Algebra
Expressions, equations, inequalities, sequences, graphs, and functions.
In this topic
Algebraic Expressions
An algebraic expression uses letters to represent unknown values. Terms are the parts of an expression separated by + or -.
Like terms have the same letter(s) raised to the same power. You can only add or subtract like terms.
Simplifying: 3x + 5y - 2x + y = x + 6y
Key Points
- Like terms: same letter, same power — can be collected
- Multiplying: 3a × 2b = 6ab
- Expanding brackets: multiply every term inside by the term outside
- Factorising: the reverse of expanding — take out common factors
Example Questions
2Expand and simplify: 3(2x + 4) - 2(x - 3)
6x + 12 - 2x + 6 = 4x + 18
[2 marks]
2Factorise completely: 6x²y + 9xy²
3xy(2x + 3y)
[2 marks]
1Simplify the expression: 3x + 4x - 2 + 5.
Combine like terms: 3x + 4x = 7x and -2 + 5 = 3. So, the answer is 7x + 3.
[1 mark]
1Expand the expression: 2(3x + 4).
Distribute 2: 2 * 3x = 6x and 2 * 4 = 8. So, the answer is 6x + 8.
[1 mark]
2Factor the expression: x² + 5x + 6.
Find two numbers that multiply to 6 and add to 5. These numbers are 2 and 3. So, the answer is (x + 2)(x + 3).
[2 marks]
2Solve for x: 2(x - 3) = 4.
First, divide both sides by 2: x - 3 = 2. Then, add 3 to both sides: x = 5.
[2 marks]
3Show that 4x² + 8x can be factored to 4x(x + 2).
Factor out the common factor 4x: 4x² + 8x = 4x(x + 2).
[3 marks]
3If a = 2 and b = 3, evaluate the expression: 2a² - 3b + 4.
First, substitute: 2(2)² - 3(3) + 4 = 2(4) - 9 + 4 = 8 - 9 + 4 = 3.
[3 marks]
4Simplify the expression: (2x + 3)(x - 4) - (x + 5)(x - 2). Show all steps.
First expand both: (2x + 3)(x - 4) = 2x² - 8x + 3x - 12 = 2x² - 5x - 12; (x + 5)(x - 2) = x² - 2x + 5x - 10 = x² + 3x - 10. Now subtract: 2x² - 5x - 12 - (x² + 3x - 10) = 2x² - 5x - 12 - x² - 3x + 10 = x² - 8x - 2.
[4 marks]
5A quadratic expression is given as f(x) = x² - 6x + 8. Find the x-intercepts of the graph of this expression. Show all your workings.
To find x-intercepts, set f(x) = 0: x² - 6x + 8 = 0. Factor to (x - 2)(x - 4) = 0. Therefore, the x-intercepts are x = 2 and x = 4.
[5 marks]
Expanding Brackets
Single brackets: multiply each term inside by the term outside. 3(x + 4) = 3x + 12
Double brackets (FOIL method): (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15
Triple brackets: expand two first, then expand the result with the third.
Key Points
- Single bracket: multiply each term
- Double bracket: First, Outside, Inside, Last (FOIL)
- Difference of two squares: (a+b)(a-b) = a² - b²
- Perfect square: (a+b)² = a² + 2ab + b²
Example Questions
2Expand (2x - 3)(x + 5)
2x² + 10x - 3x - 15 = 2x² + 7x - 15
[2 marks]
2Expand and simplify (x + 4)²
x² + 8x + 16
[2 marks]
1Expand the expression: 3(x + 4)
3(x + 4) = 3x + 12
[1 mark]
1Expand the expression: 2(3x - 5)
2(3x - 5) = 6x - 10
[1 mark]
2Expand the expression: (x + 3)(x - 2)
(x + 3)(x - 2) = x² - 2x + 3x - 6 = x² + x - 6
[2 marks]
3Expand the expression: (2x - 1)(3x + 4)
(2x - 1)(3x + 4) = 6x² + 8x - 3x - 4 = 6x² + 5x - 4
[3 marks]
4Expand and simplify: (x + 5)(x - 3) - (x + 1)(x + 2)
(x + 5)(x - 3) = x² - 3x + 5x - 15 = x² + 2x - 15; (x + 1)(x + 2) = x² + 3x + 2; Now, (x² + 2x - 15) - (x² + 3x + 2) = x² + 2x - 15 - x² - 3x - 2 = -x - 17
[4 marks]
3Show that when you expand (x + 4)(x + 1), the result is x² + 5x + 4.
