SHS2 Mathematics · Semester 1 Week 13: Applications of Expressions, Equations and Inequalities

Lesson notes

Learning Objectives

Indicator: 2.2.1.LI.2 - Analyse two linear equations in two variables and solve them using the elimination and substitution methods.

By the end of the lesson, learners can:

By the end of the lesson, learners can make a variable the subject of a linear equation, recalling and applying the skills of substitution and change of subject.
By the end of the lesson, learners can solve a pair of simultaneous linear equations using the substitution method, presenting each step clearly with proper algebraic notation.
By the end of the lesson, learners can solve a pair of simultaneous linear equations using the elimination method, choosing the most efficient variable to eliminate.
By the end of the lesson, learners can verify their solutions by substituting the values of both variables back into both original equations.
By the end of the lesson, learners can compare the substitution and elimination methods and state which method is more efficient for a given pair of equations.

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Curriculum details

Strand: Algebraic Reasoning (Strand 2)
Sub-strand: Applications of Expressions, Equations and Inequalities (2.1)
Content standard: 2.2.1.CS.1 - Demonstrate knowledge and understanding of the concept of simultaneous equations involving two variables and apply it to solve every day-life problem. 2.2.1.LO.1 Solve linear equations in two variables using the elimination, substitution and graphical methods. 2.2.1.LO.2 Analyse, model, and solve simultaneous linear equations involving real-life problems.
Indicator: 2.2.1.LI.2 - Analyse two linear equations in two variables and solve them using the elimination and substitution methods.
Suggested placement: Semester 1, Week 13 (Week 13 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
Source document note: The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:

content standard text - p.191: this sub-strand prints 2 learning outcomes (2.2.1.LO.1, 2.2.1.LO.2) against one content standard, which is what its scope-and-sequence row also promises, so 2.2.1.LO.2 has no content standard of its own; it is carried on this row with its competencies rather than dropped
exemplars - p.196: this exemplar sets a two-dimensional construct - a stacked fraction, an index or a column vector - which the text layer flattens into separate lines, so the parts are present but their vertical arrangement is lost; read the page
Curriculum reference: NaCCA curriculum document, p. 194

Exemplars (from the NaCCA curriculum)

Using talk-for-learning, review with the whole class the basic concepts of substitution and change of subject.
Example 1: If 𝑣 = 𝑢 + 𝑎𝑡 make u e subject and find the value of 𝑢 when 𝑎 = 2, 𝑣 = 10, and 𝑡 = 4
Solution: Subtract 𝑎𝑡 from both sides 𝑣 − 𝑎𝑡 = 𝑢 + 𝑎𝑡 − 𝑎𝑡 𝑢 = 𝑣 − 𝑎𝑡
Substitute the values of a, v, t into 𝑢 = 𝑣 − 𝑎𝑡 𝑢 = 10 − (2 × 4) 𝑢 = 2
Example 2: The volume of a cuboid is the product of the length, breadth, and height of the cuboid. Make breadth the subject of the equation.
The expected answer is 𝑣/(𝑙 × ℎ)
In small groups, engage learners to discuss to come out with the process of solving simultaneous equations using the substitution method. Learners must talk about the need to collaborate and acknowledge the diversity among themselves.
Example: When using the substitution method, choose any of the equations and make any of the variables the subject and substitute the expression of the variable you made the subject into the second equation and solve the value of the variable in the obtained equation. Now, substitute the value of the solved variable into any of the two equations to find the other variable. The value of the two variables is the point of intersection for the two equations.
Collaborative learning: In a small mixed gender/ability group, solve two linear equations using the substitution method.
Example: Solve 𝑥 + 𝑦 = 3..........(1) and 5x + y = 15..........(2) simultaneously using the substitution method.
Solution Steps.
- Chose any of the equations and make any of the variables the subject (I.e., 𝑥 + 𝑦 = 3, 𝑥 = 3 − 𝑦)
- Substitute the expression of x into the second equation and solve the value of the variable in the obtained equation. (I.e., 5(3 − 𝑦) = 15 therefore 𝑦 = 0)
- Substitute the value obtained for y into any of the equations to find for . (I.e., taken equation 1, 𝑥 = 3).
The solution of the two equations is 3 and 0.
Collaborative Learning: In small groups, engage learners to discuss and come out with the process of solving simultaneous equations using the elimination method. Learners must discuss the need to persevere as they strive to think critically to come up with solutions to their assigned tasks.
Example: When using the elimination method, write the two equations in the standard form and use the additive property of equality to eliminate one of the variables in both equations and solve for the remaining variable. Now, substitute the value of the solved variable into any of the two equations and solve the other variable. The value of the two variables is the point of intersection for the two equations.
Collaborative learning: In a small mixed-gender/ability group, solve two linear equations using the elimination method.
Example: Find the value of x and y that satisfies the following equations 𝑦 + 3x = 12 ... ... (1) and 2𝑥 − 𝑦 = 13 ... ... (2) using the elimination method.
Solution Steps:
- Write the two equations in the standard form and use the additive property of equality to eliminate one of the variables in both equations. Note that the coefficient of the 𝑦-term is additive inverse, so when we add the two equations together, the 𝑦-terms add to 0, and we have one equation with one variable.
i.e., 3x+y = 12 ... . (1) 2𝑥 − 𝑦 = 13 ... (2) 3𝑥 + 𝑦 = 12 2𝑥 − 𝑦 = 13 5𝑥 = 25
- Make 𝑥 the subject (I.e., = 5 ) and substitute the value into any of the two equations. (i.e., taken equation (1), 3(5) + 𝑦 = 12) the value of 𝑦 = −3)
- Therefore, the values of 𝑥 and 𝑦 that satisfy both equations are 5 and -3, respectively.
Problem-based Learning: In mixed-gender groups, solve direct and indirect problems consisting of simultaneous equations using both substitution and elimination methods. Engage learners to discuss the need to treat each gender equally and with respect, leading to the promotion of respect for divergent views and inclusivity in the mathematics learning environment and beyond.
Teaching and Learning Resources:
- * Geodot * Rubber bands * Computer
- * GeoGebra * Google search * Mobile phone
- * Calculator * YouTube videos, etc. * Geodot,
- * Rubber bands * Cardboards * Measuring instruments
Assessment (2.2.1.AS.2). The document marks no depth-of-knowledge level for this indicator.