Simplifying algebra
Algebra involves the use of letters in mathematics. These letters are unknowns and can represent either a single unknown number or a range of unknown numbers.
Sometimes algebraic expressions can be simplified – this means that we collect all the similar terms together. We would never say in conversation “I have 3 apples plus 2 apples”. Instead we would say, “I have 5 apples”. Similarly in algebra we can say:
3\({a}\) + 2\({a}\) = 5\({a}\)
However, if I had 5 bananas and 2 apples, it would be impossible to write this in a simpler way.
In algebra:
5\({b}\) + 2\({a}\) = 5\({b}\) + 2\({a}\)
This cannot be written in a simpler way. When simplifying using addition or subtraction, it is helpful to think of different letters as being completely different things – much like bananas and apples. It is important to note that 5\({b}\) means '5 lots of \({b}\)' or '5 × \({b}\)'.
Here are some more examples of how we can simplify:
7\({b}\) - 4\({b}\) = 3\({b}\)
12\({b}\) + 4 - 3\({b}\) = 4 + 9\({b}\)
2\({z}\) + 3\({y}\) - 7\({z}\) + 6\({y}\) = 9\({y}\) - 5\({z}\)
3\({ab}\) + 2\({a}\) + 7 = 7 + 3\({ab}\) + 2\({a}\)
There are four things to note about the above examples:
- the sign (+ or -) belongs to the term that comes after it
- when giving our simplified answer we always give it in alphabetical order
- a term containing, for example \({ab}\), cannot be added to terms with an \({a}\) or terms with a \({b}\) but must instead be kept separate
- numbers on their own cannot be added to terms containing a letter
Question
Simplify 5\({x}\) + 4\({y}\) - 2\({z}\) + 3\({x}\) + \({z}\) - 6\({y}\)
Collecting like terms gives 5\({x}\) + 3\({x}\) + 4\({y}\) - 6\({y}\) + \({z}\) - 2\({z}\)
Simplifying leaves 8\({x}\) - 2\({y}\) - \({z}\)
We can also simplify algebraic expressions that involve multiplication. The rules here are very different to the rules for addition and subtraction.
Consider the following:
5\({a}\) × 7\({b}\)
Firstly, we remember that 5\({a}\) = 5 × \({a}\) and 7\({b}\) = 7 × \({b}\)
This leaves us with:
5\({a}\) × 7\({b}\) = 5 × a × 7 × b
This gives the result:
5 × 7 × \({a}\) × \({b}\) = 35\({ab}\)
Sometimes we will have to simplify expressions in the form:
\({a^3}\) × \({a^5}\) or \({d^8}\) × \({d^2}\)
In general \({x^a}\) × \({x^b}\) = \({x^{(a+b)}}\)
This means that when we multiply two terms with indices, the result is that the indices are added.
Examples
\({a^7}\) × \({a^4}\) = \({a}^{7+4}\) = \({a}^{11}\)
\({f^3}\) × \({f^4}\) = \({f^7}\)
\({z^2}\) × \({z^3}\) × \({z^5}\) = \({z}^{10}\)
Or when we have two or more different letters involved:
\({a^3}\) × \({b^4}\) × \({a^2}\) × \({b^7}\) = \({a^3}\) × \({a^2}\) × \({b^4}\) × \({b^7}\) = \({a^5}\) \({b^{11}}\)
\({x^2}\) × \({y^2}\) × \({x^4}\) × \({z^3}\) = \({x^6}\)\({y^2}\)\({z^3}\)
Or when we have a mixture of indices and coefficients:
5\({a^3}\) × 3\({a^2}\) = 5 × 3 × \({a^3}\) × \({a^2}\) = 15\({a^5}\)
Question
Simplify 8\({b}\) × 3\({b}\) × 2\({c}\)
Firstly, we rearrange the expression to give:
8\({b}\) × 3\({b}\) × 2\({c}\) = 8 × 3 × 2 × \({b}\) × \({b}\) × \({c}\)
Evaluating the numbers leaves:
48\({b}\) × \({b}\) × \({c}\)
Remembering that \({b}\) × \({b}\) is \({b^2}\) gives:
48\({b^2}{c}\)
Question
Simplify 6\({b^2}\) × 3\({a^2}\)\({b^3}\)
6\({b^2}\) × 3\({a^2}\)\({b^3}\) = 6 × 3 × \({a^2}\) × \({b^2}\) × \({b^3}\)
Evaluating combining terms and remembering that \({b^2}\) × \({b^3}\) = \({b^5}\)
6\({b^2}\) × 3\({a^2}\)\({b^3}\) = 6 × 3 × \({a^2}\) × \({b^2}\) × \({b^3}\) = 18\({a^2}\)\({b^5}\)
More guides on this topic
- Equations of lines – WJEC
- Equations of curves - Intermediate & Higher tier – WJEC
- Equations and formulae – WJEC
- Factorising - Intermediate & Higher tier – WJEC
- Quadratic expressions - Intermediate & Higher tier – WJEC
- Sequences – WJEC
- Functions - Higher only – WJEC
- Inequalities - Intermediate & Higher tier – WJEC
- Simultaneous equations - Intermediate & Higher tier – WJEC