(x + 4)(x + 1) = x² + x + 4x + 4 = x² + 5x + 4. Therefore, the statement is true.
[3 marks]
4Expand the expression: (3x + 2)(2x - 5) and find the coefficient of x.
(3x + 2)(2x - 5) = 6x² - 15x + 4x - 10 = 6x² - 11x - 10. The coefficient of x is -11.
[4 marks]
5Expand the expression: (x - 2)(x + 3)(x - 1) completely and simplify. Show all working.
(x - 2)(x + 3) = x² + 3x - 2x - 6 = x² + x - 6; Now expand (x² + x - 6)(x - 1); = x³ - x² + x² - x - 6x + 6 = x³ - 6x + 6. Therefore, the final result is x³ - 5x + 6.
[5 marks]
Factorising
Factorising is the reverse of expanding — putting an expression back into brackets.
Types of factorising: 1. Common factor: 6x + 9 = 3(2x + 3) 2. Quadratic (x² + bx + c): find two numbers that multiply to c and add to b 3. Difference of two squares: a² - b² = (a+b)(a-b) 4. Quadratic (ax² + bx + c where a ≠ 1): use grouping method
Key Points
- Always look for a common factor first
- For x² + bx + c: find two numbers that multiply to c, add to b
- Check by expanding your answer
Example Questions
2Factorise x² + 7x + 12
(x + 3)(x + 4) — since 3 × 4 = 12 and 3 + 4 = 7
[2 marks]
1Factorise 25x² - 16
(5x + 4)(5x - 4) — difference of two squares
[1 mark]
2Factorise the expression x² + 5x + 6.
(x + 2)(x + 3)
[2 marks]
2Factorise the expression 3x² - 12.
3(x² - 4) = 3(x - 2)(x + 2)
[2 marks]
3Show that x² + 4x + 4 can be written as (x + 2)².
x² + 4x + 4 = (x + 2)(x + 2)
[3 marks]
2Factorise the quadratic expression 2x² + 8x.
2x(x + 4)
[2 marks]
3Factorise the expression x² - 9 completely.
(x - 3)(x + 3)
[3 marks]
4Factorise the cubic expression x³ - 3x² - 4x + 12 completely.
(x - 2)(x + 3)(x - 2) = (x - 2)²(x + 3)
[4 marks]
4Prove that x² + 6x + 9 can be expressed as a perfect square.
x² + 6x + 9 = (x + 3)², because (x + 3)(x + 3) = x² + 6x + 9.
[4 marks]
5Factorise the expression 4x² - 25 completely and explain your steps.
4x² - 25 = (2x)² - 5². This is a difference of squares, so it can be factored as (2x - 5)(2x + 5).
[5 marks]
Solving Equations
An equation has an equals sign and can be solved to find the value of the unknown.
To solve an equation, do the same operation to both sides to isolate the variable.
Linear equations: ax + b = c → solve by inverse operations Equations with unknowns on both sides: collect terms on one side
Key Points
- Whatever you do to one side, do to the other
- Undo operations in reverse order (reverse BIDMAS)
- With fractions: multiply both sides by the denominator
- Always check your solution by substituting back
Example Questions
2Solve: 5x - 3 = 2x + 12
3x = 15, so x = 5
[2 marks]
2Solve: (x + 3)/4 = 5
x + 3 = 20, so x = 17
[2 marks]
2Solve the equation 3x + 5 = 20.
3x = 20 - 5; 3x = 15; x = 15/3; x = 5.
[2 marks]
2What is the value of x in the equation 2(x - 3) = 4?
2x - 6 = 4; 2x = 4 + 6; 2x = 10; x = 10/2; x = 5.
[2 marks]
3Solve the equation 5x - 4 = 2x + 8.
5x - 2x = 8 + 4; 3x = 12; x = 12/3; x = 4.
[3 marks]
3Show that x = 2 is a solution of the equation x² + 4x - 12 = 0.
Substitute x = 2: (2)² + 4(2) - 12 = 4 + 8 - 12 = 0. Therefore, x = 2 is a solution.
[3 marks]
4Solve the equation 4(x + 1) = 2(x + 7) + 6.
4x + 4 = 2x + 14 + 6; 4x + 4 = 2x + 20; 4x - 2x = 20 - 4; 2x = 16; x = 8.
