Inverses

When you get an algebraic equation each letter stands for a number. In order to solve an equation you need to figure out the value of the missing letter.

One way to solve an equation is to use inverses. Basically this means that you undo the equation.

Here are a few inverses:

– n?

+ x an

For example: solve the equation 3b +7 = 13

First of all you subtract 7 from each side (the inverse of plus 7):

3b + 7 – 7 = 13 – 7

3b = 6

Now, you divide each side by 3 (the inverse of multiply 3):

3b/3 = 6/3

b = 6/3 = 2

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Brackets

If you get an equation with brackets and need to solve it remember to multiple out the brackets first.

For example, solves 2(h + 4) = 16

First of all, multiply out the brackets:

2h + 8 = 16

Now, subtract 8 from both sides:

2h = 8

Then finally, divide each side by 2:

h = 4

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Solving inequalities

When you solve an equality you don’t just get one answer but a whole range of answers.

For example the equation 4x + 2 = 10 is not the same as 4x + 2 ? 10.

For the first equation there is only one answer. For the inequality equation the answer could be one of a lot of answers.

For instance: solve 4x + 2 ? 10.

First of all, make x the subject as you would with a normal equation:

4x + 2 ? 10

4x ? 8

x ? 2

This means that the equation is satisfied when x is larger than 2.

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Double inequalities

You might be asked to solve a double inequality. This is where you get three parts to the equation.

Just remember that whatever you do to one part you must do to all the others too.

For example: solve -8 ? 3x + 2 ? 11

Minus 2 to from each part:

-6 ? 3x ? 9

Now, divide all parts by 3:

-2 ? x ? 3

So, the inequality is satisfied when x is between -2 and 3, including 3.

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Unknowns on both sides

You might also get a question in which you have unknowns on both sides.

For example: 5c + 6 = c + 8. Find the value of c.

First of all, minus 6 from both sides (the smaller number by the larger):

5c = c + 2

Now, subtract c from both sides:

4c = 2

Now, divide both sides by 4 and you’ll get your answer:

c = 2/4 = 1/2

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Trial and improvement

If you’re not given values for an algebraic equation you might need to try and estimate what they are. This is known as trial and improvement.

For example, answer the following equation to one decimal place: x² + 6x = 25.

So, you need to figure out what number could possibly replace x.

Your first try can be a guess. So, try 4:

4² + (6 x 4) = 16 + 24 = 40

So, this answer is too big. This means you should try with a smaller number, for example x = 2:

2² + (6 x 2) = 4 + 12 = 16

So answer is too small. This means you need to try a number in-between like x = 3:

3² + (6 x 3) = 9 + 18 = 27

This answer is a little bit too large. So the third try could be: x = 2.5:

2.5² + (6 x 2.5) = 6.25 + 15 = 21.25

This is a bit too small. Fourth try: x = 2.8:

2.8² + (6 x 2.8) = 7.84 + 16.8 = 24.64

This is the closest yet. Fifth try: x = 2.85:

2.85² + (6 x 2.85) = 8.1225 + 17.1 = 25.2225

This means x is greater than 2.8 but smaller than 2.85 so, to one decimal place, the answer is 2.8.

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Equations and fractions (Higher tier)

For the higher tier you’ll be expected to solve equations which contain fractions.

There are two main methods you can use:

• – multiply out the fraction first
• – multiply out the fraction at the end

For example, solve 6 – x/4 = 3

If you multiply out the fraction first then you multiply everything by 4:

24 – x = 12

Then add the x:

24 = 12 + x

Lastly subtract the 12:

12 = x

Or, you can multiply out the fraction at the end:

6 – x/4 = 3

6 = 3 + x/4

3 = x/4

12 = x

Always remember to do the same to both sides.