Finding the value of n

Finding the value of n

Pencil and paper
Overview
Using this Resource
Connecting to the Curriculum
Marking Student Responses
Working with Students
Further Resources
This task is about finding the value of an unknown number in equations..

Show how to find the value of n in the following equations. [5n means 5 × n.]

a) n – 4 = 13

  
 
 
n = _____
 
b)
 
25 = n – 6

 
 
 
n = _____
 
c)
 
12 = 3n

 
 
 
n = _____
 
d)
 
8n = 4
 
 
 

 n = _____
 
e)
 
4n + 2 = 22

 
 
 
n = _____
 
f)
 
15 = 7 + 2n

 
 
 
n = _____
Task administration: 
This task is completed with pencil and paper only.
Level:
4
Curriculum info: 
Maths, Number and Algebra, Equations and expressions
Key Competencies: 
Using language, symbols, and texts
Keywords: 
equations, linear equations
Description of task: 
Students show how they find the value of an unknown number in a range of linear equations.
Curriculum Links: 
This resource can help to identify students' ability to use inverse operations to simple linear relationships.
 
Key competencies
This resource involves communicating how a student solved a simple linear equation. This relates to the Key Competency: Using language, symbols and text.
Learning Progression Frameworks
This resource can provide evidence of learning associated with
Using symbols and expressions to think mathematically, sets 4-5
within the Mathematics Learning Progressions Frameworks.
Read more about the Learning Progressions Frameworks.
Answers/responses: 
    Y8 (05/2008)
a) Working showing: n = 13 + 4 or
other correct methods of solution.
17		 moderate

easy
b) Working showing 25 + 6 = n  or
other correct methods of solution.
31		 moderate

easy
c) Working showing 12 ÷ 3 = n or
other correct methods of solution.
4		 difficult

easy
d) Working showing n = 4 ÷ 8 or
other correct methods of solution.
½, 0.5		 very difficult

difficult
e) Working showing 4n = 22 – 2; n = 20 ÷ 4 or
other correct methods of solution.
5		 difficult

easy
f) Working showing: 15 – 7 = 2n; 8 ÷ 2 = n or
other correct methods of solution.
4		 difficult

moderate

Based on a representative sample of 227 students.

NOTE:

  1. Do not accept expressions that just show that the answer they gave is correct (e.g. 17 – 4 = 13), as these do not show how they arrived at their answer.
  2. Do accept trial and improvement methods. We found these to be very uncommon.
Diagnostic and formative information: 
The questions are in three pairs that lend themselves to different strategies.

  1. One step problems involving addition or subtraction.
  2. One step problems involving multiplication or division.
  3. Two step problems involving all four arithmetic operators.

Equations are also written in different forms to support:

  1. the idea that they are not restricted to using the form a + b = c; or
  2. that the "as yet unknown" (in these cases n), is always on the left-hand side of equations.

Equations are about balance of what is on each side of the equals sign (see balance, in Algebraic Thinking: Equality).

  Common error Likely calculation Likely misconception
a)
b)
e)
f)		 9
19
6
11		 13 – 4
25 – 6
(22 + 2) ÷ 4
(15 + 7) ÷ 2		 Inappropriate inverse operator (+ or –)
Student uses an incorrect inverse operator to solve the equation (subtraction instead of addition).
c)
d)		 36
32		 12 × 3
4 × 8		 Inappropriate inverse operator (× or ÷)
Student uses an incorrect inverse operator to solve the equation (multiplication instead of division).
e)
f)		 16
6		 22 – 2 – 4
15 – 7 – 2		 Incorrect inverse operator
Student uses an incorrect inverse operator to solve the equation (subtraction instead of division)
c)
d)
e)
f)		 12
4
20
8		 22 – 2
15 – 7
22 – 2
15 – 8		 Ignores the coefficient of the unknown (n)
The student does not take account of the effect of the number of n's in the equation.
e) 2 8 ÷ 4 = 2 Fails to manipulate equation to the form n = …
Does not recognise that division by 4 leads to the equation 2n = 1.

In questions e) and f), students' answers may reflect a combination of the errors above.

Next steps: 
Inappropriate inverse operator (+ or –)
Students need to explore:

  • the concept of the additive inverse (a – a = 0 for any value of a);
  • the concept of additive identity (a + 0 = a, for any value of a).  and
  • that adding or subtracting the same number to each side of an equation leave it "still in balance".

Click on the link or use the keyword additive identity or on the link Algebraic Thinking Concept Map: additive identity for resources relating to these properties.

Alternatively solving this sort of equation can be explored using the model of reverse addition shown in Reversing Addition and When Subtraction Becomes Addition, Book 8: Teaching Number Sense and Algebraic Thinking (PDF).

Inappropriate inverse operator (× or ÷)
Students need to explore:

  • the concept of the multiplicative inverse (a ÷ a = 1 for any value of a);
  • the concept of multiplicative identity (a × 1 = a, for any value of a).  and
  • that multiplying or dividing each side of an equation by the same number leaves it "still in balance".

Alternatively solving this sort of equation can be explored using a model similar to that shown in the two activities above from Book 8: Teaching Number Sense and Algebraic Thinking (PDF) but with several lots of n being drawn. This model starts to break down with fractional lots of n, or with negative lots of n.

Incorrect inverse operator
The student needs to explore both of the above concepts, seeing the link between addition and subtraction, and between multiplication and division.

Ignores the coefficient of the unknown (n)
The student needs to see that an equation such as 12 = 3n is equivalent to 3 × ? = 12.
They may need to explore the same issues as for Inappropriate inverse operator (× or ÷) (see above)

Fails to manipulate equation to the form n = …
The student needs to see that finding the value of n is the same as completing the equation n = ???
The student may need to explore the first two issues of inappropriate inverse operators
(+, –, ×, or ÷ ) above.