Linear Equations • Topic 1 of 4

One Variable

A one-variable linear equation has the form ax + b = 0, and its only solution is x = −b/a as long as a ≠ 0. The method is mechanical: clear fractions by multiplying through by the LCM of the denominators, expand brackets, gather all variable terms on one side and constants on the other, then divide. The CAT angle is rarely the algebra itself; it is the cleanup. Two edge cases are tested directly. If the variable cancels and you are left with a true statement like 5 = 5, the equation has infinitely many solutions (an identity). If you are left with a false statement like 5 = 7, there is no solution. Keep an eye on sign errors when you move terms across the equals sign and on dividing by a coefficient that could be negative — flipping a sign late is the single most common slip. For speed, isolate the variable mentally rather than writing every line.

✅ Solved examples

1. Solve 3x + 7 = 2x + 15.
Bring variables left, constants right: 3x − 2x = 15 − 7 ⇒ x = 8.
2. Solve (x − 3)/4 = (x + 1)/6.
Cross-multiply: 6(x − 3) = 4(x + 1) ⇒ 6x − 18 = 4x + 4 ⇒ 2x = 22 ⇒ x = 11.
3. Solve 5(x − 2) − 3(2x − 1) = 4 − x.
Expand: 5x − 10 − 6x + 3 = 4 − x ⇒ −x − 7 = 4 − x ⇒ −7 = 4, which is false ⇒ no solution.
4. For what value of k does 4x + 3 = 2x + k have the solution x = 5?
Substitute x = 5: 4(5) + 3 = 2(5) + k ⇒ 23 = 10 + k ⇒ k = 13.

✏️ Practice — try these, take hints as needed

1. Solve 7x − 4 = 3x + 16.
Gather variable terms on one side.
7x − 3x = 16 + 4.
4x = 20.
x = 5
2. Solve (2x + 1)/3 = (x − 1)/2.
Cross-multiply across the equals sign.
2(2x + 1) = 3(x − 1).
4x + 2 = 3x − 3.
x = −5
3. Solve 6(x + 2) = 2(3x + 6).
Expand both sides.
6x + 12 = 6x + 12.
Variable cancels, statement is true.
Infinitely many solutions (identity)
4. Solve 0.5x + 1.2 = 0.2x + 3.
Multiply through by 10 to clear decimals.
5x + 12 = 2x + 30.
3x = 18.
x = 6
5. If 3(x − 1) + 2 = 2(x + 4), find x.
Expand: 3x − 3 + 2 = 2x + 8.
3x − 1 = 2x + 8.
x = 8 + 1.
x = 9

📝 Topic test — 8 questions

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