# Understand and Use Algebraic Terms 2

In this worksheet, students will learn to recognise identities and use their unique properties to fine the value of terms in expressions.

Key stage:  KS 4

GCSE Subjects:   Maths

GCSE Boards:   Pearson Edexcel, OCR, Eduqas, AQA

Curriculum topic:   Algebra

Curriculum subtopic:   Notation, Vocabulary and Manipulation, Algebraic Formulae

Difficulty level:

### QUESTION 1 of 10

​What is an identity?​

Identities are a special type of equation.

They are always true, no matter what values are substituted in.

As this is such a special case, in identities we replace an equal sign with an identity sign which looks:

Let's see this in action now.

e.g. (a + b)² = a² + 2ab + b²

This is an identity, because whatever values we substitute for a and b, we always get a statement that is true.

Let's test this with a = 2 and b = 3...

On the left-hand side (LHS): (2 + 3)² = 5² = 25

On the right-hand side (RHS): 2² + 2×2×3 + 3² = 4 + 12 + 9 = 25

25 = 25

So (a + b)² = a² + 2ab + b² is definitely an identity.

We could also expand and simplify.

This would give us the same expression on the left as on the right of the identity sign.

How to use an identity?​

At GCSE level, you might be asked to use the properties of identities to find missing coefficients.

e.g. 5ax + 8 + 3(x - d) 18x + 14

Find the values of a and d.

The highlighted sign is an identity sign, so this means that the LHS and the RHS will simplify to exactly the same expressions.

Let's expand and simplify the LHS first:

5ax + 8 + 3(x - d) = 5ax + 8 + 3x - 3d

Now let's remove x from the RHS, by factorising the left-hand terms with x:

5ax + 3x + 8 - 3d = x(5a + 3) + 8 - 3d

Therefore, 5a + 3 = 18 and 8 - 3d = 14.

So a = 3, d = -2

In this activity, we will learn to recognise identities and use their unique properties to fine the value of terms in expressions.

Identify if each option below represents an identity or an equation.

 Equation Identity 8x + 12 = 5x + 6 8x + 12 = 4(2x + 3) 8x + 12 = 8x - 2

Is the mathematical object below an identity?

3(5x + 2) = 10x + 5x + 6

Yes

No

Using the identity below, find the value of a and b.

2(2x + 3) + 3(5x + 2) ≡ ax + b

Yes

No

Consider this expression:

x(x + 4) ≡ x<sup>2</sup> + 4x

For how many values of x is x(x + 4) equal to x<sup>2</sup> + 4x?

0

1

2

All

Using the identity below, find the value of and b.

(x + 3)(x + 5) ≡ x<sup>2</sup> + ax + b

 Value a = b =

True or false?

An identity is true for all values of the variable.

True

False

Using the identity below, find the value of a and b.

x(x - 5) + a  ≡ x<sup>2</sup> + bx + 28

 Value a = b =

Show that 5 + 3(x - 2) = 3x -1 is an identity.

Fill in the blanks below to show every step of the solution.

 Value a = b =

Tick the equations below which are not identities.

(x + 2)2 = x2 + 4

5x(x + 7) = x + 2

24s - 6t = 8(3s - t) + 2t

a(b + c) = ab + c

The identity a(x + 2) - x ≡ 3x + 8 is true for all values of x.

Find the value of a.

• Question 1

Identify if each option below represents an identity or an equation.

 Equation Identity 8x + 12 = 5x + 6 8x + 12 = 4(2x + 3) 8x + 12 = 8x - 2
EDDIE SAYS
The first and the third options are equations. This is because they can be rearranged to find the value of x which makes them true and they have an equals sign but are not necessarily identical on each side. In the first option, the value of x = -2. Go on...check it for yourself to be sure. The other equation does not have a solution. The second option is an identity. If you expand the brackets on the right hand side, you get 8x + 12 which is exactly the same as the value on the left-hand side. For an equation to be an identity, both sides of the equals sign must be exactly equal.
• Question 2

Is the mathematical object below an identity?

3(5x + 2) = 10x + 5x + 6

Yes
EDDIE SAYS
Let's start by expanding and simplifying the left-hand side of this equation: 3(5x + 2) = 15x + 6 When we expand and simplify both sides of this equation, the expressions are the same. This means that we are dealing with an identity. Did you spot that?
• Question 3

Using the identity below, find the value of a and b.

