𝑎 + 2𝑎 + 5 = 5𝑎 + 5

We know the sum of these ages is 40, so we set this expression equal to 40 and solve the equation.

            5𝑎 + 5 = 40
– 5                               – 5
            5𝑎 = 35
÷ 5                              ÷ 5
            𝑎 = 7

Remember, Cooper was 𝑎 years old, so we can work from this to find each person’s age.

Cooper is 7 years old.
Bobby is 14 years old.
Arjun is 19 years old.

Question 3: Reflection

When reflecting a shape, you will be given a straight line to reflect a shape in. Usually, you will need to draw this line in first.

Reflect the shape below in the line 𝑦 = 𝑥.

First, we need to draw the line 𝑦 = 𝑥. This will go through the points (0, 0), (1, 1), (2, 2) and so on.

Now, we think about how the reflection will affect each vertex (corner) of the shape. If we consider the perpendicular distance to the line, the reflected vertex will be the same distance on the other side of the line.

Joining these three vertices together gives us our final shape:

Describe the single transformation that maps shape A onto shape B.

The key here is to find the straight line that is the same distance from shape A and from shape B.

This is the line with equation 𝑥 = 5.

Question 4: Writing Ratios in the Form 1:𝑛

Writing ratios in the form 1:𝑛 uses the same method as simplifying ratios but, instead of dividing by a common factor, we divide to reduce one part of the ratio to 1. Unlike with simplifying, the parts of the ratio don’t have to be integers.

Write 18:27 in the form 1:𝑛.

We want to reduce the first part of the ratio to 1 so we divide each part of the ratio by 18.

18 ÷ 18 = 1

27 ÷ 18 = 1.5

So, 18:27 becomes 1:1.5. 

Question 5: Pythagoras’ Theorem

For a right-angled triangle, if we square the two shorter sides and add the results together, this gives the square of the longer side. This is known as Pythagoras’ theorem and is commonly written as where is the longest side (the hypotenuse).

We can use Pythagoras’ theorem to find the missing side of a right-angled triangle if we know two of the sides.

Let’s say we want to find the hypotenuse of this triangle:

The hypotenuse is the longest side and always opposite the right angle – in this case it’s marked . So, we know that we should get an answer larger than 0.2cm and 1.2cm. We can use this knowledge to check our answer.

We start by simply substituting these values into Pythagoras’ theorem. It doesn’t matter which way round and are as long as is the hypotenuse. We’ll use and .

(2d.p.)

As we expected, our value is larger than 1.2cm.

Let’s say we want to find the length of the side marked instead:

This time, we are looking for one of the shorter sides. We know that it will be less than 3.8cm because the hypotenuse is always the longest side. We are going to use a slightly different form of Pythagoras’ Theorem. When we are looking for a shorter side, we subtract:

Since is the longest side, and .

(2d.p.)

As we expected, our value is smaller than 3.8cm.

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