Before we begin: What number could replace the box so that $2 \times \square + 5 = 17$ is true? Write down your thinking and explain how you found your answer.
A linear equation is a statement that two expressions are equal, where the highest power of the variable is 1. Solving means finding the value of the variable that makes the statement true. The key principle: whatever you do to one side, you must do to the other.
Balance means both sides stay equal. Inverse operations undo each other. Check by substituting back.
- Solve one-step equations using inverse operations
- Solve two-step equations by reversing BIDMAS
- Expand brackets before isolating the variable
- Handle variables on both sides of the equation
- Check solutions by substitution
Wrong: Moving a term to the other side without changing its sign. $x + 5 = 12$ becomes $x = 12 + 5$.
Right: When moving a term across the equals sign, change its operation: $x + 5 = 12$ becomes $x = 12 - 5$.
Wrong: Dividing before dealing with addition. In $2x + 6 = 14$, dividing by 2 first gives $x + 6 = 7$.
Right: Apply inverse operations in reverse order of operations (BIDMAS backwards): subtract 6 first, then divide by 2.
Whatever you do to one side of an equation, you must do to the other. Add 5 to both sides. Multiply both sides by 3. Subtract x from both sides. The equation stays balanced, and the solution stays the same.
x + 5 = 12: subtract 5 from both sides. x = 7.
To isolate a variable, undo operations in the reverse order they were applied. Addition is undone first (subtraction), then multiplication (division), then powers (roots). Reverse BIDMAS keeps your steps safe.
2x + 6 = 14: undo +6 first, then times 2.
When an equation contains brackets, expand them first. Use the distributive law to multiply every term inside the bracket by the factor outside. Only then collect like terms and isolate the variable.
3(2x - 4) becomes 6x - 12. Distribute first, then solve.
When the variable appears on both sides, move all variable terms to one side and all constants to the other. Subtract the smaller variable term to avoid negatives if possible. Then isolate as usual.
7x + 5 = 3x + 21: subtract 3x, then 5, then divide.
Watch Me Solve It · 3 examples
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1Identify the operation on x$\frac{x}{4} = 7$ means x has been divided by 4.To undo division by 4, multiply both sides by 4.
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2Multiply both sides by 4$4 \times \frac{x}{4} = 4 \times 7$The 4 on the left cancels with the denominator, leaving x alone.
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3Simplify and check$x = 28$Check: $\frac{28}{4} = 7$. Correct.
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1Write the operation chain$x \xrightarrow{\times 5} 5x \xrightarrow{-8} 27$To isolate x, reverse the chain: undo -8 first, then undo times 5.
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2Add 8 to both sides$5x - 8 + 8 = 27 + 8$The -8 and +8 cancel on the left, leaving 5x alone.
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3Divide both sides by 5$\frac{5x}{5} = \frac{35}{5}$Undo multiplication by 5 with division by 5.
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4Simplify and check$x = 7$Check: $5(7) - 8 = 35 - 8 = 27$. Correct.
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1Expand both sides$3(2x - 4) = 6x - 12$ and $2(x + 5) = 2x + 10$Distribute the 3 and the 2 across every term inside each bracket.
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2Write the expanded equation$6x - 12 = 2x + 10$Now we have variables on both sides. Move them to one side.
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3Subtract 2x from both sides$6x - 2x - 12 = 2x - 2x + 10$Subtracting the smaller coefficient avoids negatives.
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4Add 12 to both sides$4x - 12 + 12 = 10 + 12$Move the constant to the right by adding 12 to both sides.
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5Divide by 4 and check$\frac{4x}{4} = \frac{22}{4}$ so $x = 5.5$Check: LHS = $3(11 - 4) = 3(7) = 21$. RHS = $2(5.5 + 5) = 2(10.5) = 21$. Correct.
Brain Trainer · 4 problems
Four equations ranging from one-step to variables on both sides. Show each step, then reveal the answer to check your working.
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1 Solve $x + 7 = 15$.
Subtract 7 from both sides: $x = 8$. Check: $8 + 7 = 15$. -
2 Solve $3x - 5 = 16$.
Add 5: $3x = 21$. Divide by 3: $x = 7$. Check: $3(7) - 5 = 21 - 5 = 16$. -
3 Solve $2(x + 4) = 18$.
Divide by 2: $x + 4 = 9$. Subtract 4: $x = 5$. Check: $2(5 + 4) = 2(9) = 18$. -
4 Solve $5x + 3 = 2x + 12$.
Subtract $2x$: $3x + 3 = 12$. Subtract 3: $3x = 9$. Divide by 3: $x = 3$. Check: $5(3) + 3 = 18$ and $2(3) + 12 = 18$.
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Solve $2x + 8 = 5x - 4$. Show all steps and check your answer.
Q7. Ava is three times as old as her brother Ben. In four years, Ava will be twice as old as Ben. Let Ben's current age be $b$ years.
(a) Write an equation for this situation. (1 mark)
(b) Solve the equation to find Ben's current age. (2 marks)
Q8. Solve $3(2x - 1) = 2(x + 5)$. Show all steps including expansion, collection and checking.
Quick Check
1. B Subtract 5: $2x = 12$. Divide by 2: $x = 6$.
2. C Multiply by 3: $x = 27$.
3. A Divide by 4: $x - 2 = 5$. Add 2: $x = 7$.
4. B Subtract $x$: $2x + 7 = 15$. Subtract 7: $2x = 8$. Divide by 2: $x = 4$.
5. C$x + (x + 3) + (2x - 1) = 22$. $4x + 2 = 22$. $4x = 20$. $x = 5$.
Show Your Working Model Answers
Q6 (3 marks): Subtract $2x$: $8 = 3x - 4$ [1]. Add 4: $12 = 3x$ [1]. Divide by 3: $x = 4$ [0.5]. Check: $2(4) + 8 = 16$ and $5(4) - 4 = 16$ [0.5].
Q7 (3 marks): (a) $3b + 4 = 2(b + 4)$ [1]. (b) $3b + 4 = 2b + 8$. Subtract $2b$: $b + 4 = 8$. Subtract 4: $b = 4$ [2]. Ben is 4 years old.
Q8 (3 marks): Expand: $6x - 3 = 2x + 10$ [1]. Subtract $2x$: $4x - 3 = 10$ [0.5]. Add 3: $4x = 13$ [0.5]. Divide by 4: $x = 3.25$ [0.5]. Check: LHS = $3(5.5) = 16.5$, RHS = $2(8.25) = 16.5$ [0.5].
The Fraction Maze
Solve $\frac{2x + 1}{3} = \frac{x - 4}{2}$. Hint: cross-multiply to eliminate the fractions first, then solve the resulting linear equation. Show every step.
Reveal solution
Cross-multiply: $2(2x + 1) = 3(x - 4)$.
Expand: $4x + 2 = 3x - 12$.
Subtract $3x$: $x + 2 = -12$.
Subtract 2: $x = -14$.
Check: LHS = $(2(-14)+1)/3 = (-27)/3 = -9$. RHS = $(-14-4)/2 = -18/2 = -9$. Both match.
Balance Rule
Whatever you do to one side, do to the other
Reverse BIDMAS
Undo operations in the opposite order they were applied
Brackets First
Expand before collecting or isolating variables
Variables to One Side
Subtract the smaller coefficient to avoid negatives
One Step Per Line
Clear working earns method marks
Always Check
Substitute your answer back into the original equation
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