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Standard form of a linear equation · 3 marks

Complete the following table after expressing the given linear equations in standard form.
Linear equation in two variablesStandard FormCoefficient of xCoefficient of yConstant term
y − 15 = √2x
3y − 2x = 0
5x = 3y
x = 8
3y = 1

Answer: √2x − y + 15 = 0 (√2, −1, 15); −2x + 3y + 0 = 0 (−2, 3, 0); 5x − 3y + 0 = 0 (5, −3, 0); x + 0×y − 8 = 0 (1, 0, −8); 0×x + 3y − 1 = 0 (0, 3, −1). Numbers in brackets: coefficient of x, coefficient of y, constant term.

Step-by-step solution

Idea: Bring every term to the left so the right side is 0, then read the coefficients. A variable that does not appear has coefficient 0, and if there is no number term the constant is 0.

  1. y − 15 = √2x: move √2x to the left: −√2x + y − 15 = 0. Multiply by −1 to make the x-coefficient positive: √2x − y + 15 = 0. Coefficients: √2, −1; constant 15.1 mark
  2. 3y − 2x = 0: write the x-term first: −2x + 3y + 0 = 0. Coefficients: −2, 3; constant 0.½ mark
  3. 5x = 3y: move 3y to the left: 5x − 3y + 0 = 0. Coefficients: 5, −3; constant 0.½ mark
  4. x = 8: no y, so x + 0×y − 8 = 0. Coefficients: 1, 0; constant −8.½ mark
  5. 3y = 1: no x, so 0×x + 3y − 1 = 0. Coefficients: 0, 3; constant −1. The completed table:
    Linear equation in two variablesStandard FormCoefficient of xCoefficient of yConstant term
    y − 15 = √2x√2x − y + 15 = 0√2−115
    3y − 2x = 0−2x + 3y + 0 = 0−230
    5x = 3y5x − 3y + 0 = 05−30
    x = 8x + 0×y − 8 = 010−8
    3y = 10×x + 3y − 1 = 003−1
    ½ mark
y − 15 = √2x → √2x − y + 15 = 0 (√2, −1, 15); 3y − 2x = 0 → −2x + 3y + 0 = 0 (−2, 3, 0); 5x = 3y → 5x − 3y + 0 = 0 (5, −3, 0); x = 8 → x + 0·y − 8 = 0 (1, 0, −8); 3y = 1 → 0·x + 3y − 1 = 0 (0, 3, −1).

Check: Multiplying a whole equation by −1 gives an equally correct answer with every sign changed. For example 2x − 3y = 0 (2, −3, 0) is also right for 3y − 2x = 0. The table must just be consistent along each row.

Answer to write in the exam

y − 15 = √2x ⇒ √2x − y + 15 = 0: a = √2, b = −1, c = 15

3y − 2x = 0 ⇒ −2x + 3y + 0 = 0: a = −2, b = 3, c = 0

5x = 3y ⇒ 5x − 3y + 0 = 0: a = 5, b = −3, c = 0

x = 8 ⇒ x + 0×y − 8 = 0: a = 1, b = 0, c = −8

3y = 1 ⇒ 0×x + 3y − 1 = 0: a = 0, b = 3, c = −1

Common mistakes that cost marks

  • Reading the coefficient of y in y − 15 = √2x as +1 after moving terms, when the form chosen is √2x − y + 15 = 0. Signs must match the form you wrote.
  • Leaving the constant column blank for 3y − 2x = 0 and 5x = 3y. There is no number term, so the constant is 0.
  • Writing coefficient 8 for x = 8. The coefficient of x is 1; −8 is the constant.

How this can come in the exam

MCQ (1 mark)

In standard form, the equation x = −34 has

  1. a = 0, b = 1, c = 34
  2. a = 1, b = 0, c = 34
  3. a = 1, b = 0, c = −34
  4. a = 34, b = 0, c = 1
Show answer

(B) a = 1, b = 0, c = 34
x + 34 = 0, i.e. 1×x + 0×y + 34 = 0.

Assertion–Reason (1 mark)

Assertion (A): y = 4 is a linear equation in two variables.
Reason (R): It can be written as 0×x + 1×y − 4 = 0, and a and b are not both zero.

  1. Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
  2. Both Assertion (A) and Reason (R) are true, but R is not the correct explanation of A.
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Show answer

(A) Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both true, and R is exactly why A holds.

Try one yourself

Write 2y = √3x − 1 in standard form and give the coefficients and constant.

Show answer

√3x − 2y − 1 = 0: coefficient of x = √3, of y = −2, constant = −1.

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