Linear Equations
Solving linear equations is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). A linear equation has its unknown to the first power, and finding the value that balances both sides is prior knowledge the course assumes.
You will revise solving two-step equations, handling the variable on both sides, expanding brackets and clearing simple algebraic fractions, then checking each solution by substituting back — building fluency with inverse operations.
Every question with a fully worked solution.
- Linear Equations - Video - Solving linear Equations Involving fractions Watch
Theory
In Year 11 Mathematical Methods (QCAA), a linear equation is an equation whose unknown appears only to the first power, such as \(5x-7=23\). This page reviews how to solve linear equations in one variable — two-step equations, the variable on both sides, brackets, and algebraic fractions — using inverse operations to make \(x\) the subject, then checking the solution.
A linear equation in one variable is a statement that two expressions are equal, where the unknown appears only to the first power (no \(x^2\), no \(\sqrt{x}\), no \(\dfrac{1}{x}\)). Examples are \(3x+4=19\) and \(2(x-1)=x+5\). To solve it we find the value of the unknown that makes the equation true.
We solve by doing the same inverse operation to both sides until the variable stands alone. An inverse operation undoes another: subtraction undoes addition, and division undoes multiplication. Whatever you do to one side you must do to the other, so the balance is kept.
When the equation contains brackets we expand first; when it contains fractions we multiply every term by a common denominator to clear them.
A linear equation can always be arranged into the standard form
whose unique solution is found by transposing:
To clear fractions with denominators \(p\) and \(q\), multiply every term by the lowest common denominator:
How to solve a linear equation
- Simplify each side: expand any brackets and clear fractions by multiplying every term by the lowest common denominator.
- Collect the variable terms on one side and the numbers on the other, using inverse \(+/-\).
- Solve by dividing both sides by the coefficient of the variable.
- Check by substituting your answer back into the original equation.
Undo the \(-7\) — add \(7\) to both sides:
| \(5x-7\) | \(=\) | \(23\) |
| \(5x-7+7\) | \(=\) | \(23+7\) |
| \(5x\) | \(=\) | \(30\) |
Undo the \(\times 5\) — divide both sides by \(5\):
| \(\dfrac{5x}{5}\) | \(=\) | \(\dfrac{30}{5}\) |
| \(x\) | \(=\) | \(6\) |
Check — substitute \(x=6\) into the left side:
| \(5(6)-7\) | \(=\) | \(30-7\) |
| \(=\) | \(23\;\checkmark\) |
Solution: \(x=6\).
Collect the variable on one side — subtract \(4x\) from both sides:
| \(4x+3\) | \(=\) | \(7x-9\) |
| \(3\) | \(=\) | \(3x-9\) |
Undo the \(-9\) — add \(9\) to both sides:
| \(3+9\) | \(=\) | \(3x\) |
| \(12\) | \(=\) | \(3x\) |
Divide both sides by \(3\):
| \(\dfrac{12}{3}\) | \(=\) | \(x\) |
| \(4\) | \(=\) | \(x\) |
Reading right to left gives \(x=4\).
Check — both sides at \(x=4\):
| \(4(4)+3\) | \(=\) | \(19\) |
| \(7(4)-9\) | \(=\) | \(19\;\checkmark\) |
Solution: \(x=4\).
Expand both brackets:
| \(5(x-2)\) | \(=\) | \(3(x+4)\) |
| \(5x-10\) | \(=\) | \(3x+12\) |
Collect the variable — subtract \(3x\) from both sides:
| \(5x-10-3x\) | \(=\) | \(12\) |
| \(2x-10\) | \(=\) | \(12\) |
Add \(10\) to both sides:
| \(2x\) | \(=\) | \(12+10\) |
| \(2x\) | \(=\) | \(22\) |
Divide both sides by \(2\):
| \(x\) | \(=\) | \(11\) |
Check — both sides at \(x=11\):
| \(5(11-2)\) | \(=\) | \(5(9)=45\) |
| \(3(11+4)\) | \(=\) | \(3(15)=45\;\checkmark\) |
Solution: \(x=11\).
Clear the fractions — multiply every term by the LCD \(10\):
| \(10\cdot\dfrac{x+4}{2}-10\cdot\dfrac{x-2}{5}\) | \(=\) | \(10(3)\) |
| \(5(x+4)-2(x-2)\) | \(=\) | \(30\) |
Expand each bracket:
| \(5x+20-2x+4\) | \(=\) | \(30\) |
| \(3x+24\) | \(=\) | \(30\) |
Subtract \(24\) from both sides, then divide by \(3\):
| \(3x\) | \(=\) | \(30-24\) |
| \(3x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(2\) |
Check — substitute \(x=2\):
| \(\dfrac{2+4}{2}-\dfrac{2-2}{5}\) | \(=\) | \(\dfrac{6}{2}-\dfrac{0}{5}\) |
| \(=\) | \(3-0=3\;\checkmark\) |
Solution: \(x=2\).
Common pitfalls
Frequently asked questions
What is a linear equation?
An equation whose unknown appears only to the first power, such as \(5x-7=23\). Its graph is a straight line, and it has exactly one solution.
How do you solve a two-step linear equation?
Undo the addition or subtraction first, then undo the multiplication or division. For \(5x-7=23\): add \(7\) to get \(5x=30\), then divide by \(5\) to get \(x=6\).
How do you solve an equation with the variable on both sides?
Collect the variable terms on one side and the numbers on the other. For \(4x+3=7x-9\), subtract \(4x\) and add \(9\) to get \(12=3x\), so \(x=4\).
How do you get rid of fractions in an equation?
Multiply every term by the lowest common denominator. In \(\dfrac{x+4}{2}-\dfrac{x-2}{5}=3\), multiply through by \(10\) to clear both fractions.
How do you check the solution of a linear equation?
Substitute your answer back into the original equation and confirm the left side equals the right side.
How many solutions does a linear equation have?
A linear equation \(ax+b=0\) with \(a\neq0\) has exactly one solution, \(x=-\dfrac{b}{a}\).