Identities and Algebraic Arguments: Edexcel GCSE Foundation maths revision notes

Equations and identities

An equation is true for only some values of the letter. For example, 2x+5=17 is true only when x=6. An inequality, such as x-2>5, uses <, \le, > or \ge in place of =, so it is not an identity either.

An identity is true for every value of the letter. For example, 2(x+4)=2x+8 is true whatever number x is, because expanding the bracket on the left gives exactly the right side.

In the exam, an identity is written with an ordinary = sign, just like an equation. You tell them apart by checking whether the two sides are the same for every value of the letter.

Trying one number is not enough to show an identity. 3x+2=2x+3 is true when x=1, since both sides are 5. But when x=2 the left side is 8 and the right side is 7, so it is an equation, not an identity.

In the exam you may be asked to pick out the identity from a list of statements, for 1 mark, or to write down an example that shows a statement is wrong, for 1 or 2 marks. You also expand and simplify expressions, sometimes to show that you get a given answer. The hardest questions give you an expression, or ask you to write one for a shape, and then ask you to show that it is always even, or always a multiple of a number such as 4, with a reason, for up to 4 marks.

Worked example 1: Equation or identity

Here are two statements. (a) 3(x+4)=3x+12 (b) 3x+12=27. Which one is an identity? For the other one, find the value of x that makes it true.

(a) Expand the left side: 3\times x=3x and 3\times4=12, so 3(x+4) is 3x+12
Both sides are exactly the same for every value of x, so (a) is an identity.
(b) Subtract 12 from both sides: 3x=15
Your turn. Solve 3x=15

Your answer

x=15\div3=5
Answer: (a) is the identity. (b) is an equation, true only when x=5

Showing that two expressions are the same

A question may say "Show that" one expression is the same as another. You have to get from the first expression to the second, one clear line at a time.

Start with the side that has brackets. Expand every bracket.
Collect like terms: the letter terms together and the number terms together.
Stop when you reach exactly the expression on the other side, and say so.

Don't try to solve it like an equation, and don't put numbers in for the letter. There is no value of x to find: you are showing the two sides match for every value.

Worked example 2: Show that two expressions are the same

Show that 4(2x-1)-2(x-3)=6x+2 is an identity.

Start with the left side. Expand the first bracket: 4(2x-1)=8x-4
Your turn. The second bracket is multiplied by -2. Expand -2(x-3)

Your answer

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

-2(x-3)=-2x+6
So the left side is 8x-4-2x+6
Collect like terms: 8x-2x=6x and -4+6=2
Answer: the left side simplifies to 6x+2, which is the right side, so 4(2x-1)-2(x-3)=6x+2 is an identity.

Showing that something is always a multiple

Some questions give an expression such as 10n+15, where n is a whole number, and ask whether it is always a multiple of a number, such as 5. Trying values is a good start: when n=1 it is 25, and when n=2 it is 35. Both are multiples of 5, but that is only two values of n.

To show it is always a multiple of 5, take out 5 as a factor: 10n+15=5(2n+3). 2n+3 is a whole number, so 5(2n+3) is 5 times a whole number, which is always a multiple of 5.

You can also say it in words: 10n and 15 are both multiples of 5, so when you add them the answer is a multiple of 5 too.

If one value of n gives a number that is not a multiple, the answer is no. That one value is enough.

Worked example 3: Is it always a multiple?

n is a whole number. Ayesha says, "10n+6 is always a multiple of 8." Is Ayesha correct? Give a reason for your answer.

Try a value first. When n=1, 10n+6=10+6=16, and 16=8\times2
That value works, but one value is not enough. Try n=2 as well.
Your turn. Work out 10n+6 when n=2

Your answer

10\times2+6=20+6=26
26\div8=3 remainder 2, so 26 is not a multiple of 8.
Answer: Ayesha is not correct, because when n=2, 10n+6=26, which is not a multiple of 8.

Showing that a statement is wrong

Some questions give you a statement, such as "All multiples of 5 are odd", and ask you to write down an example to show that it is wrong.

One example that does not work is enough. 10 is a multiple of 5, but 10 is even, so the statement is wrong.

Make sure your example fits the statement exactly. If it is about multiples of 5, use a multiple of 5. Then write the calculation and say why it breaks the rule.

Examples can show that a statement is wrong, but they can't show that it is always true. There are too many numbers to try them all. To show something is always true, you need algebra.

Worked example 4: An example that shows a statement is wrong

Owen says, "Every number that ends in 6 is a multiple of 3." Write down an example to show that Owen is wrong.

Try numbers that end in 6, starting with the smallest. 6=3\times2, so 6 is a multiple of 3 and doesn't help.
The next number that ends in 6 is 16. Is it a multiple of 3?
Your turn. What is the remainder when 16 is divided by 3?

Your answer

16\div3=5 remainder 1, so 16 is not a multiple of 3
Answer: 16 ends in 6, but 16 is not a multiple of 3, so Owen is wrong.

Odd and even numbers in algebra

Let n stand for any integer (a whole number). Then 2n is always even, because it is 2 times a whole number. Adding 1 to an even number gives an odd number, so 2n+1 is always odd.

The dots show why, with n=5. Then 2n=10 dots make two equal rows, so every dot has a partner. And 2n+1=11 dots leave one dot over.

The integer after n is n+1.

These facts about whole numbers give you the reasons you need: - odd plus odd is even, and even plus even is even - odd plus even is odd - an even number times any whole number is even - odd times odd is odd

To show that an expression is always even, write it as 2\times(a whole number), for example 6n+4=2(3n+2).

Worked example 5: Showing a result is always odd

a and b are both odd numbers. Mia says, "a+b+1 is always odd." (a) Choose two odd numbers and check Mia's statement. (b) Explain why Mia is right.

(a) Choose two odd numbers, for example a=7 and b=11
Your turn. Work out a+b+1 when a=7 and b=11

Your answer

7+11+1=19, which is odd
(b) One example is not enough, so use the facts about odd and even numbers. a and b are both odd, and odd plus odd is even, so a+b is always even.
Adding 1 to an even number gives an odd number, so a+b+1 is always odd.
Answer: (a) 7+11+1=19, which is odd. (b) a+b is even, because odd plus odd is even, and an even number plus 1 is odd, so a+b+1 is always odd.

Worked example 6: A perimeter that is always a multiple of 4

A rectangle is (3n+5) cm long and (n+1) cm wide, where n is a positive integer. Show that the perimeter of the rectangle, in cm, is always a multiple of 4.

The perimeter is two lengths and two widths: 2(3n+5)+2(n+1)
Your turn. Expand and simplify 2(3n+5)+2(n+1)

Your answer

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

6n+10+2n+2=8n+12
Take out the common factor 4: 8n+12=4(2n+3)
2n+3 is a whole number, so 4(2n+3) is 4 times a whole number.
Answer: the perimeter is 4(2n+3) cm, so it is always a multiple of 4.

Common mistakes

Trying a few numbers and saying it is always true. Numbers can show a statement is wrong, but only algebra shows it works for every value.
Multiplying only the first term in a bracket. 5(2x-3) is 10x-15: the 5 multiplies the -3 as well.
Solving a "show that" like an equation. There is no value to find: expand and simplify until you reach the other expression.
Giving an example that doesn't fit the statement. If the statement is about even numbers, your example must use even numbers, and the working must be right.
Choosing an inequality as the identity. A statement with <, >, \le or \ge is never an identity: look for the one that is true for every value.
Stopping before the last line. Write the result as 2\times(a whole number) or 4\times(a whole number), then say what that shows.