Learning Objectives
  • Solve linear equations with unknowns on one or both sides.
  • Solve equations containing brackets and fractions.
  • Form equations from word problems.
  • Check solutions by substitution.
Key Terms
Equation
A statement that two expressions are equal.
Solution
A value that makes an equation true.
Inverse operation
An operation that reverses another operation.
Linear equation
An equation in which the highest power of the unknown is 1.
Balance method
Performing the same operation on both sides of an equation.
The Balance Principle

An equation is like a balanced scale. Whatever operation is performed on one side must also be performed on the other. The aim is to isolate the unknown while preserving equality.

Worked Example: One-Sided Equation

Question: Solve 5x-7=28.

  1. Add 7 to both sides: 5x=35.
  2. Divide both sides by 5: x=7.

Answer: x=7

Unknowns On Both Sides

Collect variable terms on one side and constants on the other. Choose moves that reduce the chance of negative coefficients, but any correct balanced sequence is valid.

Worked Example: Unknown On Both Sides

Question: Solve 7x+4=3x+28.

  1. Subtract 3x from both sides: 4x+4=28.
  2. Subtract 4: 4x=24.
  3. Divide by 4.

Answer: x=6

Equations With Brackets

Expand brackets carefully before collecting terms, unless dividing by a common factor first is more efficient.

Worked Example: Brackets

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

  1. Expand: 6x-15+4=2x+14.
  2. Simplify: 6x-11=2x+14.
  3. Subtract 2x: 4x-11=14.
  4. Add 11: 4x=25.
  5. Divide by 4.

Answer: x=\frac{25}{4}=6.25

Equations With Fractions

Multiply every term by the LCM of the denominators to remove fractions. Brackets are needed when a numerator contains more than one term.

Worked Example: Fractional Equation

Question: Solve \frac{x-2}{3}+\frac{x+1}{4}=5.

  1. The LCM of 3 and 4 is 12.
  2. Multiply every term by 12: 4(x-2)+3(x+1)=60.
  3. Expand: 4x-8+3x+3=60.
  4. Simplify: 7x-5=60.
  5. So 7x=65.

Answer: x=\frac{65}{7}.

Forming Equations From Words

Define the unknown, translate the relationships into algebra, solve the equation and interpret the answer in context. A clear definition such as “Let x be the smaller number” prevents ambiguity.

Worked Example: Consecutive Integers

Question: The sum of three consecutive integers is 87. Find the integers.

  1. Let the smallest integer be x. The others are x+1 and x+2.
  2. Form x+(x+1)+(x+2)=87.
  3. Simplify: 3x+3=87.
  4. So 3x=84 and x=28.

Answer: The integers are 28, 29 and 30.

Worked Example: Perimeter Equation

Question: A rectangle has length 5 cm more than its width. Its perimeter is 54 cm. Find its dimensions.

  1. Let the width be x cm, so the length is x+5 cm.
  2. Use 2x+2(x+5)=54.
  3. Simplify: 4x+10=54, so 4x=44 and x=11.

Answer: Width 11 cm and length 16 cm.

Checking A Solution

Substitute the solution into the original equation, not a rearranged line. Both sides should give the same value. A check can detect arithmetic or sign errors.

8th Edition Chapter Map
  • Linear equations in one unknown
  • Equations with fractional coefficients and fractional equations
  • Applications in real-world contexts
  • Mathematical formulae
Solving Equations Systematically

An equation states that two expressions are equal. A valid transformation performs the same operation on both sides. The aim is to isolate the unknown while preserving equality. It is usually efficient to expand brackets, clear fractions if necessary, collect variable terms on one side, collect constants on the other and then divide by the coefficient.

Worked Example: Brackets On Both Sides

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

  1. Expand: 8x-17=5x+21.
  2. Subtract 5x: 3x-17=21.
  3. Add 17: 3x=38.
  4. x=\frac{38}{3}.
Fractional Equations

Multiplying every term by the lowest common multiple of the denominators clears fractions. State restrictions when a denominator contains the unknown, because values making a denominator zero are not allowed.

Worked Example: Linear Denominators

Solve \frac{x+1}{3}-\frac{x-2}{4}=2. Multiply throughout by 12:

4(x+1)-3(x-2)=24, so 4x+4-3x+6=24, giving x=14.

Forming Equations From Context

Define the unknown clearly, translate each relationship, solve and interpret the answer in the original context. A diagram or table can help when quantities involve consecutive numbers, perimeter, ages or costs.

Worked Example: Perimeter

A rectangle is 5 cm longer than it is wide and has perimeter 46 cm. Let the width be w cm. Then length =w+5 and 2w+2(w+5)=46. Thus 4w=36, w=9 and the length is 14 cm.

Using And Constructing Formulae

Substitution evaluates a formula for known values. Constructing a formula means expressing a general relationship. Keep units consistent. At this stage, the focus is evaluation and straightforward formation rather than advanced rearrangement.

Worked Example: Constructing A Formula

A taxi fare consists of a fixed charge of 180 plus 65 per kilometre. For d kilometres, C=180+65d. For 12 km, C=180+65(12)=960.

Checking And Interpreting Solutions

Substitute the result into the original equation, especially after working with fractions. Reject answers that violate a denominator restriction or make no sense in context, such as a negative number of objects. Preserve exact fractional answers unless a decimal is requested.

Extended Practice

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

15-6x=5x+4, so 11=11x and x=1.

B. Solve \frac{2x-1}{5}=\frac{x+4}{3}.

3(2x-1)=5(x+4), hence 6x-3=5x+20 and x=23.

C. Three consecutive integers have sum 96. Find them.

Let them be n,n+1,n+2. Then 3n+3=96, so n=31. The integers are 31, 32 and 33.

Examination Guidance
  • Show balanced operations clearly; unexplained jumps can lose method marks.
  • When multiplying an equation to remove fractions, multiply every term.
  • Define variables before forming an equation from a word problem.
  • Check whether the final value is sensible in context, especially for lengths, ages and numbers of objects.
Common Mistakes
  • Changing the sign of a term merely because it is “moved”; the sign changes because an inverse operation is performed on both sides.
  • Expanding -3(x-2) as -3x-6 instead of -3x+6.
  • Multiplying only the fractional terms by the LCM and forgetting a whole-number term.
  • Giving x without interpreting what x represents in a word problem.
Knowledge Check

1. Solve 9x+5=50.

Answer

x=5.

2. Solve 8x-3=5x+24.

Answer

x=9.

3. Solve 4(2x+1)=3(x+6).

Answer

x=\frac{14}{5}.

4. Solve \frac{x}{5}-\frac{x-2}{3}=4.

Answer

x=-25.

5. Two consecutive even integers have sum 74. Find them.

Answer

36 and 38.

Mixed Review With Full Solutions

1. Solve 5(2x-1)-3(x+4)=2(2x+7).

10x-5-3x-12=4x+14, so 7x-17=4x+14, 3x=31, x=31/3.

2. Solve \frac{x-2}{6}+\frac{x+1}{4}=3.

Multiply by 12: 2(x-2)+3(x+1)=36. Thus 5x-1=36 and x=37/5.

3. The sum of a number and three times the next integer is 109. Find the integers.

Let the first be n. Then n+3(n+1)=109, so 4n=106. This gives n=26.5, not an integer, showing that no pair of consecutive integers satisfies the statement.

4. A mobile plan costs 900 plus 4.5 per text. Write a formula and find the cost for 260 texts.

C=900+4.5t. For t=260, C=900+1170=2070.

PreviousNext