Formulae

Creating formulae.

An algebraic formula is a bit like an equation. It has an equals sign and variables; these are letters that represent values that can be changed. It might also have a constant, which is a number that always stays the same.

Let's look at how to create formulae for real life situations.

Vina receives £20 from her parents for her birthday. She then earns £30 each week from her Saturday job. She puts all the money she receives into a savings account.

Part A of this question asks: write a formula for the amount of money, 𝑆, in pounds, Vina will have saved after 𝑥 weeks.

To answer this, you can use a bar model where a 20 bar represents the initial £20 Vina receives from her parents and separate 30 bars represent the extra £30 Vina earns each week from her Saturday job.

So, after 1 week, Vina will have a total of £20 add £30. Then, after 2 weeks, Vina will have a total of £20 add two £30. After 3 weeks, £20 add three £30, and so on. The number of 30 bars is equal to the number of weeks. So, after 𝑥 weeks there will be 𝑥 30 bars.

The total amount of money Vina has saved is equal to the sum of all the bars. So, after 𝑥 weeks, Vina will have £20 add 𝑥 times £30, or 20 add 30𝑥. In the question, 𝑆 is used to represent this amount. So, the formula is 𝑆 equals 20 add 30𝑥.

Part B of the question asks you to work out how much money, in pounds, Vina would have saved after 12 weeks.

𝑥 represents the number of weeks. So, substitute 𝑥 equals 12 into the formula. 30 times 12 is 360 and 360 add 20 is 380. So, Vina will have saved £380 after 12 weeks.

Let's look at another question.

Bruno adds £56 to his travel card. Every day he buys a £2 bus ticket using this money. Write a formula for how much money, 𝑀, in pounds, Bruno has left on his travel card after 𝑥 days.

Pause the video and have a go.

Be careful here. This time, the amount he’s spending has been taken away from the initial £56.

Starting with a 56 bar, after 1 day, Bruno will have 56 subtract 1 times £2, which is £54. Then, after 2 days he will have 56 subtract 2 times £2, or £52. After 3 days, 56 subtract 3 times £2, or £50, and so on. So, after 𝑥 days, he will have 56 subtract £2𝑥 left.

𝑀 is how much money Bruno has left. So, the formula is 𝑀 equals 56 subtract 2𝑥.

While studying for your exams, have you ever wondered, 'how will algebra and formulae ever help me in real life?' Here's how they can be useful.

Let's say your favourite band are performing at a concert you want to go to with your friends. Booking online tickets cost £25, and merch t-shirts cost £10 each. There is also a £5 admin fee, no matter how many tickets and t-shirts you order.

How much money do you need to buy four tickets and three t-shirts for you and your friends? You can create a formula to help you calculate this. The total cost we are trying to work out is C. The admin fee is £5 and that's a fixed cost, no matter what. So, that's the constant.

The tickets and t-shirts have a fixed cost, but the number you can buy can change. These changing values are called the variables. We'll call the number of tickets, 𝑥, and the number of t-shirts, 𝑦. Each ticket costs £25, so that's why we add on 25𝑥 to the formula.

Then t-shirts cost £10 each, so we add on 10𝑦. Our formula is C = 5 + 25𝑥 + 10𝑦. We want 4 tickets, so the variable, 𝑥 = 4. And we want 3 t-shirts, so the variable, 𝑦, is 3.

So, C = 5 + (25 × 4) + (10 × 3). The total money you'll need is C = 5 + 100 + 30, which equals £135.

Maybe you're going with 7 friends and have £250 between you. Once you've paid for the tickets and the admin fee, how many of you can buy t-shirts? Let's use our formula, C = 5 + 25𝑥 + 10𝑦 to model this new situation.

We know that we have £250 to spend, so we need to replace the C in our formula with 250. We can also replace the variable, 𝑥, with the number of tickets we need, which is 8, including your own ticket. So, our formula is now 250 = 5 + 25 × 8 + 10𝑦, which is 250 = 205 + 10𝑦.

To find the number of t-shirts we can buy, we need to work out what 𝑦 equals.

First, we subtract 205 from both sides to give 45 = 10𝑦. Then, divide both sides by 10 to get 𝑦 = 4.5. This means you can buy four t-shirts and have a little bit of money left over.

So, formulae aren't just for the classroom. Putting them to use in real life can help you make smart decisions when it comes to things you enjoy.

Rearranging equations with fractions – Higher tier.

It is sometimes useful to rearrange equations or formulae to make a different variable the subject. Remember, variables are letters that represent values that can be changed.

For example, the area of a rectangle is given by the length times the width. You can substitute known length and width values into this formula to calculate the area. But what if you know the area and length but not the width? Then, it would be useful to have the formula in the form 𝑤 equals, where 𝑤 is now the subject. To make 𝑤 the subject, you would need to divide both sides of the equation by 𝑙.

Remember, all equations have an equals sign, which means they need to be balanced like a set of scales.

So, you must divide both sides by 𝑙, giving 𝐴 over 𝑙 equals 𝑙𝑤 over 𝑙. On the right-hand side, 𝑙 cancels out. Leaving 𝐴 over 𝑙 equals 𝑤, which you can rewrite as 𝑤 equals 𝐴 over 𝑙.

This process is called rearranging. Let's look at an example question.

Make 𝑥 the subject of 𝑦 equals 2𝑥 subtract 5, all over 𝑥.

To do this, you need to rearrange the equation so that it's in the form 𝑥 equals instead of 𝑦 equals. In its current form, the equation is tricky to work with. So, let's start by addressing the fraction part.

2𝑥 subtract 5 is divided by 𝑥, so to remove the fraction you need to use the inverse operation, which is multiplying by 𝑥. The left-hand side simplifies to 𝑥𝑦 and on the right-hand side, multiplying by 𝑥 cancels with the denominator. So, the equation becomes 𝑥𝑦 equals 2𝑥 subtract 5.

The next step is to get all the 𝑥 terms on one side of the equation, and all the non-𝑥 terms on the other side. To do this, subtract 2𝑥 from both sides, giving 𝑥𝑦 subtract 2𝑥 equals 2𝑥 subtract 5 subtract 2𝑥.

2𝑥 subtract 2𝑥 equals 0, so this gives 𝑥𝑦 subtract 2𝑥 equals –5. Then, 𝑥𝑦 and –2𝑥 both have a common factor of 𝑥 which means you can factorise the left-hand side or in other words, put the expression into brackets. Taking out the factor of 𝑥 gives 𝑥 all multiplied by 𝑦 subtract 2 equals negative five. The 𝑥 is now on its own outside the bracket, so you're nearly there.

To finish and get 𝑥 completely on its own on the left-hand side, you need to use the opposite operation to multiplying by 𝑦 subtract 2 which is dividing by 𝑦 subtract 2. 𝑦 subtract 2 cancels on the left-hand side, so you have 𝑥 equals –5 all over 𝑦 subtract 2, where 𝑥 is now the subject.