Solve a Linear Equation in One Variable

Understand the concept

An equation states that two expressions are equal. A solution is a value of the unknown that makes this statement true. A linear equation in one variable contains the unknown only to the first power after simplification.

The governing principle is balance. Adding or subtracting the same expression on both sides preserves equality. Multiplying both sides by the same quantity also preserves equality, but division must use a nonzero quantity. Writing the operation explicitly is a useful safeguard when learning to rearrange equations.

For an equation ax + b = c with a not equal to zero, subtract b from both sides to obtain ax = c − b. Then divide both sides by a to obtain x = (c − b)/a. Instead of memorising this form alone, understand what each step undoes: subtraction undoes addition, and division undoes multiplication by a nonzero number.

When brackets appear, distribute a multiplier to every term inside the bracket. For example, 3(x − 2) means 3x − 6, not 3x − 2. If the unknown appears on both sides, collect its terms on one side by adding or subtracting the same term from both sides.

Not every equation has one solution. If simplification produces 0x = 5, no value of x can satisfy it. If it produces 0x = 0, every value satisfies the resulting identity, subject to any restrictions in the original expression. Do not divide by the zero coefficient in either case.

Finally, substitute the proposed solution into the original equation. Both sides should evaluate to the same value. In a word problem, also check that the value makes sense for the quantity being represented, such as a nonnegative whole number when counting complete items.

Worked example

A service has a fixed charge of 50 units plus 12 units per item. The total charge is 146 units. Let x be the number of items.

Write the relationship: 50 + 12x = 146. Subtract 50 from both sides: 12x = 96. Divide by 12: x = 8. Check in the original relationship: 50 + 12 × 8 = 146, so eight items satisfy the condition.

For a bracket example, solve 3(x − 2) + 4 = 19. Expand to get 3x − 6 + 4 = 19, or 3x − 2 = 19. Add 2 to both sides to obtain 3x = 21, so x = 7. Substitution gives 3(7 − 2) + 4 = 19.

Practice and answer

1. Solve 5x − 7 = 18.
Answer: add 7 to get 5x = 25, then divide by 5. x = 5.

2. Solve 2(x + 3) = x + 11.
Answer: 2x + 6 = x + 11. Subtract x and then 6 to obtain x = 5. Both original sides equal 16.

3. How many solutions does 4x + 2 = 4x + 9 have?
Answer: subtracting 4x gives 2 = 9, a contradiction, so there is no solution.

4. How many solutions does 2(x + 1) = 2x + 2 have?
Answer: both sides are the same expression after expansion, so every real x is a solution.

What you will learn

• Translate a simple verbal relationship into an equation.
• Preserve equality while rearranging terms.
• Check solutions and recognise contradictions or identities.