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Standard form of a linear equation · 4 marks

Write each of the following equations in standard form, ax + by + c = 0, and indicate the values of a, b and c in each case:

  1. (i) 5x + 2y = 2.7
  2. (ii) x2 − 4 = 32y
  3. (iii) 4 = 1.5x − 1.3y
  4. (iv) 3y = 5
Answer: (i) 5x + 2y − 2.7 = 0: a = 5, b = 2, c = −2.7 (ii) 12x − 32y − 4 = 0: a = 12, b = −32, c = −4 (iii) 1.5x − 1.3y − 4 = 0: a = 1.5, b = −1.3, c = −4 (iv) 0×x + 3y − 5 = 0: a = 0, b = 3, c = −5

Step-by-step solution

Idea: Standard form has every term on the left and 0 on the right. Move each term across the equals sign, changing its sign, then read off the number in front of x (a), in front of y (b) and the constant (c). A missing variable has coefficient 0.

(i) 5x + 2y = 2.7

  1. Move 2.7 to the left side; it becomes −2.7: 5x + 2y − 2.7 = 0.½ mark
  2. Compare with ax + by + c = 0: a = 5, b = 2, c = −2.7. (Multiplying the whole equation by −1 gives −5x − 2y + 2.7 = 0, with a = −5, b = −2, c = 2.7, which is also correct.)½ mark
5x + 2y − 2.7 = 0; a = 5, b = 2, c = −2.7

(ii) x2 − 4 = 32y

  1. Move 32y to the left side; it becomes −32y: x2 − 32y − 4 = 0.½ mark
  2. x2 means 12x, so a = 12, b = −32, c = −4.½ mark
x2 − 32y − 4 = 0; a = 12, b = −32, c = −4

(iii) 4 = 1.5x − 1.3y

  1. Here the variables are on the right. Move 4 to the right instead, then swap sides: 1.5x − 1.3y − 4 = 0.½ mark
  2. a = 1.5, b = −1.3, c = −4. (Equally correct: −1.5x + 1.3y + 4 = 0 with a = −1.5, b = 1.3, c = 4.)½ mark
1.5x − 1.3y − 4 = 0; a = 1.5, b = −1.3, c = −4

(iv) 3y = 5

  1. There is no x-term, so write it with coefficient 0: 0×x + 3y − 5 = 0.½ mark
  2. a = 0, b = 3, c = −5.½ mark
0×x + 3y − 5 = 0; a = 0, b = 3, c = −5
(i) 5x + 2y − 2.7 = 0; a = 5, b = 2, c = −2.7 (ii) (1/2)x − (3/2)y − 4 = 0; a = 1/2, b = −3/2, c = −4 (iii) 1.5x − 1.3y − 4 = 0; a = 1.5, b = −1.3, c = −4 (iv) 0·x + 3y − 5 = 0; a = 0, b = 3, c = −5

Check: Put a solution back in. In (i), x = 0.5, y = 0.1 gives 2.5 + 0.2 = 2.7 and also 2.5 + 0.2 − 2.7 = 0 ✓. In (iv), y = 53 gives 3 × 53 − 5 = 0 ✓.

Answer to write in the exam

(i)

5x + 2y = 2.7

⇒ 5x + 2y − 2.7 = 0

∴ a = 5, b = 2, c = −2.7

(ii)

x2 − 4 = 32y

⇒ x2 − 32y − 4 = 0

∴ a = 12, b = −32, c = −4

(iii)

4 = 1.5x − 1.3y

⇒ 1.5x − 1.3y − 4 = 0

∴ a = 1.5, b = −1.3, c = −4

(iv)

3y = 5

⇒ 0×x + 3y − 5 = 0

∴ a = 0, b = 3, c = −5

Common mistakes that cost marks

  • Forgetting to change the sign when a term crosses the equals sign, e.g. writing c = +2.7 in (i).
  • In (ii), giving b = 32 instead of −32: the y-term moved from the right side, so its sign changed.
  • In (iv), saying the equation has no value of a. The x-term is missing, so a = 0.

How this can come in the exam

MCQ (1 mark)

When 2y = 7 − 3x is written as ax + by + c = 0 with a = 3, the value of c is

  1. 7
  2. −7
  3. 2
  4. −2
Show answer

(B) −7
3x + 2y − 7 = 0, so c = −7.

Short answer (2 marks)

Write y3 = 2x + 1 in the form ax + by + c = 0 with integer coefficients and state a, b, c.

Show answerMultiply by 3: y = 6x + 3, so 6x − y + 3 = 0 (1 mark). a = 6, b = −1, c = 3 (1 mark).

Try one yourself

Write 7 = 2x − 0.5y in standard form and give a, b, c.

Show answer

2x − 0.5y − 7 = 0: a = 2, b = −0.5, c = −7.

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