Algebra

Solving linear equations

Keep the balance: how to isolate the unknown in any first-degree equation.

About 8 minutes

A linear equation contains the unknown only to the first power, as in 3x - 1 = x + 4. Solving it means finding the value of $x$ that makes both sides equal. The whole method rests on one principle: an equation stays true if you do the same thing to both sides.

The balance principle

Think of the equals sign as the pivot of a balance. You may add, subtract, multiply or divide both sides by the same number (not dividing by zero) and the balance holds. Each step is chosen to undo something that has been done to $x$ .

A reliable routine

1. Expand any brackets and clear fractions by multiplying both sides by a common denominator.

2. Collect the terms containing $x$ on one side and the constants on the other.

3. Divide by the coefficient of $x$ .

4. Check by substituting the answer into the original equation.

ax + b = cx + d \;\Rightarrow\; x = \frac{d - b}{a - c},\quad a \ne c

Special cases

If the $x$ terms cancel and a true statement remains, such as 0 = 0, every number is a solution: the equation is an identity. If a false statement remains, such as 0 = 3, there is no solution.

Worked examples

Solve 3x - 1 = x + 4.

Subtract $x$ from both sides: 2x - 1 = 4.
Add 1 to both sides: 2x = 5.
Divide by 2: x = \frac52.
Check: 3 \cdot \frac52 - 1 = 6.5 and \frac52 + 4 = 6.5.

Answer: x = \frac{5}{2}

Solve \frac{x}{2} + \frac{1}{3} = \frac{5}{6}.

Multiply every term by 6: 3x + 2 = 5.
Subtract 2: 3x = 3.
Divide by 3: x = 1.

Answer: x = 1

Common mistakes

Changing only one side — for example subtracting 3 on the left but not on the right.
Sign errors when expanding a bracket with a minus in front: -(x - 4) = -x + 4.
Multiplying only some of the terms when clearing fractions.
Dividing by the coefficient before collecting like terms.

Practice problems

1.Solve 2x + 3 = 7
2.Solve 5 - 2x = 11
3.Solve 4(x - 2) = 3x + 1
4.Solve 2(x + 1) = 2x + 5

Practise this topic with checked answers

Tools for this lesson

Linear Equation SolverSolve linear equations in one variable exactly, with each algebraic step shown and fractions kept exact.Equation CalculatorSolve linear, quadratic and polynomial equations exactly, and other single-variable equations numerically, including complex roots.System of Equations SolverSolve systems of linear equations with exact Gauss–Jordan elimination; detects no-solution and infinitely-many-solution cases.

Next lesson: Quadratic equations

A reliable routine

1. Expand any brackets and clear fractions by multiplying both sides by a common denominator.

2. Collect the terms containing

x

on one side and the constants on the other.

3. Divide by the coefficient of

x

4. Check by substituting the answer into the original equation.

ax + b = cx + d \;\Rightarrow\; x = \frac{d - b}{a - c},\quad a \ne c

ax + b = cx + d \;\Rightarrow\; x = \frac{d - b}{a - c},\quad a \ne c

Worked examples

Solve $3x - 1 = x + 4$ 3x - 1 = x + 4.

Subtract $x$ from both sides: $2x - 1 = 4$ 2x - 1 = 4.
Add 1 to both sides: $2x = 5$ 2x = 5.
Divide by 2: $x = \frac52$ x = \frac52.
Check: $3 \cdot \frac52 - 1 = 6.5$ 3 \cdot \frac52 - 1 = 6.5 and $\frac52 + 4 = 6.5$ \frac52 + 4 = 6.5.

Answer: $x = \frac{5}{2}$ x = \frac{5}{2}

Solve $\frac{x}{2} + \frac{1}{3} = \frac{5}{6}$ \frac{x}{2} + \frac{1}{3} = \frac{5}{6}.

Multiply every term by 6: $3x + 2 = 5$ 3x + 2 = 5.
Subtract 2: $3x = 3$ 3x = 3.
Divide by 3: $x = 1$ x = 1.

Answer: $x = 1$ x = 1

Common mistakes

Changing only one side — for example subtracting 3 on the left but not on the right.

Sign errors when expanding a bracket with a minus in front:

-(x - 4) = -x + 4

-(x - 4) = -x + 4.

Multiplying only some of the terms when clearing fractions.

Dividing by the coefficient before collecting like terms.

Practice problems

1.Solve

2x + 3 = 7

2x + 3 = 7

2.Solve

5 - 2x = 11

5 - 2x = 11

3.Solve

4(x - 2) = 3x + 1

4(x - 2) = 3x + 1

4.Solve

2(x + 1) = 2x + 5

2(x + 1) = 2x + 5

Practise this topic with checked answers