2 Algebraic expressions
Algebraic expressions are just statements about numbers. However, letters are used as place holders for some of the numbers. There are many reasons this is useful, it could be because we would like to uncover the structure of something, or, because we don’t know the specific numbers to use yet.
2.1 Substitution
In order to evaluate an algebraic expression we have to substitute the letters for numbers. After the numbers are written in place of the letters we must take care to evaluate the statement in the correct order. BIDMAS is often used to remember the order:
- Brackets Work out anything in brackets first.
- Indices Powers are next, something like \(3^2\).
- Division and Multiplication these two have equal priority. If there is a ‘tie’ work left to right. However if you see a large division they have implicit brackets in them. For example \(\frac{2 + 10}{2 \times 3}\) should be thought of as \(\frac{(2 + 10)}{(2 \times 3)}\).
- Addition and Subtraction like multiplication and division these are equal priority. If there is a tie work left to right.
One more thing to know before we start making substitutions is that the multiplication symbol \(\times\) is often not used in algebraic expressions. Letters and numbers that are next to each other are multiplied together. For example \(3a\) means \(3 \times a\). You can show two numbers multiplied together like this \(2 \times 3 = (2)(3) = 6\).
Here are some examples:
If \(a=2\) and \(b=-3\) then we can evaluate \(5a + 4b\) like this:
\[ 5(2) + 4(-3) \]
When things are written next to each other this means multiplication.
\[ 5\times 2 + 4 \times -3 \]
Using BIDMAS to do the multiplication first and remembering that a positive number multiplied by a negative gives a negative number.
\[ 10 +- 12 = -2 \]
Substituting \(n = 3\) and \(x = 2\) into \(5x^n\). By replacing the letters with numbers we have:
\[ 5(2)^3 \]
Remembering that when things are next to each other it means multiplication, which gives:
\[ 5 \times 2^3 \]
Following BIDMAS we must deal with the powers first. Since \(2^3 = 2 \times 2 \times 2 = 8\) we have:
\[ 5 \times 8 = 40 \]
Finally consider \(\frac{2p+q}{r}\) where \(p=6\), \(q=3\) and \(r=5\). Replacing the letters with numbers we have:
\[ \frac{2(6)+3}{5} = \frac{2 \times 6+3}{5} \]
Remembering that there are implicit brackets in fractions, the numerator needs to be evaluated first.
\[ \frac{(2 \times 6+3)}{5} = \frac{(12+3)}{5}=\frac{15}{5} \]
Now the fraction can be evaluated.
\[ \frac{15}{5} = 3 \]
You can practice these techniques with the following questions. The numbers change each time to try them as much as you like.
2.2 Simplification
Algebraic expressions are made up of terms. Similar terms can be combined to create a simplified expression, this processes is called collecting like terms. For example \(2a + 3a\) can be simplified to \(5a\) by collecting the \(a\) terms. Here’s another example with a bit more going on:
\[ 5x + 7y - 3x + 3y = \overbrace{5x - 3x}^{x\text{ terms}} + \overbrace{7y + 3y}^{y\text{ terms}} = 2x + 10y \]
Notice that the like terms were grouped first to make it easier to simplify. Also, each term owns the positive or negative symbol ahead of it.
Terms can be more complex too. Although it’s tempting to find something to simplify there are no like terms in this expression: \(3xy + 6x^2 + 2x - 5y\). Only the exact same multiples can be simplified. For example:
\[ 6x^2 + 2x - 5x^2 -8x = \overbrace{6x^2 -5x^2}^{x^2\text{ terms}} + \overbrace{2x-8x}^{x\text{ terms}}=x^2-6x \]
Notice that the two different types of term are \(x\) and \(x^2\). Also, I could have written \(1x^2\) but we normally don’t bother with the \(1\). It’s also important to note that capitalisation matters; \(x\) is different from \(X\).
Take care when simplifying multiples of different letters \(3xy + 5yx\) can be simplified. This is because the order of multiplication doesn’t matter so \(3xy + 5yx = 3xy + 5xy = 8xy\). Terms are normally written in alphabetical order with the highest powers first.
- \(x \times x = x^2\)
- \(x + x = 2x\)
- \(x\) is different from \(X\)
- \(1x\) is written as \(x\)
Have a go at simplifying with these questions.