The unit is designed as a simple introduction to systems of linear equations. Students solve problems in which they meet two constraints to find a single solution. Both constraints can be expressed as linear equations.
Systems of equations are extremely useful for modelling real world situations. Often two or more conditions exist in a situation that must be satisfied. We can often express conditions using representations such as tables, graphs and equations. These representations provide powerful tools for solving problems.
As an example, consider simultaneously meeting these two conditions:
Condition (i) can be represented by the following table, graph and equation.
c = 2p or p = c/2 where p represents the number of pigs and c the number of chickens.
Condition (ii) can be represented in the same ways.
c + p = 90 or c = 90 – p or p = 90 – c
Satisfying both conditions involves finding values for the number of pigs and number of chickens that work. In table form this involves searching for a common pair in both tables. In graph form this involves finding an intersection of both lines.
An algebraic method is to solve the two equations simultaneously. There are different ways to do this. Here is a substitution method:
Putting 2p in for c in equation (ii) gives:
2p + p = 90 so 3p = 90 and p = 30.
Since c + p = 90, c must equal 60.
The problems in this unit involve using representations to meet common conditions.
Representing relations in algebraic equations involves two important and connected types of knowledge, related to the language conventions (semiotics), and to the nature of variables. When we write c = 2p, or c = 90 – p + 2 the equations are meaningless to anyone else unless we clearly define what the variables, c and p, represent. Note that both c and p refer to quantities that vary and are not fixed objects, such as a chicken or a pig. Quantities are a combination of count and measurement unit. In this case c expresses many animals. Animals are the unit in this problem. 2p means the number of pigs multiplied by two, not twenty-something.
Semiotics, the meaning of symbols and signs, is central to algebra. Transfer between semiotic forms is difficult at times. For example, a statement such as “there are twice as many chickens and pigs” seems innocuous and it is easy to generate a table of values that satisfy the statement. However, recording the statement as an algebraic equation requires a student to accept letters as variables, not as objects that can be counted. 2p = c or c = 2p is correct but appears ordinally different to the spoken form. Some spoken languages are more consistent with algebra and would express the relation as “To get the number of chickens multiply the number of pigs by two.”
Working with variables also requires acceptance of lack of closure, that is thinking with symbols (c and p in this case) without specifically knowing the values they hold. For example, knowing that c = 2p can be substituted into c + p = 90 while conserving its structure, irrespective of whatever the value of c or p, is itself a generalisation.
The equals sign represents a statement of ‘transitive balance’ meaning that the balance is conserved if equivalent operations are performed on both sides of the equation. Knowledge of which operations conserve equality and those which disrupt it are important generalisations about the properties of numbers under those operations, e.g. distributive property of multiplication.
Students can be supported through the learning opportunities in this unit by differentiating the nature and complexity of the tasks, and by adapting the contexts. Ways to support students include:
Task can be varied in many ways including:
The contexts for this unit can be adapted to suit the interests and cultural backgrounds of your students. Animals on the farm provide the contexts for all problems in the unit and appeal to a range of students. Other contexts can also be used, such as buying items at a shop, e.g. socks at $3 per pair and hats at $12 each. Catering at a marae or tulaga fale provides a useful context around managing a budget while still providing ample food for manuhiri (visitors). People in an extended whānau also provides an interesting context, with variables such as the number of adults and children.
Te reo Māori vocabulary terms such as ōrite (equal), kīanga taurangi (algebraic expression), kīanga ōrite (equivalent expression) and whārite rārangi (linear equation) could be introduced in this unit and used throughout other mathematical learning.
It is anticipated that students at Level 4 and 5 understand, and are proficient with, multiplicative thinking. Students are expected to know about simple ratios though this unit reintroduces some key ideas. Some proficiency at solving linear equations would be beneficial, so working through the units on linear algebra before this unit would be helpful.
Begin with this new version of a very old problem. The problem solving pathway in e-ako maths includes a version of this problem called "Sheep and chickens". This might be used as an extension task for students who enjoy the algebraic equation approach.
On Mr MacDonald’s farm there are only pigs and chickens. He counts 24 heads and 80 legs. How many of each kind of animal is there? |
After discussing important conditions in the problem, encourage the students to work in small co-operative groups. Allow access to supportive tools such as calculators and computers. Recording on paper will be important.
Look for the following common approaches:
Trial and improvement
This approach typically involves choosing a pair of possible values for the number of pigs and chickens, and making alterations until the other condition is met. For example, 12 pigs and 12 chickens might be tried. This assumption meets the number of heads condition but 4 x 12 + 2 x 12 = 72. So that pair of values fails the number of legs condition. However, systematic adjustment of increasing the number of pigs and reducing the number of chickens will eventually give the answer. Note that this strategy is very protracted if the numbers involved are large.
