Number and Algebra Sec 2 E-Mathematics

Equations and Inequalities

Equations and Inequalities

Key Concepts

1. Linear inequalities

  • An inequality compares two expressions that may not be equal.
  • Common inequality signs:
    • > means greater than
    • < means less than
    • β‰₯ means greater than or equal to
    • ≀ means less than or equal to
  • A linear inequality contains a variable with power 1 only, such as:
    • x+3>7x + 3 > 7
    • 2yβˆ’5≀92y - 5 \leq 9

How to solve a linear inequality

  • Solve it in a similar way to a linear equation:
    • add the same number to both sides
    • subtract the same number from both sides
    • multiply both sides by the same positive number
    • divide both sides by the same positive number
  • Important rule:
    • when you multiply or divide both sides by a negative number, you must reverse the inequality sign
  • Example:
    • If βˆ’2x>6 -2x > 6
    • divide both sides by βˆ’2-2
    • then x<βˆ’3x < -3

Why the sign reverses

  • Multiplying or dividing by a negative changes the order of numbers on the number line.
  • Example:
    • 3>13 > 1
    • multiply both sides by βˆ’1-1
    • βˆ’3<βˆ’1-3 < -1, so the sign changes

2. Representing solutions on a number line

  • A number line is a straight line used to show values in order.
  • Solutions to inequalities can be shown clearly on a number line.

Symbols used on number lines

  • Open circle:
    • means the endpoint is not included
    • used for < or >
  • Closed/filled circle:
    • means the endpoint is included
    • used for ≀ or β‰₯
  • Arrow:
    • shows the solution continues forever in that direction

Examples

  • x>2x > 2
    • open circle at 2
    • shade or draw arrow to the right
  • xβ‰€βˆ’1x \leq -1
    • closed circle at -1
    • shade or draw arrow to the left

3. Simultaneous linear equations

  • Simultaneous equations are two equations with the same variables.
  • The values of the variables must satisfy both equations at the same time.
  • Example:
    • x+y=7x + y = 7
    • xβˆ’y=1x - y = 1

What is a solution?

  • A solution is a pair of values, such as x=4x = 4, y=3y = 3, that makes both equations true.

4. Solving simultaneous equations by substitution

  • Substitution means replacing one variable with an equivalent expression.
  • Best used when:
    • one equation is already written as a variable in terms of the other
    • for example, y=2x+1y = 2x + 1

Steps for substitution

  1. Make one variable the subject in one equation.
  2. Substitute that expression into the other equation.
  3. Solve the resulting single-variable equation.
  4. Substitute back to find the other variable.
  5. Check both values in the original equations.

Example idea

  • If y=2x+1y = 2x + 1 and x+y=7x + y = 7
  • replace yy in the second equation with 2x+12x + 1

5. Solving simultaneous equations by elimination

  • Elimination means removing one variable by adding or subtracting the equations.
  • Best used when:
    • the coefficients of one variable are already the same
    • or can be made the same easily

Steps for elimination

  1. Write the equations in aligned form:
    • ax+by=cax + by = c
    • dx+ey=fdx + ey = f
  2. Multiply one or both equations if necessary so that one variable has equal coefficients.
  3. Add or subtract the equations to eliminate one variable.
  4. Solve for the remaining variable.
  5. Substitute back to find the other variable.
  6. Check in both original equations.

Example idea

  • 2x+y=92x + y = 9
  • 2xβˆ’y=32x - y = 3
  • Add the equations:
    • 4x=124x = 12
    • x=3x = 3

6. Solving simultaneous equations graphically

  • Each linear equation can be drawn as a straight line graph.
  • The solution to the simultaneous equations is the point where the two lines intersect.

Important ideas

  • Intersect means cross.
  • The coordinates of the intersection point give the values of xx and yy.
  • If the lines cross at (2,3)(2, 3), then the solution is:
    • x=2x = 2
    • y=3y = 3

Possible graph situations

  • One intersection point:
    • one solution
  • Parallel lines:
    • no solution
  • Same line:
    • infinitely many solutions

Accuracy

  • Graphical solutions are often approximate unless the intersection is exactly on grid points.
  • Use a ruler and label axes clearly.

7. Quadratic equations

  • A quadratic equation is an equation in which the highest power of the variable is 2.

