Solving Linear Equations
Linear equations are one of the most important topics in algebra.
In this lesson, you’ll learn how to solve equations by using inverse operations to find the value of an unknown.
What are linear equations?
A linear equation in one unknown can be simplified to ax + b = 0, where a and b are numbers and a is not zero. The unknown has power 1 after simplification; there are no x² or higher-power terms.
For example, 3x − 4 = 11 is linear, but x² + 3 = 12 is quadratic. The goal is to find the value of the unknown that makes the equation true.
Examples:
x + 5 = 12
2x = 10
3x − 4 = 11
In each case, there is only one value of x that satisfies the equation.
Example Question:
We will work through this equation step by step:
2x + 3 = 11
Solve for x.
Exam Tip
Always keep the equation balanced.
Whatever operation you perform on one side of the equation, you must perform on the other side as well.
Look for:
• Brackets or fractions to simplify
• Unknown terms on one or both sides
• Keeping both sides equal at every step
Method – Solving Linear Equations
We solve linear equations by isolating the variable.
Follow these steps:
1. Simplify each side
Expand brackets and collect like terms. If fractions make the equation awkward, multiply every term on both sides by a suitable common denominator.
2. Gather the unknown terms
If x appears on both sides, add or subtract the same x term on both sides so the unknown terms are together.
3. Isolate the unknown
Add or subtract to remove the constant, then divide both sides by the coefficient of x. The coefficient may be negative.
4. Check your answer
Substitute the solution into the original equation and compare both sides. A solution can be a negative number, a fraction or a decimal.
Worked Examples
Worked Example 1 — A two-step equation
Solve:
2x + 3 = 11
Step 1: Remove the constant
Subtract 3 from both sides:
2x + 3 − 3 = 11 − 3
2x = 8
Step 2: Remove the coefficient
Divide both sides by 2:
2x ÷ 2 = 8 ÷ 2
x = 4
Step 3: Check your answer
Substitute x = 4 into the original equation:
2(4) + 3 = 11
8 + 3 = 11
11 = 11 ✔
The solution is correct.
Worked Example 2 — A negative coefficient
Solve: 5 − 2x = 13
Subtract 5 from both sides: −2x = 8.
Divide both sides by −2: x = −4.
Check: 5 − 2 × (−4) = 5 + 8 = 13.
Worked Example 3 — Brackets
Solve: 3(x + 2) = 21
Expand the bracket: 3x + 6 = 21.
Subtract 6 from both sides: 3x = 15.
Divide both sides by 3: x = 5.
Check in the original: 3(5 + 2) = 3 × 7 = 21.
Alternatively, divide both sides of 3(x + 2) = 21 by 3 first, giving x + 2 = 7. Both methods give x = 5.
Worked Example 4 — Unknowns on both sides
Solve: 5x − 4 = 2x + 11
Subtract 2x from both sides: 3x − 4 = 11.
Add 4 to both sides: 3x = 15.
Divide both sides by 3: x = 5.
Check both sides: 5 × 5 − 4 = 21 and 2 × 5 + 11 = 21.
Worked Example 5 — A fraction
Solve: (x − 1) ÷ 3 + 2 = 5. Here the whole expression x − 1 is divided by 3.
Subtract 2 from both sides: (x − 1) ÷ 3 = 3.
Multiply both sides by 3: x − 1 = 9.
Add 1 to both sides: x = 10.
Check: (10 − 1) ÷ 3 + 2 = 9 ÷ 3 + 2 = 5.
You could also multiply the original equation by 3 first, but you must multiply every term: x − 1 + 6 = 15.
Final answer (Example 1):
The solution to the equation
2x + 3 = 11
is:
x = 4
Always leave your answer in its simplest form.
Check your solution (important habit)
Substitute your answer into the original equation.
Example:
x = 4
Original equation:
2x + 3 = 11
Substitute:
2(4) + 3 = 11
11 = 11 ✔
Both sides are equal, so the answer is correct.
Fractions on both sides and special cases
Worked example — Clear two denominators
Solve (x + 1)/2 = (x + 5)/3. Each bracket is the whole numerator.
Multiply both sides by 6, a common multiple of 2 and 3.
3(x + 1) = 2(x + 5).
3x + 3 = 2x + 10.
Subtract 2x, then subtract 3: x = 7.
Check: (7 + 1)/2 = 4 and (7 + 5)/3 = 4.
With a calculator, enter (7 + 1) ÷ 2; leaving out the brackets would calculate something else.
What if all the x terms cancel?
For 2(x + 1) = 2x + 5, expansion gives 2x + 2 = 2x + 5. Subtracting 2x leaves 2 = 5, which is false. There is no solution.
For 2(x + 1) = 2x + 2, expansion gives the same expression on both sides. It is true for every x; it is an identity.
