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Solving Linear Equations

Solving an equation means finding the value that makes it true. Keeping the two sides balanced, you undo the equation step by step until the unknown stands alone.

MYP 4Standard MathsAlgebraCriteria A · C · D~10 min read

Written by Fayek Jawad Haider · MYP student

The balance method

An equation is like a set of scales: the two sides weigh the same. To keep them level, any operation you apply to one side you apply to the other. Your aim is to get the unknown by itself.

Solution
The value of the unknown that makes the two sides of an equation equal.

Undo operations in reverse order, dealing with what is added or subtracted before what multiplies. Each step should leave a simpler but equally true equation.

Equations with brackets

When an equation contains a bracket, expand it first, or divide both sides by the number in front of it. Either route works; expanding is the safer habit when signs are involved.

Worked example

Solve \(2(x + 4) = 18\).

1
Expand the bracket: \(2x + 8 = 18\).
2
Subtract \(8\) from both sides: \(2x = 10\).
3
Divide both sides by \(2\): \(x = 5\).
\(x = 5\)

Expand carefully when a minus sits outside

In \(3 - 2(x - 1)\), the \(-2\) multiplies both terms inside: \(3 - 2x + 2\). Forgetting to flip the sign of the \(-1\) is the most common way to lose the whole question.

Unknowns on both sides

If the unknown appears on both sides, gather all the unknown terms on one side and the numbers on the other. Move the smaller unknown term across so the coefficient stays positive.

Worked example

Solve \(5x - 3 = 2x + 9\).

1
Subtract \(2x\) from both sides to gather the \(x\) terms: \(3x - 3 = 9\).
2
Add \(3\) to both sides: \(3x = 12\).
3
Divide both sides by \(3\): \(x = 4\).
\(x = 4\)

Checking a solution

Substitute your answer back into the original equation. If both sides give the same number, the solution is correct. For \(x = 4\) above: left side \(5(4) - 3 = 17\), right side \(2(4) + 9 = 17\). They match, so it holds.

Always spend ten seconds checking

A quick substitution catches nearly every sign slip and shows the marker your reasoning under Criterion D. It is the cheapest mark in algebra.

Check yourself

1. Solve \(4(x - 1) = 12\). +

Expand: \(4x - 4 = 12\). Add \(4\): \(4x = 16\). Divide by \(4\): Answer: \(x = 4\).

2. Solve \(7x + 2 = 3x + 14\). +

Subtract \(3x\): \(4x + 2 = 14\). Subtract \(2\): \(4x = 12\). Divide by \(4\): Answer: \(x = 3\).

3. Solve \(2(x + 3) = x + 10\). +

Expand: \(2x + 6 = x + 10\). Subtract \(x\): \(x + 6 = 10\). Subtract \(6\): Answer: \(x = 4\).

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