Nothing makes the heart of a reluctant mathematician sink like an algebra question.

Algebra is supposed to make life easier. By learning a formula or an equation, you can solve any similar type of problem—whatever the numbers involved. However, an awful lot of students find it difficult, because letters just don’t seem to ‘mean’ as much as numbers. Here, we’ll try to make life a bit easier…

Before we start, there are three words that you need to know if you want to learn algebra:

• A variable or unknown is a letter that stands for a number, eg a, b, c, x, y or z.
• A term is either a number, a letter or a combination of a letter and a number or two or more letters, eg 7, y, 7y or xy.
• An expression is a sum involving two or more terms, eg 7y + x or 5 + xy.
• An equation is an expression that equals something, eg 7y = 14 or xy = 4x.

In addition, there are a couple of other tips that make life easier:

• When you write the letter ‘x’, you should always use the ‘curly x’, ie a backwards and forwards ‘c’, to avoid confusion with the times operator.
• You should never use the times symbol in algebra. You should put letters and numbers side-by-side to show that they need to be multiplied together, eg 7y means ‘7 times y’ and yz means ‘y times z’.
• You should never put a 1 before a letter, eg 1a is just written as a.
• You should generally put your variables in alphabetical order in the final answer, eg ab not ba and 2a + 3b not 3b + 2a.
• The squared symbol only relates to the number or letter immediately before it, eg 3m² means 3 x m x m, NOT (3 x m) x (3 x m).

Great! Now we can go over the main kinds of algebra questions.

Gathering Terms

X’s and y’s look a bit meaningless, but that’s the point. They can stand for anything. The simplest form of question you’ll have to answer is one that involves gathering your terms. That just means counting how many variables or unknowns you have (like x and y). I like to think of them as pieces of fruit, so an expression like…

2x + 3y – x + y

…just means ‘take away one apple from two apples and add one banana to three more bananas’. That leaves you with one apple and four bananas, or x + 4y.

If it helps, you can arrange the expression with the first kind of variables (in alphabetical order) on the left and the second kind on the right like this:

2x – x + 3y + y

x + 4y

Just make sure you bring the operators with the variables that come after them so that you keep exactly the same operators, eg two plus signs and a minus sign in this case.

Here are a few practice questions:

1. 3x + 4y – 2x + y
2. 2m + 3n – m + 3n
3. p + 2q + 3p – 3q
4. 2a – 4b + a + 4b
5. x + y – 2x + 2y

Multiplying out Brackets

This is one of the commonest types of question. All you need to do is write down the same expression without the brackets. To take a simple example:

2(x + 3)

In this case, all you need to do is multiply everything inside the brackets by the number outside, which is 2, but what do we do about the ‘+’ sign? We could just multiply 2 by x, write down ‘+’ and then multiply 2 by 3:

2x + 6

However, that gets us into trouble if we have to subtract one expression in brackets from another (see below for explanation) – so it’s better to think of the ‘+’ sign as belonging to the 3. In other words, you multiply 2 by x and then 2 by +3. Once you’ve done that, you just convert the ‘+’ sign back to an operator. It gives exactly the same result, but it will work ALL the time rather than just with simple sums!

Here are a few practice questions:

1. 2(a + 5)
2. 3(y + 2)
3. 6(3 + b)
4. 3(a – 3)
5. 4(3 – p)

Solving for x

Another common type of question involves finding out what x stands for (or y or z or any other letter). The easiest way to look at this kind of equation is using fruit again. In the old days, scales in a grocery shop sometimes had a bowl on one side and a place to put weights on the other.

To weigh fruit, you just needed to make sure that the weights and the fruit balanced and then add up all the weights. The point is that every equation always has to balance – the very word ‘equation’ comes from ‘equal’ – so you have to make sure that anything you do to one side you also have to do to the other. Just remember the magic words: BOTH SIDES!

There are three main types of operations you need to do in the following order:

1. Multiplying out any brackets
2. Adding or subtracting from BOTH SIDES
3. Multiplying or dividing BOTH SIDES by the x coefficient (ie the number next to the variable)

Once you’ve multiplied out any brackets (see above), what you want to do is to simplify the equation by removing one expression at a time until you end up with something that says x = The Answer. It’s easier to start with adding and subtracting and then multiply or divide afterwards (followed by any square roots). To take the same example as before:

2(x + 3) = 8

Multiplying out the brackets gives us:

2x + 6 = 8

Subtracting 6 from BOTH SIDES gives us:

2x = 2

Dividing BOTH SIDES by 2 gives us the final answer:

x = 1

Simple!

Here are a few practice questions:

1. b + 5 = 9
2. 3y = 9
3. 6(4 + c) = 36
4. 3(a – 2) = 24
5. 4(3 – p) = -8

Number Triangles

Number triangles (see article) are a helpful way of rearranging the relationship between three terms involving multiplication or division. To take a simple example:

• speed = distance ÷ time
• time = distance ÷ speed
• distance = speed x time

That’s a lot to remember! However, if you use a number triangle, you can put distance (d) at the top and speed (s) and time (t) at the bottom. If you then put your finger over the one you need, you get the formula, eg if you put your finger over speed (s), you get distance (d) over time (t).