The earlier definition ax + b = 0 with a ≠ 0 describes the usual one-solution case. If the x terms cancel, inspect the remaining statement instead of dividing by zero.
You can also estimate a linear solution graphically: plot each side as a y-value and read the x-coordinate where the lines meet. For 2x + 1 = 6, the lines y = 2x + 1 and y = 6 meet when x = 2.5. The Linear Graphs lesson develops this method.
Form an equation and interpret the answer
Worked example — Perimeter in context
A rectangle has width x cm, length (x + 4) cm and perimeter 28 cm. Its two widths and two lengths give 2x + 2(x + 4) = 28.
Expand: 4x + 8 = 28. Subtract 8 and divide by 4: x = 5. The width is 5 cm and length 9 cm; both are positive, and 2 × 5 + 2 × 9 = 28.
Worked example — Check what a solution means
A £5 booking fee plus £4 per ticket is said to total £19. If n is the ticket count, 5 + 4n = 19 gives n = 3.5. That solves the equation, but a whole-ticket count must be a non-negative integer. There is no valid ticket count for that stated total; do not round 3.5 to a different solution.
Check your understanding — A £3 delivery charge plus £5 per box is said to cost £20. Form an equation and decide whether a whole number of boxes fits.
Let n be the number of boxes. 3 + 5n = 20 gives n = 17/5 = 3.4.
The count must be a non-negative integer, so no whole number of boxes gives exactly £20. Three cost £18 and four cost £23.
Think Like a Mathematician
Solving equations is like undoing instructions.
If an equation says:
“Multiply by 2 and add 3”
then solving it means:
“Subtract 3 and divide by 2”
The order matters.
For a chain of operations on the unknown, undo them in reverse order. If the unknown occurs on both sides, first gather the unknown terms using the same operation on both sides.
Common mistakes to avoid
• Performing an operation on only one side of the equation
• Forgetting to divide by the coefficient of x
• Arithmetic errors when adding or subtracting
• Not checking the final answer
Most mistakes happen because students rush through the inverse operations.
Try one yourself
Work through these questions in order. Write each balancing step before opening its answer.
1. Solve: 3x − 5 = 16
Add 5 to both sides: 3x = 21.
Divide both sides by 3: x = 7.
Check: 3 × 7 − 5 = 16.
2. Solve: 7 − 3x = 19
Subtract 7 from both sides: −3x = 12.
Divide both sides by −3: x = −4.
Check: 7 − 3 × (−4) = 7 + 12 = 19.
3. Solve: 4(x − 2) = 20
Expand: 4x − 8 = 20.
Add 8 to both sides: 4x = 28.
Divide both sides by 4: x = 7.
Check: 4(7 − 2) = 20.
4. Solve: 6x − 5 = 2x + 9
Subtract 2x from both sides: 4x − 5 = 9.
Add 5 to both sides: 4x = 14.
Divide both sides by 4: x = 7/2 = 3.5.
Check: 6 × 3.5 − 5 = 16 and 2 × 3.5 + 9 = 16.
5. Solve: (x + 4) ÷ 3 = 6
The whole expression x + 4 is divided by 3. Multiply both sides by 3: x + 4 = 18.
Subtract 4 from both sides: x = 14.
Check: (14 + 4) ÷ 3 = 18 ÷ 3 = 6.
6. A rectangle has width x cm, length (x + 3) cm and perimeter 30 cm. Form and solve an equation to find its width and length.
Perimeter is twice the width plus twice the length: 2x + 2(x + 3) = 30.
Expand and collect terms: 4x + 6 = 30.
Subtract 6, then divide by 4: 4x = 24, so x = 6.
Width = 6 cm; length = 9 cm.
Check: 2 × 6 + 2 × 9 = 30 cm.
Further practice — fractions and interpretation
7. Solve (x + 2)/3 = (x + 6)/4.
Multiply both sides by 12: 4(x + 2) = 3(x + 6).
4x + 8 = 3x + 18. Subtract 3x and then 8: x = 10.
Check: 12/3 = 4 and 16/4 = 4.
8. Solve 3(x − 2) = 3x + 1.
Expand: 3x − 6 = 3x + 1.
Subtract 3x: −6 = 1. This is false, so there is no solution.
9. A graph shows y = 3x + 2 crossing y = 10 at about x = 2.7. Find the exact solution of 3x + 2 = 10.
Subtract 2: 3x = 8. Divide by 3: x = 8/3.
8/3 = 2.666… agrees with a graph reading of about 2.7. The fraction keeps the exact value.
Check: 3 × (8/3) + 2 = 10.
Frequently Asked Questions
What is an inverse operation?
An inverse operation reverses another operation.
Examples:
Addition ↔ Subtraction
Multiplication ↔ Division
Why do we perform the same operation on both sides?
This keeps the equation balanced and preserves equality.
Can equations have fractions?
Yes.
The same principles apply, although the calculations may be more challenging.