Leaving Cert Mathematics Ordinary Paper 1 free practice questions with answers

Question 1 (30 marks) [Algebra and equations]

(a) Solve the following equation, where x ∈ ℝ:

(a) Answer: x = 13

• Multiply every term by the lowest common multiple of 3 and 4, which is 12.
• 12
• Expand and collect: 7x − 7 = 84.
• x = (84 − (−7)) / 7 = 13.

Linear equations: remove brackets or clear fractions first, collect x terms on one side and numbers on the other, then divide; check by substituting back.

(b) Solve the following equation in x: x² + 8x − 2 = 0 Give each answer correct to 2 decimal places.

(b) Answer: x = −8.24 or x = 0.24

• Compare with ax² + bx + c = 0: a = 1, b = 8, c = −2.
• Write the formula x = (−b ± √(b² − 4ac)) / 2a and substitute: x = (−8 ± √((8)² − 4(1)(−2))) / 2.
• Work out the part under the root: (8)² − 4(1)(−2) = 72, and √72 = 8.4853.
• x = (−8 + 8.4853) / 2 = 0.24 and x = (−8 − 8.4853) / 2 = −8.24.

Quadratic formula: copy a, b and c with their signs, find b² − 4ac first, and keep 4 decimals until the final rounding.

(c) Express as a single fraction in its simplest form:

(c) Answer:

• The common denominator of 5 and 6 is 30.
• Write each fraction over 30: 6
• Combine into one fraction and expand the brackets:
• Check for a common factor in the numerator and denominator: there is none, so the answer is in simplest form.

Algebraic fractions: find a common denominator, multiply each numerator by the missing factor (use brackets), then combine and simplify. x − 4

3 + x + 3

4 = 7 x − 4

3 + 12x + 3

4 = 84, so 4(x − 4) + 3(x + 3) = 84.

5x − 4

5

+ 3x − 5

6

45x − 49

30

5x − 4

30 + 53x − 5

30 .

30x − 24 + 15x − 25

30

= 45x − 49

30

.