Year 10 Paper 2: Algebra and Equations
Covers algebraic manipulation, surds and indices, simultaneous equations and inequalities, functions and their graphs, and coordinate geometry.
Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [3 marks]
Surds and Indices
Rationalise the denominator of 6/sqrt(3), giving your answer in its simplest form.
Question 2 [3 marks]
Simultaneous Equations and Inequalities
Solve the inequality 4x + 3 > 7x - 12, giving your answer in the form x < k.
Question 3 [4 marks]
Algebraic Manipulation
Expand and simplify (3x + 1)(x - 2) - (x - 1)^2.
Question 4 [4 marks]
Coordinate Geometry
Points A(-3, 4) and B(5, -2) are given.
Find the midpoint of AB and the length of AB.
Question 5 [4 marks]
Surds and Indices
Solve 3^(2x - 1) = 27^(x - 2), giving the exact value of x.
Question 6 [5 marks]
Functions and their Graphs
The function h is defined by h(x) = 2x^2 - 5 for all real x.
Find the values of x for which h(x) = 3x + 4.
Question 7 [5 marks]
Algebraic Manipulation
Simplify fully (2x^2 - 3x - 2)/(4x^2 - 1).
Question 8 [5 marks]
Functions and their Graphs
The function h is defined by h(x) = 2x^2 - 3 for all real x.
Find the values of x for which h(x) = 5x - 6.
Question 9 [5 marks]
Simultaneous Equations and Inequalities
Solve the inequality 4(2x - 3) < 3(3x + 4), giving your answer in the form x > k.
Question 10 [5 marks]
Algebraic Manipulation
Solve the equation 4/(x - 1) - 2/(x + 2) = 1.
Show that your equation reduces to a quadratic before solving it, and give both solutions.
Question 11 [5 marks]
Functions and their Graphs
The function h is defined by h(x) = x^2 + 8x + 19 for all real x.
Express h(x) in the form (x + a)^2 + b, where a and b are integers.
Hence state the minimum value of h(x) and the value of x at which it occurs.
Question 12 [6 marks]
Algebraic Manipulation
Write as a single fraction in its simplest form: (x + 1)/(x^2 - 4) - 2/(x - 2).
Question 13 [6 marks]
Functions and their Graphs
f(x) = 5/(x - 2) + 1 for x not equal to 2.
Find f^-1(x), the inverse function of f, stating the value excluded from its domain.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying numerator and denominator by sqrt(3) | M1 |
| simplifying to 6sqrt(3)/3 | M1 |
| 2sqrt(3) | A1 |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| collecting x terms on one side | M1 |
| collecting number terms on the other side | M1 |
| x < 5 | A1 |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| expanding (3x + 1)(x - 2) to give 3x^2 - 5x - 2 | M1 |
| expanding (x - 1)^2 to give x^2 - 2x + 1 | M1 |
| combining like terms | M1 |
| 2x^2 - 3x - 3 | A1 |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the midpoint formula ((x1 + x2)/2, (y1 + y2)/2) | M1 |
| midpoint (1, 1) | A1 |
| using the distance formula sqrt((x2 - x1)^2 + (y2 - y1)^2) | M1 |
| length AB = 10 | A1 |
| Final answer: Midpoint (1, 1); length AB = 10 | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| writing 27 as 3^3 | M1 |
| writing 27^(x - 2) as 3^(3x - 6) | M1 |
| equating exponents 2x - 1 = 3x - 6 | M1 |
| x = 5 | A1 |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equation 2x^2 - 5 = 3x + 4 | M1 |
| rearranging to 2x^2 - 3x - 9 = 0 | M1 |
| factorising as (2x + 3)(x - 3) = 0 | M1 |
| x = 3 | A1 |
| x = -1.5 | A1 |
| Final answer: x = 3 or x = -1.5 | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| factorising the numerator as (2x + 1)(x - 2) | M1 |
| the correct factorisation (2x + 1)(x - 2) | A1 |
| factorising the denominator as the difference of two squares (2x - 1)(2x + 1) | M1 |
| cancelling the common factor (2x + 1) | M1 |
| (x - 2)/(2x - 1) | A1 |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| forming the equation 2x^2 - 3 = 5x - 6 | M1 |
| rearranging to 2x^2 - 5x + 3 = 0 | M1 |
| factorising as (2x - 3)(x - 1) = 0 | M1 |
| x = 1 | A1 |
| x = 1.5 | A1 |
| Final answer: x = 1 or x = 1.5 | |
| Question 9[5 marks] | |
|---|---|
| Answer or working | Marks |
| expanding the left side to 8x - 12 | M1 |
| expanding the right side to 9x + 12 | M1 |
| collecting x terms to give x on one side | M1 |
| collecting number terms on the other side | M1 |
| x > -24 | A1 |
| Question 10[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplying both sides by (x - 1)(x + 2) | M1 |
| expanding to 4(x + 2) - 2(x - 1) = (x - 1)(x + 2) | M1 |
| rearranging to x^2 - x - 12 = 0 | M1 |
| factorising as (x - 4)(x + 3) = 0 | M1 |
| x = 4 or x = -3 | A1 |
| Question 11[5 marks] | |
|---|---|
| Answer or working | Marks |
| attempting to complete the square on x^2 + 8x | M1 |
| (x + 4)^2 as the squared term | M1 |
| (x + 4)^2 + 3 | A1 |
| identifying the minimum value as 3 | B1 |
| stating this occurs at x = -4 | B1 |
| Final answer: (x + 4)^2 + 3; minimum value 3 at x = -4 | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| factorising x^2 - 4 as (x - 2)(x + 2) | M1 |
| identifying the common denominator (x - 2)(x + 2) | M1 |
| writing 2/(x - 2) as 2(x + 2)/((x - 2)(x + 2)) | M1 |
| combining the numerators to give x + 1 - 2(x + 2) | M1 |
| simplifying the numerator to -x - 3 | A1 |
| -(x + 3)/((x - 2)(x + 2)) | A1 |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| subtracting 1 from both sides to give y - 1 = 5/(x - 2) | M1 |
| taking reciprocals of both sides to give 1/(y - 1) = (x - 2)/5 | M1 |
| multiplying both sides by 5 to give 5/(y - 1) = x - 2 | M1 |
| adding 2 to both sides to give x = 2 + 5/(y - 1) | M1 |
| swapping x and y to give f^-1(x) = 2 + 5/(x - 1) | A1 |
| stating x = 1 is excluded from the domain of f^-1 | B1 |
| Final answer: f^-1(x) = 2 + 5/(x - 1), x not equal to 1 | |