Solving for x
0
Overview
Using this Resource
Connecting to the Curriculum
Marking Student Responses
Working with Students
This task is about solving equations for x.
Task administration:
This task can be completed with pencil and paper or online (with auto marking displayed to students).
Level:
5
Curriculum info:
Keywords:
Description of task:
This task is about finding the solutions for linear equations.
Answers/responses:
Y10 (04/2016)  
a)  1^{1}/_{2}  easy 
b)  2^{ 1}/_{2}  moderate 
c)  3  easy 
d)  1  easy 
e)  ^{1}/_{2}  easy 
f)  3 ^{1}/_{2}  easy 
a)  f) 
All 6 correct
5 or more correct
34 correct
12 correct*

difficult
moderate
easy
very easy

Based on a representative sample of 114 Y10 students.
* These may include some guesses made by students.
Diagnostic and formative information:
Common error  Likely reason  Next steps  
a)
b)
d)

4^{ 1}/_{2}
1
2

Incorrect use of additive inverse
Studentsmay not change the sign when moving units to the other side of the equation: e.g.:
a) Transforms 2x + 3 = 6 into 2x = 6 + 3 rather than 2x = 6 – 3
b) Transforms 3 – 4x = ⁻7 into ⁻4x = ⁻7 + 3 rather than ⁻4x = ⁻7 – 3
d) Transforms 6 + 3x = 9 into 3x = 9 + 6 rather than 3x = 9 – 6

Students need to understand the role of the additive identity / additive inverse when solving problems, coupled with the idea that adding or subtracting the same number from each side of an equation keeps the equation equivalent e.g.,
2x + 3 = 6 ⇔ 2x + (3 – 3) = 6 – 3 (i.e., subtracting 3 from each side of the equation leaves it in equivalent)⇔ 2x = 3 because (3 – 3) = 0, and taking zero off a number leaves it the same.
This understanding should always precede the rule of thumb "Change the side, change the sign.", which can obsure the understanding of how the additive inverse and additive identity can be used to solve equations. The resource Solving equations II could be useful for this.

b)
c)
f)

5
5
4^{ 1}/_{2}

Incorrectly crossing place value boundaries
Students may make arithmetic errors when crossing the tens or ones boundary, e.g.,
b) Calculates 3 – 4 × 5 as 3 – 20 = ⁻7 rather than ⁻17
c) Calculates ⁻4 + 5 × 5 as ⁻4 + 25 = 11 rather than 21
d) Calculates 6 – 2.5 as 4^{1}/_{2} rather than 3^{1}/_{2}

Students need more practice at crossing boundaries when negative numbers, or decimal numbers are involved. You could use the resource Working with negative numbers.

For more on solving linear equations go to the section: Additive identity / Applying the property in the Algebraic thinking concept map