Figurative diagram
Students using this strategy often begin with numbers of pigs and chickens that satisfy the number of heads condition, though sometimes they draw 48 animals of one kind. Some students draw figures that satisfy the number of legs condition. It is important that the drawings are symbolic and not literal (i.e. not life-like pigs or chickens) for two reasons. Life-like drawings are time consuming and indicate that the farmyard context assumes more importance than the conditions of the problem. A drawing might look like this, with 12 pigs and 12 chickens:
Chickens can easily be turned into pigs by adding two extra legs until a solution is found that meets the number of legs condition.
Making a table
Often students’ approach the problem using trial and error (rather than improvement). They try combinations of pigs and chickens in an unsystematic way. A table helps them to organise their data but also allows for noticing of patterns that otherwise are missed.
An example of a table based strategy is given below:
Number of Pigs | Number of chickens | Number of pigs’ legs | Number of chickens’ legs | Total number of legs | ||
0 | 24 | 0 | 48 | 48 | ||
1 | 23 | 4 | 46 | 50 | ||
2 | 22 | 8 | 44 | 52 | ||
… | … | … | … | … |
The table can be extended until a solution is found. Note that use of a spreadsheet makes this strategy highly efficient.
Equations
This strategy is unusual for students at Level 4 unless they have exposure to writing and solving linear equations (link to first algebra learning object unit). First, students need to identify the variables in this problem. While it makes sense to use p and c as symbols, it is very important that students regard these as variables not fixed objects. P is not a pig nor is c a chicken. P represents possible numbers of pigs and c possible numbers of chickens.
Second, students need to write the conditions using these variables. Conventions are involved here, notably that 4p means 4 x p and 2c means 2 x c, and equals means a state of balance or sameness.
Third, solving for p or c involves trusting that these variables can remain ‘unclosed’ and conserved under a sequence of steps. This is a significant shift from arithmetic thinking which aims to ‘close’ the answer as immediately as possible.
Equation (i) can be reorganised as p = 24 – c or c = 24 – p. Either of these equalities can be substituted into equation (ii) so the equation is in one variable:
4(24 – c) + 2c = 80 or 4p + 2(24 – p) = 80
Discussion
Solutions:
The "Sheep and chickens" problem solving e-ako provides guidance to developing a general algebraic solution to the pigs and chicken problem.
Introduce the next farmyard problem using slide 2 of the PowerPoint.
On Young Maree MacDonald’s farm the ratio of pigs to sheep to chickens is 2:3:5. Maree has 640 animals in total. How many of each kind of animal are there? |
Solutions:
Problem One
Problem Two
Problem Three
Problem Four
This problem is easier if you think of parts consisting of 12 animals, and the whole made of 45 parts of 12.
Animal | Number | Fraction | ||
Goats | 72 | 6/45 = 2/15 | ||
Llamas | 108 | 9/45 = 1/5 | ||
Pigs | 240 | 24/45 = 4/9 | ||
Sheep | 96 | 8/45 | ||
Cows | 24 | 2/45 |
Show the student today’s starting problem on Slide 4 of the PowerPoint. Ask them to identify the important conditions in the problem:
Jessica Jones buys 60 animals at the market. She only buys cows and pigs. Cows are $120 each, three times the price of pigs. The total cost is $4 800. How many of each animal does she buy? |
The second problem for this lesson is available on Slide 5 of the PowerPoint.
Farmer Fred goes to market. He buys 100 animals for $100. Lambs cost $10 each. Piglets cost $3 each. Chicks cost $0.50 each (50 cents). How many of each animal does he buy? |
The solution is five lambs, one piglet and 94 chicks.
In this lesson students are encouraged to connect their strategies and knowledge of ratios to solve problems. Slide 6 of the PowerPoint poses the problem of expressing a ratio as an equation.
The ratio of pigs to sheep is 1:2. If p = number of pigs and s = number of sheep, write an equation for this relationship. |
Before asking for suggestions make a table of possible numbers for p and s:
Number of pigs (p) | Number of sheep (s) |
1 | 2 |
10 | 20 |
3 | 6 |
5 | 10 |
7 | 14 |
0 | 0 |
Students are likely to suggest two equations, s = 2p or p = 2s. Check to see which of the equations works with the table values. An important idea is that p and s refer to numbers of animals not an individual animal, pig or sheep. So, the equation s = 2p works but seems counter-intuitive with the way the ratio is said, “For every pig there are two sheep” or “There are twice as many sheep as pigs.”
Solutions:
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/down-farm at 9:09pm on the 26th February 2024