  • General form:

    ax2+bx+c=0 ax^2 + bx + c = 0

    where:

    • aa, bb, cc are constants
    • aβ‰ 0a \neq 0

Examples

  • x2βˆ’5x+6=0x^2 - 5x + 6 = 0
  • 2x2+3xβˆ’2=02x^2 + 3x - 2 = 0

Roots or solutions

  • The values of xx that make the equation true are called:
    • roots
    • solutions
    • sometimes zeros

8. Solving quadratic equations by factorisation

  • Factorisation means writing an expression as a product of simpler expressions.

  • Example:

    x2βˆ’5x+6=(xβˆ’2)(xβˆ’3) x^2 - 5x + 6 = (x - 2)(x - 3)

Zero-product property

  • If AB=0AB = 0, then:
    • A=0A = 0, or
    • B=0B = 0
  • This is how factorised quadratics are solved.

Steps for factorisation

  1. Write the equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  2. Factorise the quadratic expression.
  3. Set each factor equal to zero.
  4. Solve each simple equation.

Example

x2βˆ’5x+6=0 x^2 - 5x + 6 = 0
(xβˆ’2)(xβˆ’3)=0 (x - 2)(x - 3) = 0

So:

  • xβˆ’2=0β‡’x=2x - 2 = 0 \Rightarrow x = 2
  • xβˆ’3=0β‡’x=3x - 3 = 0 \Rightarrow x = 3

9. Solving quadratic equations by quadratic formula

Not in Sec 2 2026: The quadratic formula and the discriminant are not in the 2026 Sec 2 E-Mathematics syllabus. These are Sec 3/4 topics. Solving quadratic equations by factorisation (Section 8) is the only quadratic-solving method required at Sec 2. The notes below are kept for reference and future preparation only.

  • Some quadratic equations are difficult or impossible to factorise easily.
  • In such cases, use the quadratic formula:
x=βˆ’bΒ±b2βˆ’4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Meaning of symbols

  • aa, bb, cc are the coefficients in:

    ax2+bx+c=0 ax^2 + bx + c = 0
  • The symbol Β± means there are usually two possible answers:

    • one using ++
    • one using βˆ’-

Steps for using the quadratic formula

  1. Write the equation in the form ax2+bx+c=0ax^2 + bx + c = 0.
  2. Identify aa, bb, and cc.
  3. Substitute carefully into the formula.
  4. Simplify step by step.
  5. Give both answers if there are two.

Discriminant

  • The expression under the square root is:

    b2βˆ’4ac b^2 - 4ac
  • It is called the discriminant.

  • It tells you the number of real solutions:

    • if b2βˆ’4ac>0b^2 - 4ac > 0: two real solutions
    • if b2βˆ’4ac=0b^2 - 4ac = 0: one repeated real solution
    • if b2βˆ’4ac<0b^2 - 4ac < 0: no real solution

Important Definitions

  • Inequality: a mathematical statement showing that two values or expressions are not necessarily equal.

  • Linear inequality: an inequality involving a variable of power 1 only.

  • Solution set: all values that satisfy an equation or inequality.

  • Number line: a straight line used to represent numbers in order.

  • Open circle: a symbol on a number line showing that a value is not included in the solution.

  • Closed circle: a symbol on a number line showing that a value is included in the solution.

  • Simultaneous equations: two or more equations involving the same variables, solved together.

  • Substitution: a method of solving equations by replacing one variable with an equivalent expression.

  • Elimination: a method of solving simultaneous equations by removing one variable.

  • Graphical solution: the solution obtained from the point where graphs intersect.

  • Quadratic equation: an equation with highest power of the variable equal to 2.

  • Factorisation: writing an expression as a product of factors.

  • Factor: a quantity that is multiplied by another quantity.

  • Root: a value of the variable that makes an equation equal to zero.

  • Quadratic formula: a formula used to solve quadratic equations:

    x=βˆ’bΒ±b2βˆ’4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
  • Discriminant: the part b2βˆ’4acb^2 - 4ac in the quadratic formula that shows the nature of the roots.

  • Coefficient: the number multiplying a variable.

  • Constant: a fixed value without a variable.