2(2x + 3) + 3(5x + 2) ≡ ax + b

EDDIE SAYS
Do not worry! To answer this question you just need to equate the coefficients. Simplify the right-hand side first. Did you get: 4x + 6 + 15x + 6? Remember that the RHS needs to be exactly the same as the LHS, so: 4x + 6 + 15x + 6 ≡ ax + b We can factorise 4x, 15x and ax by x to remove this term. So a = 19 What is left gives us b (6 + 6), so b = 12 Don't worry if you didn't get this one first time, there is another practise question coming up!
• Question 4

Consider this expression:

x(x + 4) ≡ x<sup>2</sup> + 4x

For how many values of x is x(x + 4) equal to x<sup>2</sup> + 4x?

All
EDDIE SAYS
This was a bit of a trick question. Did you notice this ≡ in between the expressions? This is an identity sign, which means that the statement is true for ALL values of x. When we expand the left-hand side, we get the same expression as the right-hand side, so they are identical. Check this out now if you want to be sure.
• Question 5

Using the identity below, find the value of and b.

(x + 3)(x + 5) ≡ x<sup>2</sup> + ax + b

 Value a = b =
EDDIE SAYS
This is an identity so we know both sides of the sign are identical. We need to simplify the expressions and then compare coefficients. Expand and simplify the expression on the LHS first: (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15 Now check the coefficients and compare the RHS with the LHS: x² + 8x + 15 ≡ x² + ax + b The two x²'s cancel each other out. Can you see now that a = 8 and b = 15?
• Question 6

True or false?

An identity is true for all values of the variable.

True
EDDIE SAYS
Identities are mathematical statements which hold true for ALL values of the variable. e.g. 2(k+2) ≡ 2k + 4 is an identity, because if we substitute any number for k, we will always get a statement which is true. Give this a try yourself with different values of k to check that it can never be uneven!
• Question 7

Using the identity below, find the value of a and b.

x(x - 5) + a  ≡ x<sup>2</sup> + bx + 28

 Value a = b =
EDDIE SAYS
We need to simplify the expressions and then compare coefficients again. Expand and simplify the expression on the LHS first: x(x - 5) + a = x² - 5x + a Now check the coefficients and compare the RHS with the LHS: x² - 5x + a ≡ x2 + bx + 28 The two x²'s cancel each other out. Can you see now that a = 28 and b = -5?
• Question 8

Show that 5 + 3(x - 2) = 3x -1 is an identity.

Fill in the blanks below to show every step of the solution.

EDDIE SAYS
To prove that an equation IS an identity, you need to show that the expression on the right-hand side of the equals sign is exactly the same as the expression on the left-hand side. Let's follow our working through: 5 + 3(x - 2) = 3x -1 5 + 3x - 6 = 3x -1 3x - 1 ≡ 3x -1 LHS = RHS so this is a proven identity. Does that make sense?
• Question 9

Tick the equations below which are not identities.

5x(x + 7) = x + 2
a(b + c) = ab + c
EDDIE SAYS
Try expanding and simplifying both sides of each of the equations above. If you get the same answer on both sides, it is an identity! If not, then it's not an identity. Let's work out one of each to check our method. (x + 2)2 = x2 + 4 Work out the bracket to reach: x2 + 4 = x2 + 4 so this is an identity. Let's work out: 5x(x + 7) = x + 2 > 5x2 + 35x = x + 2 Both of these equations cannot be true so this is not an identity. Great work completing this activity!
• Question 10

The identity a(x + 2) - x ≡ 3x + 8 is true for all values of x.

Find the value of a.

4
EDDIE SAYS
You need to simplify the expression of the LHS as much as possible and then compare coefficients. Firstly, expand the brackets to reach: ax + 2a - x ≡ 3x + 8 Then gather together all the terms with x and factorise: ax - x + 2a ≡ 3x + 8 x(a - 1) + 2a ≡ 3x + 8 Now compare coefficients. The coefficient of x must be 3, so a - 1 = 3. This tells us that a = 4. We could check the other part of the equation too so we are 100% sure! 2a should equal 8. This also gives us a = 4, so we know for sure that we are correct.
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