Real mathematicians, unlike statisticians and engineers, very rarely do much with numbers. They use symbols to represent not one particular number, but any number. They spend their time in very abstract worlds.
Sometimes this can be useful to the rest of us, too. We can find out what might happen in variety of situations by using symbols in place of numbers.
The numerals 1, 2, 3, 4, etc. all have a fixed value. Letter symbols such as a, b, c, d, and the favourites x and y represent variables which can take any value. Mathematicians don't just use the Roman letters we use in English. They also use Greek letters such as α, β, γ, δ (alpha, beta, gamma, delta) and a huge number of special symbols. If any lecturer uses a Greek letter you don't recognise, don't be shy, ask them what it is. In the same way, if they use a strange symbol like ∝ or ⊕, ask them what it is and what it means.
In Brush up your maths, we shall stick to Roman characters and familiar symbols used in arithmetic.
The BODMAS rules mentioned in Numbers
apply to algebra also.
For example, x2 + 3(2y + 4) means:
1. multiply y by 2
2. add 4
3. square x
4. multiply the contents of the brackets by 3
5. add the quantities found in steps 3 and 4.
Of course, we can only do all these things if we replace x and y by numbers.
Sometimes and algebraic expression contains more than one term which have the same structure. For example, 3x + 4x means multiply x by 3, multiply x by 4, and then add the results. This would give 7 xs altogether: 3x + 4x = 7x. We can replace 3x + 4x by 7x. We call this collecting together like terms.
Consider the expression 3x + 2y + 5x + 4y. We have two terms in x and two terms in y. We can collect these together:
3x + 2y + 5x + 4y = (3 + 5)x + (2 + 4)y = 8x + 6y.
In the same way, we can collect together terms which are the same power of a variable. Consider the expression 3x2 + 2x + 5x2 + 4x. We have two terms in x2 and two terms in x. We can collect these together:
3x2 + 2x + 5x2 + 4x = (3 + 5)x2 + (2 + 4)x = 8x2 + 6x.
We cannot collect together different powers. We cannot combine x2 and x.
We might simplify this a bit more by noticing that in 8x2 + 6x, x is present in both terms. We could write
8x2 + 6x = x(8x + 6).
We might also notice that the terms in the bracket have a common factor of 2 and write
8x2 + 6x = x(8x + 6) = 2x(4x + 3)
Whether we want to do this will depend on the application. Sometimes we take an expression like 2x(4x + 3) and multiply out the bracket. This means that we multiply each term in the bracket by the term outside it:
2x(4x + 3) = 2x × 4x + 2x × 3 = 8x2 + 6x
as before.
Sometimes we have two brackets to multiply, like this:
(2x + 3)(x – 7)
We multiply each term in the first bracket by each term in the second bracket:
(2x + 3)(x – 7) = 2x2 – 14x + 3x – 21
Then we collect together like terms:
2x2 – 14x + 3x – 21 = 2x2 + (3 – 14)x – 21 = 2x2 – 11x – 21
Exercise: algebraic expressions
What would the expression 3x + 7 be when x = 3?
Simplify 5x2 + 7x – 2(x2 + x).
Check answer to algebraic expressions exercise.
Algebraic expressions have factors, just like numbers. For example, 4x2y has factors 1, 2, 4, x, 2x, 4x, x2, 2x2, 4x2, y, 2y, 4y, xy, 2xy, 4xy, x2y, 2x2y, 4x2y.
We can use these when we have one expression divided by another to simplify. For example, we can write numerator and denominator as multiples of their smallest factors:
4x2y
2 × 2 × x × x × y
2 × x 2x
–––– =
–––––––––––––
= ––––– = –––
2xy2
2 × x × y × y
y y
We have two 2s on the top and one on the bottom, so we can cancel out one of the 2s on top and the 2 on the bottom. We have two xs on the top and one on the bottom, so we can cancel out one of the xs on top and the x on the bottom. We have one y on the top and two ys on the bottom, so we can cancel out the y on top and one of the ys on the bottom.
We can also cancel out more complex parts of expressions. For example
(x2 – y)2 (x – 15)
(x2 – y) × (x2 – y) ×
(x – 15)
(x2 – y)
–––––––––––––
= –––––––––
–––––––––––––
= –––––––
(x2 – y) (x – 15)2
(x2 – y) × (x – 15) × (x – 15)
(x – 15)
We can cancel one (x2 – y) and one (x – 15) from top and bottom.
Exercise: factors and cancelling
Simplify the following:
25p3q2
––––––
35pq
(r – s)(2r + s)2
–––––––––––
(r + s)(2r + s)
Check answer to factors and cancelling exercise.
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Last updated: 16 April, 2008.