Worked Examples

Example 1: Solving a linear inequality and representing it on a number line

Solve:

3xβˆ’5≀7 3x - 5 \leq 7

Step 1: Add 5 to both sides

3x≀12 3x \leq 12

Step 2: Divide both sides by 3

x≀4 x \leq 4

Final answer

x≀4 x \leq 4

Number line representation

  • draw a closed circle at 4
  • shade to the left
  • add an arrow to the left

Example 2: Solving simultaneous equations by substitution

Solve:

y=2x+1 y = 2x + 1
x+y=7 x + y = 7

Step 1: Substitute y=2x+1y = 2x + 1 into the second equation

x+(2x+1)=7 x + (2x + 1) = 7

Step 2: Simplify

3x+1=7 3x + 1 = 7

Step 3: Solve for xx

3x=6 3x = 6
x=2 x = 2

Step 4: Substitute back to find yy

y=2(2)+1=5 y = 2(2) + 1 = 5

Final answer

x=2,y=5 x = 2,\quad y = 5

Check

  • First equation: y=2x+1β‡’5=2(2)+1=5y = 2x + 1 \Rightarrow 5 = 2(2) + 1 = 5 βœ”
  • Second equation: x+y=7β‡’2+5=7x + y = 7 \Rightarrow 2 + 5 = 7 βœ”

Example 3: Solving simultaneous equations by elimination

Solve:

2x+y=9 2x + y = 9
2xβˆ’y=3 2x - y = 3

Step 1: Add the equations

(2x+y)+(2xβˆ’y)=9+3 (2x + y) + (2x - y) = 9 + 3
4x=12 4x = 12

Step 2: Solve for xx

x=3 x = 3

Step 3: Substitute into one original equation

Using 2x+y=92x + y = 9:

2(3)+y=9 2(3) + y = 9
6+y=9 6 + y = 9
y=3 y = 3

Final answer

x=3,y=3 x = 3,\quad y = 3

Example 4: Solving simultaneous equations graphically

Solve:

y=x+1 y = x + 1
y=βˆ’x+5 y = -x + 5

Step 1: Plot the first line

For y=x+1y = x + 1, choose points:

  • if x=0x = 0, y=1y = 1 β†’ (0,1)(0,1)
  • if x=2x = 2, y=3y = 3 β†’ (2,3)(2,3)

Step 2: Plot the second line

For y=βˆ’x+5y = -x + 5, choose points:

  • if x=0x = 0, y=5y = 5 β†’ (0,5)(0,5)
  • if x=2x = 2, y=3y = 3 β†’ (2,3)(2,3)

Step 3: Draw the lines

  • use a ruler
  • label each line

Step 4: Read the intersection point

The lines intersect at:

(2,3) (2,3)

Final answer

x=2,y=3 x = 2,\quad y = 3

Example 5: Solving a quadratic equation by factorisation

Solve:

x2βˆ’7x+12=0 x^2 - 7x + 12 = 0

Step 1: Factorise

Find two numbers that multiply to 1212 and add to βˆ’7-7:

  • βˆ’3-3 and βˆ’4-4

So:

x2βˆ’7x+12=(xβˆ’3)(xβˆ’4) x^2 - 7x + 12 = (x - 3)(x - 4)

Step 2: Use the zero-product property

(xβˆ’3)(xβˆ’4)=0 (x - 3)(x - 4) = 0

So:

xβˆ’3=0orxβˆ’4=0 x - 3 = 0 \quad \text{or} \quad x - 4 = 0

Step 3: Solve

x=3orx=4 x = 3 \quad \text{or} \quad x = 4

Final answer

x=3Β orΒ x=4 x = 3 \text{ or } x = 4

Example 6: Solving a quadratic equation by quadratic formula (Extension β€” not required for NA)

Solve:

2x2+xβˆ’3=0 2x^2 + x - 3 = 0

Step 1: Identify aa, bb, and cc

  • a=2a = 2
  • b=1b = 1
  • c=βˆ’3c = -3

Step 2: Write the quadratic formula

x=βˆ’bΒ±b2βˆ’4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Step 3: Substitute values

x=βˆ’1Β±12βˆ’4(2)(βˆ’3)2(2) x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-3)}}{2(2)}

