Equations and formulae

1. If n is an even integer, which expression represents the product of n and the next consecutive even integer?

Answer C: n(n + 2)

Because n is already defined as even, the next even integer is n + 2. Their product is therefore n(n + 2). The expression n(n + 1) uses consecutive integers, not consecutive even integers.

2. A rectangle has width x cm, length x + 4 cm and perimeter 36 cm. Which equation models the perimeter?

Answer B: 2x + 2(x + 4) = 36

The two widths contribute 2x and the two lengths contribute 2(x + 4), so 2x + 2(x + 4) = 36. The product x(x + 4) would represent area rather than perimeter.

3. Solve 7x - 5 = 30.

Answer B: x = 5

Adding 5 gives 7x = 35. Dividing both sides by 7 gives x = 5, and checking produces 7(5) - 5 = 30.

4. Solve 5 - 2x = 3(x + 7).

Answer A: x = -16/5

Expansion gives 5 - 2x = 3x + 21. Therefore -5x = 16 and x = -16/5. Substitution makes both sides equal to 57/5.

5. Solve x/3 - (x - 2)/4 = 5.

Answer C: x = 54

Multiplying through by 12 gives 4x - 3(x - 2) = 60. Hence 4x - 3x + 6 = 60, so x = 54. In the original equation, 18 - 13 = 5.

6. Solve x/(2x + 1) = 4.

Answer B: x = -4/7

Multiplication gives x = 8x + 4, so -7x = 4 and x = -4/7. This is not the excluded value -1/2, and substitution gives a quotient of 4.

7. Solve x/(x + 2) = 3/(x - 6).

Answer A: x = (9 ± √105)/2

Cross-multiplication gives x(x - 6) = 3(x + 2), so x^2 - 9x - 6 = 0. The quadratic formula gives x = (9 ± √105)/2. Neither root equals -2 or 6, so both are valid.

8. Solve 2/(x + 2) + 3/(2x - 1) = 1.

Answer A: x = 3 or x = -1

Clearing denominators leads to x^2 - 2x - 3 = 0, which factorises as (x - 3)(x + 1) = 0. Thus x = 3 or -1; neither equals the excluded values -2 or 1/2.

9. Before solving 4/(x - 5) = x, which restriction must be recorded?

Answer C: x ≠ 5

The denominator x - 5 is zero when x = 5, so x ≠ 5. The restriction comes from the original fraction and does not require x to be positive or greater than 5.

10. Solve 2x + 3y = 17 and 3x - 2y = 6.

Answer B: x = 4, y = 3

Doubling the first equation and tripling the second gives 4x + 6y = 34 and 9x - 6y = 18. Adding yields x = 4, then substitution gives y = 3.

11. Solve y = 2x - 1 and 3x + y = 19.

Answer A: x = 4, y = 7

Substitution gives 3x + 2x - 1 = 19, so 5x = 20 and x = 4. Then y = 2(4) - 1 = 7, and 3(4) + 7 = 19.

12. Three adult tickets and two child tickets cost $34. Two adult tickets and five child tickets cost $41. What are the ticket prices?

Answer B: Adult $8, child $5

Eliminating c gives 11a = 88, so the adult price is $8. Substitution gives 24 + 2c = 34, hence the child price is $5. Both original totals are reproduced.

13. Solve x^2 - 9x + 20 = 0.

Answer C: x = 4 or 5

The quadratic factorises as (x - 4)(x - 5) = 0. Setting either factor equal to zero gives x = 4 or x = 5.

14. Solve 2x^2 - 5x - 3 = 0.

Answer B: x = 3 or -1/2

The factorisation is (2x + 1)(x - 3) = 0. Therefore 2x + 1 = 0 gives x = -1/2, while x - 3 = 0 gives x = 3.

15. Use completing the square to solve x^2 - 6x - 7 = 0.

Answer B: x = 7 or -1

The equation becomes (x - 3)^2 = 16. Thus x - 3 = 4 or -4, giving x = 7 or x = -1. Both values satisfy the original quadratic.

16. Write 2x^2 + 8x - 3 in completed-square form.

Answer A: 2(x + 2)^2 - 11

2x^2 + 8x - 3 = 2(x^2 + 4x) - 3 = 2[(x + 2)^2 - 4] - 3 = 2(x + 2)^2 - 11. Expanding verifies the original expression.

17. Solve 3x^2 + 2x - 4 = 0 exactly.

Answer A: x = (-1 ± √13)/3

The formula gives [-2 ± √(4 + 48)]/6 = (-2 ± √52)/6. Since √52 = 2√13, simplifying gives x = (-1 ± √13)/3.

18. Solve x^2 + 4x + 1 = 0 exactly.

Answer A: x = -2 ± √3

The quadratic formula gives x = [-4 ± √12]/2. Because √12 = 2√3, division by 2 gives the exact roots x = -2 ± √3.

19. A rectangle has width x cm, length x + 3 cm and area 40 cm^2. What is its width?

Answer B: 5 cm

The equation factorises as (x + 8)(x - 5) = 0, giving -8 and 5. A width cannot be negative, so the valid width is 5 cm; the length is 8 cm.

20. Make t the subject of v = u + at, where a ≠ 0.

Answer A: t = (v - u)/a

Subtracting u gives v - u = at. Dividing both sides by a gives t = (v - u)/a, provided a is non-zero.

21. Make x the subject of y = (3x - 5)/2.

Answer B: x = (2y + 5)/3

Multiplying by 2 gives 2y = 3x - 5. Adding 5 gives 2y + 5 = 3x, so x = (2y + 5)/3.

22. Make x the subject of P = (ax + b)/(cx + d), assuming Pc - a ≠ 0 and the original denominator is non-zero.

Answer B: x = (b - Pd)/(Pc - a)

From P(cx + d) = ax + b, obtain Pcx - ax = b - Pd. Factoring gives x(Pc - a) = b - Pd, so x = (b - Pd)/(Pc - a).

23. Make r the subject of A = πr^2, where r is a radius.

Answer B: r = √(A/π)

Dividing gives r^2 = A/π. Algebraically there are two square-root branches, but a radius is non-negative, so r = √(A/π).

24. Make L the subject of T = 2π√(L/g), where g > 0 and T ≥ 0.

Answer A: L = gT^2/(4π^2)

T/(2π) = √(L/g), so T^2/(4π^2) = L/g after squaring. Multiplying by g gives L = gT^2/(4π^2).