# Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 3 • Question 6

a) Solve the inequality 5𝑥 − 4 ≤ 31. b) Solve the inequality 12 < 4(𝑥 + 3). c) Represent the solution set of both inequalities on a number line.

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### Video Transcript

a) Solve the inequality five 𝑥 minus four is less than or equal to 31. b) Solve the inequality 12 is less than four multiplied by 𝑥 plus three. c) Represent the solution set of both inequalities on a number line.

Let’s begin by solving the inequality five 𝑥 minus four is less than or equal to 31. The inequality symbol can be off-putting. But we can treat it like any other equation and solve by applying a series of inverse operations. There’s just one little thing that we need to be aware of: if at any point we end up multiplying or dividing by a negative number, we need to remember to reverse the direction of the sign. So the first thing we’re going to do to solve this inequality is add four to both sides. Five 𝑥 minus four plus four is simply five 𝑥 and 31 plus four is 35.

Next, we need to divide both sides of this inequality by five. Five 𝑥 divided by five is 𝑥 and 35 divided by five is seven. And we can see that the solution to the inequality five 𝑥 minus four is less than or equal to 31 is 𝑥 is less than or equal to seven. And what this means in real terms is that any value of 𝑥 that is less than or equal to seven when substituted into the expression five 𝑥 minus four will give us an output of less than or equal to 31.

We’re next going to solve the inequality 12 is less than four multiplied by 𝑥 plus three. Once again, we’ll solve this just like a normal equation, remembering that if we end up at any point multiplying or dividing by a negative number, we need to change the direction of that symbol. And in fact, there are two ways we could go about solving this. Let’s look at both methods.

For the first method, we would expand the brackets four 𝑥 plus three. And we do that by multiplying both terms inside the bracket by four. Four multiplied by 𝑥 is four 𝑥 and four multiplied by three is 12. So this inequality is the same as 12 is less than four 𝑥 plus 12. To solve, we’ll subtract 12 from both sides and we see that zero is less than four 𝑥. We then divide through by four and that gives us zero is less than 𝑥 or 𝑥 is greater than zero.

Now, let’s look at the alternative method. This method relies on the fact that we spot that this four is multiplying everything inside the bracket. And the opposite of multiplying is of course dividing. So we can divide through both sides of this inequality by four. 12 divided by four is three and four times 𝑥 plus three divided by four is just 𝑥 plus three. This time, we subtract three from both sides. And once again, we see that zero is less than 𝑥 or 𝑥 is greater than zero.

The final part of this question asks us to represent the solution set to both inequalities on the number line. Remember those are the solutions that are less than or equal to seven and greater than zero. Let’s begin at the lower end. We know that 𝑥 is greater than zero. In integer terms, that’s one, two, three, four, and so on, not including the number zero itself. This is called a strict inequality. And we’ve represented it on the number line using an open or empty dot as shown.

Next, we’ll deal with the top end: 𝑥 is less than or equal to seven. This is sometimes called a weak inequality. In integer terms, we say that 𝑥 can be seven, six, five, four, and so on. And we represent a weak inequality with a solid dot as shown.

Now, in fact, we’re interested in the solution set of both inequalities. So those are all the numbers that are greater than zero and less than or equal to seven. On the number line, that’s as shown.