Step 4: Simplify inside the square root

x=βˆ’1Β±1+244 x = \frac{-1 \pm \sqrt{1 + 24}}{4}
x=βˆ’1Β±254 x = \frac{-1 \pm \sqrt{25}}{4}
x=βˆ’1Β±54 x = \frac{-1 \pm 5}{4}

Step 5: Find both answers

Using ++:

x=βˆ’1+54=44=1 x = \frac{-1 + 5}{4} = \frac{4}{4} = 1

Using βˆ’-:

x=βˆ’1βˆ’54=βˆ’64=βˆ’32 x = \frac{-1 - 5}{4} = \frac{-6}{4} = -\frac{3}{2}

Final answer

x=1orx=βˆ’32 x = 1 \quad \text{or} \quad x = -\frac{3}{2}

Common Mistakes to Avoid

  • Forgetting to reverse the inequality sign when dividing or multiplying by a negative number.

  • Using a closed circle instead of an open circle on a number line, or vice versa.

  • Shading the wrong direction on the number line.

  • Making sign errors when rearranging equations.

  • In substitution, forgetting to put substituted expressions in brackets.

    • Example: writing 3x+2x+13x + 2x + 1 incorrectly from 3x+(2x+1)3x + (2x + 1)
  • In elimination, not multiplying the whole equation correctly before adding or subtracting.

  • Mixing up which variable was eliminated.

  • Reading the graph inaccurately because of poor scale or untidy lines.

  • Not checking whether the intersection point is exact or approximate in graphical solutions.

  • Forgetting to write the quadratic equation in the form:

    ax2+bx+c=0 ax^2 + bx + c = 0

    before factorising or using the formula.

  • Choosing factors that multiply correctly but do not add to the middle coefficient.

  • Forgetting the or when giving two roots of a quadratic.

  • Copying the quadratic formula wrongly.

  • Forgetting that if bb is negative, then βˆ’b-b becomes positive.

  • Not using brackets for negative values in the quadratic formula:

    • write (βˆ’3)2(-3)^2, not βˆ’32-3^2

Exam Tips

  • For inequalities:
    • write each algebra step clearly on a new line
    • if you reverse the sign, do it neatly and clearly
  • For number lines:
    • show the correct endpoint symbol
    • use arrows to show the solution extends infinitely
  • For simultaneous equations:
    • always label your final answer clearly as:
      • x=…x = \dots, y=…y = \dots
    • substitute back to check if time allows
  • For substitution:
    • choose the equation that already has one variable isolated if possible
  • For elimination:
    • line up equations carefully before adding or subtracting
    • if multiplying an equation, show the new equation clearly
  • For graphical solutions:
    • use a sharp pencil and ruler
    • label both lines
    • state the coordinates of the intersection point
  • For quadratics by factorisation:
    • first make sure one side is 0
    • write the factorised form completely before solving
  • For quadratic formula:
    • identify aa, bb, cc first
    • substitute with brackets, especially for negative values
  • Mark-earning phrases:
    • β€œreverse the inequality sign”
    • β€œsubstitute into”
    • β€œeliminate xx” or β€œeliminate yy”
    • β€œpoint of intersection”
    • β€œfactorise”
    • β€œusing the quadratic formula”
  • Always simplify final answers fully where possible.

Quick Summary

  • An inequality compares values using >>, <<, β‰₯β‰₯, or ≀≀.
  • Solve linear inequalities like equations, but reverse the sign when multiplying or dividing by a negative number.
  • On a number line:
    • open circle for >> or <<
    • closed circle for β‰₯β‰₯ or ≀≀
  • Simultaneous equations are solved to find values that satisfy both equations.
  • In substitution, replace one variable using an equivalent expression.
  • In elimination, add or subtract equations to remove one variable.
  • In graphical solving, the solution is the intersection point of the two lines.
  • A quadratic equation has highest power 2 and can be written as ax2+bx+c=0ax^2 + bx + c = 0.
  • Solve quadratics by factorisation using the zero-product property. This is the required method at Sec 2.
  • (Not in Sec 2 2026) The quadratic formula x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} and the discriminant b2βˆ’4acb^2 - 4ac are Sec 3/4 topics.
  • Always check signs carefully and present final answers clearly.
✏️ 30 practice questions available

30 questions from school exam papers

Q1

A lake of 2 km is represented on a map with a length of 10 cm. Find the scale of the map in the form 1 : n.