You can use number triangles to rearrange complicated equations when solving for x, eg 2 / (x – 2) = 7 becomes x – 2 = 2/7, which means x = 2 2/7.

Fractions

Some algebraic expressions look complicated when they’re really just fractions. As long as you know how to add, subtract, multiply and divide fractions, you can do the same with algebraic fractions, eg to solve (2x – 1) / 4 – (x + 1) / 5 = 2, you can just take the second fraction away from the first using cross-multiplication or a similar method.

Linear Inequalities

If the equation has no powers (eg b²) and contains a ‘less than’ or ‘more than’ operator (<, >, ≤ or ≥) instead of an equals sign (=), it’s called a ‘linear inequality’. Here, the rules are a little bit different.

1. If you have to multiply or divide both sides of a linear inequality by a negative number to get rid of the number next to the variable (the ‘x coefficient’), you need to reverse the inequality, eg if you divide both sides by -5, -5x < 25 becomes x > -5 (not x < -5).
2. If the variable ends up on the ‘wrong’ side of the inequality, you have to reverse the direction of the inequality when you flip it, eg 3 < x becomes x > 3 (not x < 3).

Multiplying Two Expressions in Brackets (‘FOIL’ Method)

When you have to multiply something in brackets by something else in brackets, you should use what’s called the ‘FOIL’ method. FOIL is an acronym that stands for:

First
Outside
Inside
Last

This is simply a good way to remember the order in which to multiply the terms, so we start with the first terms in each bracket, then move on to the outside terms in the whole expression, then the terms in the middle and finally the last terms in each bracket.

Just make sure that you use the same trick we saw earlier, combining the operators with the numbers and letters before multiplying them together. For example:

(a + 1)(a + 2)

First we multiply the first terms in each bracket:

a x a

…then the outside terms:

a x +2

…then the inside terms:

+1 x a

…and finally the last terms in each bracket:

+1 x +2

Put it all together and simplify:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

Here are a few practice questions:

1. (a + 1)(b + 2)
2. (a – 1)(a + 2)
3. (b + 1)(a – 2)
4. (p – 1)(q + 2)
5. (y + 1)(y – 3)

Factorising Quadratics (‘Product and Sum’ Method)

This is just the opposite of multiplying two expressions in brackets. Normally, factorisation involves finding the Highest Common Factor (or HCF) and putting that outside a set of brackets containing the rest of the terms, but some expressions can’t be solved that way, eg a² + 3a + 2 (from the previous example).

There is no combination of numbers and/or letters that goes evenly into a², 3a and 2, so we have to factorise using two sets of brackets. To do this, we use the ‘product and sum’ method.

This simply means that we need to find a pair of numbers whose product equals the last number and whose sum equals the multiple of a. In this case, it’s 1 and 2 as +1 x +2 = +2 and +1 + +2 = +3.

The first term in each bracket is just going to be a as a x a = a². Hence, factorising a² + 3a + 2 gives (a + 1)(a + 2). You can check it by using the FOIL method (see above) to multiply out the brackets:

(a + 1)(a + 2)

= a² + 2a + a + 2

=a² + 3a + 2

Note that some questions may not have a term in the middle, but that’s just because the two terms you get when you use the FOIL method cancel each other out, eg x² – 25 becomes (x – 5)(x + 5) because x² + 5x – 5x – 25 = x² + 0 – 25.

Here are a few practice questions:

1. y² + 9y + 20
2. y² + 10y + 9
3. p² + 5p – 24
4. p² + 8p + 16
5. z² – 121

Subtracting One Expression from Another*

Here’s the reason why we don’t just write down operators as we come across them. Here’s a simple expression we need to simplify:

20 – 4(x – 3) = 16

If we use the ‘wrong’ method, then we get the following answer:

20 – 4(x – 3) = 16

20 – 4x – 12 = 16

8 – 4x = 16

4x = -8

x = -2

Now, if we plug our answer for x back into the original equation, it doesn’t balance:

20 – 4(-2 – 3) = 16

20 – 4 x -5 = 16

20 – -20 = 16

40 = 16!!

That’s why we have to use the other method, treating the operator as a negative or positive sign to be added to the number before we multiply it by whatever’s outside the brackets:

20 – 4(x – 3) = 16

20 – 4x + 12 = 16

32 – 4x = 16

4x = 16

x = 4

That makes much more sense, as we can see:

20 – 4(4 – 3) = 16

20 – 4 x 1 = 16

20 – 4 = 16

16 = 16

Thank Goodness for that!

Here are a few practice questions:

1. 30 – 3(p – 1) = 0
2. 20 – 3(a – 3) = 5
3. 12 – 4(x – 2) = 4
4. 24 – 6(x – 3) = 6
5. 0 – 6(x – 2) = -12

 

 

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