[4 marks]
4Find the value of x in the equation 3(2x - 4) + 5 = 2(4x - 1).
6x - 12 + 5 = 8x - 2; 6x - 7 = 8x - 2; 6x + 2 = 8x + 7; -2 = 2x + 7; -9 = 2x; x = -9/2.
[4 marks]
5Solve the equation x² - 6x + 5 = 0 by factorization.
(x - 1)(x - 5) = 0; x - 1 = 0 or x - 5 = 0; x = 1 or x = 5.
[5 marks]
5Prove that x = 3 is a solution of the equation 2x² - 12x + 18 = 0.
Substituting x = 3: 2(3)² - 12(3) + 18 = 2(9) - 36 + 18 = 18 - 36 + 18 = 0. Thus, x = 3 is a solution.
[5 marks]
Solving Quadratic Equations
A quadratic equation has the form ax² + bx + c = 0. Three methods to solve:
1. Factorising: x² + 5x + 6 = 0 → (x+2)(x+3) = 0 → x = -2 or x = -3
2. Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a
3. Completing the square: x² + bx + c → (x + b/2)² - (b/2)² + c
The discriminant b² - 4ac tells you about the roots: - Positive: two real solutions - Zero: one repeated solution - Negative: no real solutions
Key Points
- Always try factorising first — it's quickest
- Quadratic formula works for ALL quadratics
- If it can't factorise, use the formula
- A quadratic can have 0, 1, or 2 solutions
Example Questions
2Solve x² - 5x + 6 = 0
(x-2)(x-3) = 0, so x = 2 or x = 3
[2 marks]
3Solve 2x² + 3x - 5 = 0 using the quadratic formula.
a=2, b=3, c=-5. x = (-3 ± √(9+40))/4 = (-3 ± 7)/4. x = 1 or x = -2.5
[3 marks]
2Solve the equation x² - 5x + 6 = 0.
To solve, we can factor: (x - 2)(x - 3) = 0. Thus, x = 2 or x = 3.
[2 marks]
2Show that 2x² + 8x + 6 = 0 can be solved by factoring.
First, divide the whole equation by 2 to simplify: x² + 4x + 3 = 0. Factoring gives (x + 1)(x + 3) = 0, so x = -1 or x = -3.
[2 marks]
2Find the roots of the equation x² + 4x + 4 = 0.
This can be factored as (x + 2)(x + 2) = 0. Thus, x = -2 (a repeated root).
[2 marks]
4Solve the quadratic equation 3x² - 12x + 9 = 0 using the quadratic formula.
Using the quadratic formula, x = [12 ± √(12² - 4(3)(9)]] / (2*3) = [12 ± √(144 - 108)] / 6 = [12 ± √36] / 6 = [12 ± 6] / 6. Thus, x = 3 or x = 1.
[4 marks]
4Explain how you would solve x² - 7x + 10 = 0 by completing the square.
To complete the square: x² - 7x = -10. Take half of -7 (which is -3.5) and square it (12.25), so x² - 7x + 12.25 = 2.25. This gives (x - 3.5)² = 2.25. Taking the square root gives x - 3.5 = ±√2.25, so x = 3.5 ± 1.5. Thus, x = 5 and x = 2.
[4 marks]
3Solve the equation 4x² - 16 = 0 and find the values of x.
First, add 16 to both sides: 4x² = 16. Divide by 4: x² = 4. Taking the square root gives x = ±2.
[3 marks]
3Given that one root of the quadratic equation x² - 8x + k = 0 is 4, find the value of k.
Using the fact that one root is 4, we substitute into the equation: 4² - 8(4) + k = 0. This gives 16 - 32 + k = 0, so k = 16.
[3 marks]
5Prove that the equation x² + 2x + 5 = 0 has no real roots.
Calculate the discriminant: D = b² - 4ac = 2² - 4(1)(5) = 4 - 20 = -16. Since the discriminant is negative, there are no real roots.
[5 marks]
Simultaneous Equations
Simultaneous equations are two (or more) equations with two (or more) unknowns, solved together.
Elimination method: Make the coefficients of one variable the same, then add or subtract. Substitution method: Rearrange one equation to express one variable in terms of the other, then substitute.
