It is often useful to create a template equation in which numbers are represented by letters. You can then substitute different values into your equation, i.e. set some of the letters to be equal to particular numbers, in order to determine the value of other numbers.
The amount of profit a business makes will vary from month to month. Therefore, it can be convenient when you are calculating total profit made, in a year for example, to create a template equation. You can then substitute monthly income and expense figures into this equation to calculate your total profit.
SGNEH is a large auditing firm. They have $E$ number of interns in one department whom they pay a wage of $£W$ per year, they also have $Z$ number of employees in this department whom they pay a wage of $£S$ per year. The other running costs of the firm (transport, buildings etc.) total $£Y$ per month.
Create an equation for the total costs of SGNEH per month.
Algebraic expressions can sometimes be simplified to expressions which are shorter, tidier and thus often clearer.
For example, if you have two or more terms that are the same, for instance $3a + 5a$, you can simplify this to $8a$. However, you cannot simplify $3ab +5c +2d$ as there are no terms that can be collected together.
Similarly, on collecting terms in the expression $-2a+3b+6a-c-2b+c$ we have:
\[-2a+3b+6a-c-2b+c=(-2a+6a)+(3b-2b)+(-c+c) = 4a+b.\] We can see that the resulting expression is easier to understand and use than the original.
Factorising is also a useful tool for tidying up expressions and can reveal more information. You factorise an expression by writing it as a product of two or more simpler expressions. You can think of it as breaking it down into simpler parts.
You can factorise an expression if there is a common factor throughout some of the terms.
For example: $3a+9b+6c$ can be factorised into $3(a+3b+2c)$, since $3$ divides $3, 6 \text{ and } 9$.
Another example: $ab-b+bc+a=b(a-1+c)+a$ as $b$ is a common factor in the first three terms.
Sometimes factorising is slightly more complicated and you have to do more work to find the factorisation. For example, $x^2 +3x+2=(x+2)(x+1)$ expresses the quadratic as a product of $2$ simpler expressions and the factorisation gives more information about the quadratic, including its roots.
It is not immediately obvious how to factorise such a quadratic. However, there are several methods you can use to find this factorisation. See the section on factorising expressions for more information.
An equation is an expression of the form $L=R$ where $L$ amd $R$ are usually algebraic expressions involving variables such as $x$ or $y$ etc. You may want to express one variable in terms of the other. To do this you would rearrange the equation so that the variable you want is by itself on one side of the equation and all other terms are on the other side.
You will find plenty of discussion and examples in the section on rearranging equations.
Now we shall demonstrate how to rearrange a demand equation so that price is expressed in terms of demand.
Demand equations are methods of calculating the different levels of demand of a product. There are many different forms of demand equations, from supply demand to simultaneous equations.
The demand equation for your company's cars is $Q =50000 - 7P$ where $Q$ is the number of units of cars in demand given the price $P$ (in £) of a car . Express price $P$ in terms of demand $Q$.
There are many ways to do this, including the method shown here.
Given $Q =50000 - 7P$, we want to express $P$ in terms of $Q$.
First add $7P$ to both sides to get \[7P+Q=50000.\] Now take away $Q$ from both sides: \[7P=50000-Q.\] The finally divide both sides by $7$ to obtain the required expression: \[P=\frac{50000-Q}{7}\] .
The demand equation for your company's cars is $Q =50000 - 7P$ and the supply demand is $Q = 1500 + 4P$. How many cars must be created for supply to match demand and at what price must they be sold?
For supply to match demand, the above equations must be equal:
\[50000 - 7P = 1500 + 4P.\]
We now need to solve the above equation. First subtract $1500$ from both sides and then add $7P$ to both sides to give:
\[48500 = 11P.\]
Then divide by $11$ to calculate the price a car must be sold at to give $P = 4409.09 $.
Then substitute this value into the supply demand equations to give: \[Q = 1500 + 4 \times 4409.09 = 1500 + 17636.36 = 19136.36\] As we want a whole number of cars (not part of a car!) we round this value of $Q$ to $19137$. So $19137$ cars need to be made for supply to match demand.
These equations can be manipulated to suit the needs of any company and their products. They are useful in finding the required price to set to break-even or make a profit.
Test yourself: Numbas test on rearranging equations
For more information on all of the topics in this section see our page on numerical reasoning.
Also try these workbooks (these are also very useful for numerical reasoning tests):