Solving Equations and Inequalities

General Mathematics — Second Secondary

Worked examples

Comparison aspect	Main idea	Keywords
•	Solve \|2x − 3\| = 5	Solve
Solution	2x − 3 = 5 gives 2x = 8 so x = 4. Or 2x − 3 = −5 gives 2x = −2 so x = −1. The solution set is {−1	gives
Check	\|2 × 4 − 3\| = \|5\| = 5 and \|2 × (−1) − 3\| = \|−5\| = 5

Solve |x − 2| < 3.
Solution −3 < x − 2 < 3. Adding 2 to all three parts gives −1 < x < 5. The solution is the interval ]−1, 5[.

Comparison aspect	Main idea	Keywords
•	Solve \|3x + 1\| ≥ 4	Solve
Solution	3x + 1 ≥ 4 gives 3x ≥ 3 so x ≥ 1. Or 3x + 1 ≤ −4 gives 3x ≤ −5 so x ≤ −5/3. The solution is ]−∞ −5/3] ∪ [1	gives
Check	At x = 2 we get \|7\| = 7 ≥ 4 which satisfies the inequality. At x = 0 we get \|1\| = 1 which is less than 4 so zero is not part of the solution	satisfiesinequalitylesszero

Solve x² − 5x + 6 ≤ 0.
Solution Factorise: (x − 2)(x − 3) ≤ 0. The roots are 2 and 3. Between the roots the product is negative, and at the roots it is zero. So 2 ≤ x ≤ 3, and the solution is [2, 3].

Solve x² − x − 6 > 0.
Solution Factorise: (x − 3)(x + 2) > 0. The roots are −2 and 3. For x < −2 the product is positive, between −2 and 3 it is negative, and for x > 3 it is positive. So the solution is x < −2 or x > 3.

Solve |2x − 1| = |x + 3|.
Solution Two absolute values are equal when the expressions are equal or when one is the negative of the other.

Key points

Case 12x − 1 = x + 3, which gives x = 4.

Case 22x − 1 = −(x + 3) = −x − 3, which gives 3x = −2 and x = −2/3.

CheckAt x = 4 both sides equal 7. At x = −2/3 the left side is |−7/3| = 7/3 and the right side is |7/3| = 7/3. Both solutions are correct.

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