Key Points
- Elimination: match coefficients, then add/subtract
- Substitution: works well when one equation is already rearranged
- Always find BOTH unknowns
- Check by substituting both values into the other equation
Example Questions
3Solve: 2x + 3y = 13 and x - y = 1
From equation 2: x = y + 1. Substitute: 2(y+1) + 3y = 13 → 5y + 2 = 13 → y = 2.2. Hmm, let me redo: 2y + 2 + 3y = 13 → 5y = 11 → y = 11/5. Actually, let's use elimination: multiply eq2 by 3: 3x - 3y = 3. Add to eq1: 5x = 16, x = 3.2, y = 2.2
[3 marks]
2Solve the simultaneous equations: x + y = 10 and x - y = 2.
From the first equation, y = 10 - x. Substitute into the second: x - (10 - x) = 2 -> 2x - 10 = 2 -> 2x = 12 -> x = 6. Then, y = 10 - 6 = 4. So, x = 6, y = 4.
[2 marks]
2Find the values of x and y given the equations 3x + 2y = 12 and x - y = 1.
From the second equation, y = x - 1. Substitute into the first: 3x + 2(x - 1) = 12 -> 3x + 2x - 2 = 12 -> 5x = 14 -> x = 14/5. Then, y = (14/5) - 1 = 9/5. So, x = 14/5, y = 9/5.
[2 marks]
3Show that the equations 2x + 3y = 18 and 4x - y = 2 have a solution.
We can solve the equations. From 2x + 3y = 18, express y: y = (18 - 2x)/3. Substitute into 4x - y = 2: 4x - (18 - 2x)/3 = 2. Solve for x: 12x - (18 - 2x) = 6 -> 14x = 24 -> x = 24/14 = 12/7. Substitute x into y = (18 - 2(12/7))/3 = 36/7. Solutions exist.
[3 marks]
3Solve the simultaneous equations: 5x - 2y = 3 and 3x + y = 11.
From the second equation, y = 11 - 3x. Substitute into the first: 5x - 2(11 - 3x) = 3 -> 5x - 22 + 6x = 3 -> 11x = 25 -> x = 25/11. Then, y = 11 - 3(25/11) = (121 - 75)/11 = 46/11. So, x = 25/11, y = 46/11.
[3 marks]
4The sum of two numbers is 30, and their difference is 6. Find the numbers.
Let the numbers be x and y. Then x + y = 30 and x - y = 6. Solving these: From x + y = 30, y = 30 - x. Substitute into x - y = 6: x - (30 - x) = 6 -> 2x - 30 = 6 -> 2x = 36 -> x = 18. Then, y = 30 - 18 = 12. So, the numbers are 18 and 12.
[4 marks]
4Solve the simultaneous equations: 2x + 4y = 16 and x - 2y = 1.
From the second equation, x = 2y + 1. Substitute into the first: 2(2y + 1) + 4y = 16 -> 4y + 2 + 4y = 16 -> 8y = 14 -> y = 14/8 = 7/4. Then x = 2(7/4) + 1 = 14/4 + 4/4 = 18/4 = 9/2. So, x = 9/2, y = 7/4.
[4 marks]
5A rectangle's length is twice its width. If the perimeter is 48 cm, find the length and width of the rectangle.
Let width be w and length be l. Thus l = 2w. The perimeter P = 2(l + w) = 48. Substitute: 2(2w + w) = 48 -> 6w = 48 -> w = 8. Then, l = 2(8) = 16. Length = 16 cm, Width = 8 cm.
[5 marks]
6Prove that the equations 2x + 5y = 20 and 3x - 2y = -6 have a unique solution.
To determine if a unique solution exists, we find the determinant of the coefficients: |2 5| = 2*-2 - 5*3 = -4 - 15 = -19, which is not zero. Therefore, a unique solution exists. Now solve: From 2x + 5y = 20, y = (20 - 2x)/5. Substitute into 3x - 2y = -6: 3x - 2((20 - 2x)/5) = -6. Solve for x and y.
[6 marks]
Sequences
A sequence is an ordered list of numbers following a pattern. The nth term formula gives you any term without listing them all.
Arithmetic (linear) sequences have a constant difference between consecutive terms. nth term = dn + (a - d), where d = common difference, a = first term.
Geometric sequences multiply by a constant ratio each time.
Key Points
- Find the common difference for arithmetic sequences
- nth term of arithmetic: a + (n-1)d
- Geometric: multiply by common ratio r each time
- Quadratic sequences have a changing difference — find the second difference
Example Questions
2Find the nth term of the sequence 5, 8, 11, 14, ...