Diagram for question 3a
1 mark
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q2

Find the length of an expressway on the map that represents an actual distance of 1480 m.

A horizontal number line with arrows on both ends and equally spaced tick marks
πŸ“Š Diagram: A horizontal number line with arrows on both ends and equally spaced tick marks
1 mark
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q3

Find the actual area of a park, in mΒ², that is represented by 0.35 cmΒ² on the map.

Diagram for question 3c
1 mark
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q4

Solve 5/(2y + 3) = 6/(1 - y)

Diagram for question 7
3 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q5

Write the number 85.898 correct to 2 decimal places.

3 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q6

Solve the simultaneous equations. a + b = 17 a + 4b = 41

3 marks
Math_Sec2NA_SA2_2023_ACS_Barker 2023
Q7

Solve 7a - 9 = 5

2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q8

Solve x/4 - 2 = 5

2 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q9

Solve (y - 7)/(2y) = 3/4

3 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q10

Solve the inequality 2x β‰₯ -6.

Diagram for question 1a
1 mark
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q11

Represent the solution on the number line given below.

A horizontal number line with arrows on both ends, marked with integers from -5 to 5 at regular intervals.
πŸ“Š Diagram: A horizontal number line with arrows on both ends, marked with integers from -5 to 5 at regular intervals.
1 mark
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q12

Hence find the smallest integer satisfying 2x β‰₯ -6.

Diagram for question 1c
1 mark
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q13

Solve these simultaneous equations. -5x + 4y = 13 3x + y = -1

3 marks
Math_Sec2NA_SA2_2023_Bartley_Sec 2023
Q14

Solve the simultaneous equations. 8p + q = 32 5p - 2q = -1

3 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q15

Hence, find the length and the breadth of the rectangle.

Diagram for question 4a(ii)
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q16

Given that 4x β‰₯ 24, solve this inequality, representing your answer on the number line below.

A horizontal number line with arrows on both ends
πŸ“Š Diagram: A horizontal number line with arrows on both ends
2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q17

Hence, find the smallest perfect cube value of x which satisfies the inequality 4x > 24.

Diagram for question 4b(ii)
1 mark
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q18

Hence solve the equation (x+4)/2 = -3/2 x + 6.

Diagram for question 5d
1 mark
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q19

Solve the following equation. (3/5)x + 5 = 20

2 marks
Math_Sec2NA_SA2_2023_Bedok_View_Sec 2023
Q20

Solve the inequality 4x β‰₯ βˆ’10 and represent the solution on the number line.

A number line with marks at -4, -3, -2, -1, 0, 1, with arrows extending left and right
πŸ“Š Diagram: A number line with marks at -4, -3, -2, -1, 0, 1, with arrows extending left and right
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q21

Hence, write down the smallest integer x that satisfies 4x β‰₯ βˆ’10.

Diagram for question 6b
1 mark
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q22

Solve the following equation: (2x + 5)/3 = (x βˆ’ 1)/2

Diagram for question 7a
2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q23

Find the number of students with weight more than 55kg.

Diagram for question 7b
3 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q24

Solve these simultaneous equations. 3x + y = 12 4x + 2y = 19

2 marks
Math_Sec2NA_SA2_2023_Broadrick_Sec 2023
Q25

Find x and y in the given simultaneous equations below. 3x - y = 6 x + y = 1

Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q26

Solve the inequality βˆ’3x+1β‰€βˆ’8.

Diagram for question 10a
2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q27

Illustrate the solution on the number line given below.

A horizontal number line with tick marks, unlabeled, for plotting the solution
πŸ“Š Diagram: A horizontal number line with tick marks, unlabeled, for plotting the solution
1 mark
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q28

Hence, write down the smallest odd number that satisfies βˆ’3x+1β‰€βˆ’8.

Diagram for question 10c
1 mark
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q29

Solve the following equation: 1/2(y+1)-6=0

2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023
Q30

Solve the following equation: (x+1)/2 = (5x-2)/4

2 marks
Math_Sec2NA_SA2_2023_Canberra_Sec 2023

Past year papers cover the full exam β€” browse by subject below.

View All Papers β€Ί