For 7(x − 6) = 7x − k to be an identity, k = □.

42. The left expands to 7x − 42. Both sides must match exactly, so k = 42.

Is (x + 7)^2 = x^2 + 49 an identity? Put in x = 1: the left side is □ and the right side is □. Fill in each blank

64, 50. Left: (1 + 7)^2 = 64; right: 1^2 + 49 = 50. One value where they differ is enough: it is not an identity. The identity is (x + 7)^2 ≡ x^2 + 14x + 49.

Which of these is not an identity? A. 5(x − 3) + 1 = 5x − 14 B. (x + 2)^2 = x^2 + 4x + 4 C. (x − 2)^2 = x^2 − 4 D. (x − 3)(x + 3) = x^2 − 9 Choose the right answer

C. (x − 2)^2 = x^2 − 4. (x − 2)^2 = (x − 2)(x − 2) = x^2 − 4x + 4, not x^2 − 4. Putting in x = 0 shows it too: 4 on the left, −4 on the right.

Polynomials & identities · Ages 13–14

Identities and equations

An identity: both sides simplify to exactly the same.

How to do it

Method

An identity: both sides simplify to exactly the same

2(x + 3) ≡ 2x + 6; 2x + 6 = 10 only when x = 2

Another way

Look for a value that fails

(x + 1)² and x² + 1 at x = 1: 4 ≠ 2

In full

An equation holds only for some values: 2x + 6 = 10 only when x = 2. An identity holds for every value of its letters, and is written with '≡': 2(x + 3) ≡ 2x + 6. To tell whether an equality is an identity, expand and simplify both sides: if they come out exactly the same, every coefficient matching, it is an identity; otherwise it is only an equation. Or test with numbers: one value that makes the sides differ shows it is not an identity, but a few values that agree prove nothing; only expanding makes sure. (x + 1)² = x² + 1 gives 1 and 1 at x = 0, but 4 and 2 at x = 1, so it is not an identity.

Worked examples

Have a go first, then open each one for the answer and the working.

Example 1
For 7(x − 6) = 7x − k to be an identity, k = .
Show the answer and working
Answer: 42
The left expands to 7x − 42. Both sides must match exactly, so k = 42.
Example 2
Is (x + 7)² = x² + 49 an identity? Put in x = 1: the left side is and the right side is .
Fill in each blank
Show the answer and working
Answer: 64, 50
Left: (1 + 7)² = 64; right: 1² + 49 = 50. One value where they differ is enough: it is not an identity. The identity is (x + 7)² ≡ x² + 14x + 49.
Example 3
Which of these is not an identity?
- A. 5(x − 3) + 1 = 5x − 14
- B. (x + 2)² = x² + 4x + 4
- C. (x − 2)² = x² − 4
- D. (x − 3)(x + 3) = x² − 9
Choose the right answer
Show the answer and working
Answer: C. (x − 2)² = x² − 4
(x − 2)² = (x − 2)(x − 2) = x² − 4x + 4, not x² − 4. Putting in x = 0 shows it too: 4 on the left, −4 on the right.

Practice questions to try

Try these 5; the answers are at the end. For a whole worksheet, see VIP worksheets below.

1.4(x + 9) = 4x + 36 is an identity.
True or false?
Answer:
2.Which of these is an identity?
- A. (x + 8)² = x² + 64
- B. 2(x + 8) = 2x + 16
- C. 2(x + 8) = 2x + 8
- D. 2x + 8 = 18
Choose the right answer
Answer:
3.(x − 7)(x + 7) = x² − 49 is an identity.
True or false?
Answer:
4.Which of these is not an identity?
- A. (x + 4)² = x² + 8x + 16
- B. 6(x − 5) + 9 = 6x − 21
- C. (x − 4)² = x² − 16
- D. (x − 5)(x + 5) = x² − 25
Choose the right answer
Answer:
5.Which kind is the equality 6(x + 7) = 2x + 42?
- A. An identity: true for every x
- B. An equation with one solution
- C. An equation with two solutions
- D. An equation with no solution
Choose the right answer
Answer:

All the answers

1. True
2. B. 2(x + 8) = 2x + 16
3. True
4. C. (x − 4)² = x² − 16
5. B. An equation with one solution

Free worksheet: Identities and equationsWorking out, multiple choice and word problems, with answers and working, to print or save as a PDF

VIP practice notes and papers

Aimed at tests and exams: questions on this topic, new every time, to print or save as a PDF.

Practice notes: The key point, worked examples step by step, then questions of the same kind to follow, hints fading
Practice paper: Set out in parts like a school paper, marked, with answers and full working

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