Common difference = 3. nth term = 3n + 2
[2 marks]
2A geometric sequence starts 2, 6, 18, 54, ... Find the 8th term.
Common ratio = 3. 8th term = 2 × 3⁷ = 2 × 2187 = 4374
[2 marks]
1Find the 5th term of the arithmetic sequence where the first term is 3 and the common difference is 2.
The nth term can be found using the formula a_n = a + (n-1)d. Here, a = 3, d = 2, and n = 5. So, a_5 = 3 + (5-1) * 2 = 3 + 8 = 11.
[1 mark]
1The first term of a geometric sequence is 4 and the common ratio is 3. What is the 4th term of this sequence?
The nth term of a geometric sequence can be found using the formula a_n = a * r^(n-1). Here, a = 4, r = 3, and n = 4. So, a_4 = 4 * 3^(4-1) = 4 * 27 = 108.
[1 mark]
2Show that the sequence defined by the formula a_n = 5n - 3 is linear.
To show that a_n = 5n - 3 is linear, we find the difference between consecutive terms: a_n - a_(n-1) = (5n - 3) - (5(n-1) - 3) = 5n - 3 - (5n - 5) = 5. Since the difference is constant, the sequence is linear.
[2 marks]
3Find the 10th term of the sequence defined as follows: 2, 5, 10, 17, ...
The differences are: 5-2 = 3, 10-5 = 5, 17-10 = 7. The second differences are constant: 5-3 = 2, 7-5 = 2. It is a quadratic sequence. The nth term is a_n = n^2 + 1. So, a_10 = 10^2 + 1 = 101.
[3 marks]
3A sequence is defined by the recursive formula a_n = a_(n-1) + 3 with a_1 = 2. Find the 6th term.
Using the recursive formula: a_2 = 2 + 3 = 5, a_3 = 5 + 3 = 8, a_4 = 8 + 3 = 11, a_5 = 11 + 3 = 14, a_6 = 14 + 3 = 17. Thus, the 6th term is 17.
[3 marks]
4Explain how you would find the sum of the first 10 terms in the arithmetic sequence defined by a_n = 2 + (n - 1) * 4.
The sum of the first n terms of an arithmetic sequence can be found using the formula S_n = n/2 * (a + l), where l is the last term. First, find the 10th term: a_10 = 2 + (10 - 1) * 4 = 38. Then, S_10 = 10/2 * (2 + 38) = 5 * 40 = 200.
[4 marks]
4Find the explicit formula for the nth term of the sequence 3, 9, 18, 30, ... and then calculate the 8th term.
The sequence can be expressed as: a_n = n(n + 2). The 8th term is a_8 = 8(8 + 2) = 8 * 10 = 80.
[4 marks]
5A quadratic sequence has terms 2, 6, 12, 20, ... Prove that the nth term can be expressed as n(n + 1).
The first differences are 4, 6, 8. The second differences are constant at 2. Hence, it's quadratic. Assume a_n = an^2 + bn + c. Using the first three terms: 2 = a(1)^2 + b(1) + c, 6 = a(2)^2 + b(2) + c, 12 = a(3)^2 + b(3) + c, we get a = 1, b = 1, c = 0. Thus, a_n = n(n + 1).
[5 marks]
Graphs
Graphs provide a visual representation of mathematical relationships. The most common types at GCSE are:
Linear graphs: y = mx + c (straight lines), where m is the gradient and c is the y-intercept. Quadratic graphs: y = ax² + bx + c (parabolas/U-shapes). Cubic, reciprocal, and exponential graphs each have distinctive shapes.
Key Points
- Gradient = rise/run = change in y / change in x
- Parallel lines have the same gradient
- Perpendicular lines: gradients multiply to give -1
- The turning point of a quadratic is its minimum or maximum
Example Questions
3Find the equation of the line passing through (1, 3) and (3, 7).
Gradient = (7-3)/(3-1) = 2. y = 2x + c. Using (1,3): 3 = 2(1) + c, c = 1. Equation: y = 2x + 1
[3 marks]
3A line is perpendicular to y = 3x + 1 and passes through (6, 2). Find its equation.
Perpendicular gradient = -1/3. y = -1/3 x + c. 2 = -1/3(6) + c → c = 4. y = -1/3 x + 4
[3 marks]
2Plot the following points on a graph: A(1, 2), B(2, 3), C(3, 5). What is the gradient of the line joining points A and B?
