# Using and Making Formulas

KS3AQAEdexcelOCRWJEC

## Using and Making Formulas

Formulas are extremely useful in maths. They tell us how to calculate new values based on information we already know.

Using formulas is relatively easy but making formulas for yourself can sometimes be tricky.

## Skill 1: Using Formulas

A formula is a rule that helps you work something out. For example, the formula for converting from kilograms ($k$) to pounds ($p$) is:

$p=2.2k$

We can use this formula to convert any value from kilograms to pounds by simply replacing the $k$ with our value. This is known as substitution.

Example: Convert $5$ kilograms into pounds.

We should rewrite the formula, then we need to substitute our value of $5$ into it:

\begin{aligned}p &= 2.2 \times k \\ p &= 2.2 \times 5 \\ p &= 11\end{aligned}

Therefore $5$ kilograms is equal to $11$ pounds.

Level 1-3 GCSE KS3

## Skill 2: Making Formulas from Words

Sometimes you may be asked to give an expression or a formula. An expression is similar to a formula, but does not feature an equals sign ($=$).

Example: A taxi company charges $£3$ as a base fee plus an additional $£0.50$ per minute.

If we want to construct a formula to represent this information, we need to define a couple of things. Let the total cost of the journey be $\textcolor{red}C$ and the time of the journey be $\textcolor{blue}t$.

We can now use these values to construct our formula:

Level 1-3 GCSE KS3

## Notes:

Having a good understanding of BIDMAS is important when constructing formulas as you need to make sure you get things in the right order.

## Example 1: Using a Formula

The formula to convert temperature from degrees Celsius ($\textcolor{blue}C$) to degrees Fahrenheit ($\textcolor{red}F$) is:

$\textcolor{red}{F} = \bigg(\dfrac{9}{5}\bigg)\textcolor{blue}{C} + 32$

Convert $25$ degrees Celsius ($\textcolor{blue}{C}$) into degrees Fahrenheit ($\textcolor{red}{F}$).

[2 marks]

We need to substitute the value of $\textcolor{blue}{25}$ into the formula:

\begin{aligned}\textcolor{red}{F} &= \bigg(\dfrac{9}{5}\bigg) \times \textcolor{blue}{25} + 32 \\ \textcolor{red}{F} &= (9\times5) +32 \\ \textcolor{red}{F} &= 45+32 \\ \color{red} F &= 77\end{aligned}

So $25$ degrees Celsius is equal $77$ degrees Fahrenheit.

Level 1-3 GCSE KS3

## Example 2: Making a Formula to Solve a Problem

A car salesperson earns a salary of $£20000$ per year and earns a bonus of $£250$ per car sold.

If the salesperson earns $£25500$ in one year, how many cars did they sell?

[3 marks]

First we need to construct a formula to represent this information. Let the number of cars sold be $n$, and the total amount the salesperson earned in the year be $E$:

$E = 20000 + 250n$

Now we can substitute in the value of $£25500$ in place of $E$:

$25500 = 20000 + 250n$

Now we have an equation which we need to rearrange to get the value of $n$:

\begin{aligned}25500 &= 20000 + 250n \\ 25500-20000 &= 250n \\ 5500 &= 250 n \\ 5500 \div 250 &= n \\ n &=22\end{aligned}

So, the salesperson sold $22$ cars.

Level 1-3 GCSE KS3

## Example Questions

Substitute the value of $p=6$ into the formula:

\begin{aligned} q &= 4p-6 \\ q &= 4\times6-6 \\ q&=24-6 \\ q &= 18 \end{aligned}

Substitute the values $h = 1.6$ and $m = 80$ into the formula:

\begin{aligned} \text{BMI} &= \dfrac{80}{1.6^2} \\[1.5em] \text{BMI} &= \dfrac{80}{2.56} \\[1.5em] \text{BMI} &= 31.25\end{aligned}

So the person has a BMI of $31.25$

There are $2$ more green sweets than there are red sweets, and there are $x$ red sweets. So the number of green sweets is:

$x+2$

So there are $3$ blue sweets, plus $x$ red sweets, plus $x+2$. Combining these gives the following expression:

$3+x+(x+2)$

Which can be simplified by collecting like terms to:

$5+2x$

The amount of sodium lost ($S$) can be found by multiplying the loss per litre by the number of litres lost ($W$):

$S=1.1\times W$

The amount of water lost ($W$) can be found by multiplying the loss per hour by the time spent exercising ($t$):

$W = 1.5 \times t$

Now we need to combine the two expressions by substituting our expression for $W$ into our expression for $S$:

\begin{aligned}S &= 1.1 W \\ S &=1.1 \times (1.5t)\\ &=S= 1.65 t \end{aligned}

Our formula needs to add the base cost of $£40$ to the hourly rate of $£9.50$. Our formula is therefore:

$C = 40 + 9.5h$

If the plumber earns $£68.50$ in a single repair job, we can substitute this into our formula and rearrange to get $h$:

\begin{aligned} 68.5 &= 40+9.5h \\ (-40)\quad 28.5 &= 9.5 h \\ (\div9.5)\quad 3 &= h \end{aligned}

So the repair job took $3$ hours.

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