APPLICATIONS OF EXPRESSIONS, EQUATIONS AND INEQUALITIES
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Subject: Mathematics
Class: SHS 2
Term: 1st Term
Week: 9
Grade code: 2.2.1.LI.2
Strand code: 2
Sub-strand code: 1
Content standard code: 2.2.1.CS.1
Indicator code: 2.2.1.LI.2
Theme: ALGEBRAIC REASONING
Subtheme: APPLICATIONS OF EXPRESSIONS, EQUATIONS AND INEQUALITIES
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This lesson introduces learners to a powerful mathematical tool for solving real-world problems: simultaneous linear equations. Often in life, we face situations with more than one unknown quantity and more than one piece of information. For example, a market woman needs to know the price of a single tin of milk and a single loaf of bread when she only knows the total cost of different combinations she sold. This lesson provides the algebraic methods—substitution and elimination—to find these unknown values precisely. Mastering this skill is fundamental for further studies in mathematics, science, economics, and engineering.
A. Starter: Review of Prerequisite Skills (10 mins)
Before we tackle two equations, let's refresh our skills with one. This is based on the concept of "Change of Subject" and "Substitution".
Concept 1: Change of Subject This means rearranging a formula or equation to make one of the variables stand alone on one side of the equals sign. Example (from NaCCA exemplar): If `v = u + at`, make `u` the subject. Solution: We want to isolate `u`. The term `at` is added to `u`. To undo this, we subtract `at` from both sides. `v - at = u + at - at` `v - at = u` Therefore, `u = v - at`
Concept 2: Substitution This means replacing a variable with its given numerical value or an equivalent algebraic expression. Example (continuing from above): Find the value of `u` when `v = 20`, `a = 3`, and `t = 4`. Solution: We use the formula where `u` is the subject: `u = v - at`. Substitute the given values: `u = 20 - (3 × 4)` `u = 20 - 12` `u = 8`