Gradient = (y2 - y1) / (x2 - x1) = (3 - 2) / (2 - 1) = 1. The gradient is 1.
[2 marks]
1The equation of a line is y = 2x + 3. What is the y-intercept of this line?
The y-intercept is 3, as this is the value of y when x = 0.
[1 mark]
2Show that the point (2, 7) lies on the line with the equation y = 3x + 1.
Substituting x = 2 into the equation: y = 3(2) + 1 = 6 + 1 = 7. Therefore, (2, 7) lies on the line.
[2 marks]
3The graph of y = x² - 4x + 3 is a parabola. Find the x-intercepts of the graph.
Set y = 0: 0 = x² - 4x + 3. Factor: (x - 1)(x - 3) = 0. The x-intercepts are x = 1 and x = 3.
[3 marks]
4A quadratic function is given by f(x) = x² - 6x + 8. Find the vertex of the parabola.
Vertex x-coordinate = -b/(2a) = 6/(2*1) = 3. Substitute x = 3 into f(x): f(3) = 3² - 6(3) + 8 = -1. Vertex is (3, -1).
[4 marks]
5The line y = mx + c intersects the line y = 2x - 1 at the point (3, 5). Find the values of m and c.
Substituting x = 3 into y = 2x - 1: y = 2(3) - 1 = 5. For y = mx + c, 5 = 3m + c. We need a second point to find m and c. Assume another point for simplicity: Choose m = 1, then c = 2. Final equations depend on chosen m and c.
[5 marks]
3The function f(x) = 2x - 3 is translated 4 units to the right and 2 units up. Write the new function.
New function is f(x) = 2(x - 4) - 3 + 2 = 2x - 8 + 2 = 2x - 6.
[3 marks]
5The graph of the function y = -x² + 4x - 3 is represented as a parabola. Find the maximum point of this parabola.
The vertex occurs at x = -b/(2a) = 4/(2* -1) = 2. Substitute x = 2 into the equation: y = -2² + 4(2) - 3 = -4 + 8 - 3 = 1. The maximum point is (2, 1).
[5 marks]
Inequalities
Inequalities use < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to).
Solve inequalities like equations, but remember: if you multiply or divide by a negative number, FLIP the inequality sign.
On a number line: - Open circle ○ means < or > (not included) - Closed circle ● means ≤ or ≥ (included)
Key Points
- Solve like equations, but flip the sign when multiplying/dividing by a negative
- Represent solutions on a number line
- Double inequalities: solve the middle part, applying to all three sections
Example Questions
2Solve 3x + 5 < 20
3x < 15, so x < 5
[2 marks]
3List the integer values of n where -3 < 2n + 1 ≤ 7
-4 < 2n ≤ 6 → -2 < n ≤ 3. Integers: -1, 0, 1, 2, 3
[3 marks]
2Solve the inequality 3x - 5 < 4. Show your working.
3x < 9; x < 3.
[2 marks]
2Solve the inequality 2x + 3 ≥ 5. What is the value of x?
2x ≥ 2; x ≥ 1.
[2 marks]
3Solve the inequality x/4 + 1 > 2. Write your answer in simplest form.
x/4 > 1; x > 4.
[3 marks]
3A number x satisfies the inequality 5 - 2x < 15. Find the range of x.
-2x < 10; x > -5.
[3 marks]
4Show that the solution to the inequality 4(x - 1) > 2x + 6 is x > 4.
4x - 4 > 2x + 6; 4x - 2x > 10; 2x > 10; x > 5.
[4 marks]
4The perimeter of a rectangle is represented by the inequality 2(l + w) < 30, where l and w are the length and width. If the length is at least 5, find the possible values for the width, w.
l + w < 15; 5 + w < 15; w < 10.
[4 marks]
5Solve the inequality 3x - 7 < 2(x + 1) and represent the solution on a number line.
3x - 7 < 2x + 2; 3x - 2x < 9; x < 9. Draw a number line with an open circle at 9 and shading to the left.
[5 marks]
6A quadratic inequality x² - 5x + 6 ≤ 0 is given. Solve this inequality and state the intervals for which it holds true.
Factor: (x - 2)(x - 3) ≤ 0. The solution is 2 ≤ x ≤ 3 (test values like 1, 2, 3, 4).
[6 marks]