**Task 2**

**Pair work (follow-up from previous Task Algebra Change of Subject 2)**

Based on the activity in task 1, students have submitted the following responses (refer Table 1).

Your task is to reflect on the solution(s) submitted by your classmates.

Identify at

*least 2 points*for consideration for 'Change of Subject'. Example Arithmetic error (wrong associative law applied)You may discuss with a partner in class.

Submit this as a

**.***comment*Indicate your name as well as your partner's name.

reference:

Worksheet 5, ACE learning.

Question Posted by Teacher

__Table 1: Answer Posted by Students__

Student D:

ReplyDeleteError: Did not change the right hand side of the equation from + to - when he changed the left hand side from - to + and did not divide 2z by p.

Tan Yu Tao, Jonathan Tan

ReplyDeleteMultiplication and division error

Arithmetic error

Student D:

ReplyDeleteHe/She did not divide the 2z by the p and also did not change the addition and subtraction symbols properly.

Student A,B,L,Q,C,E,H,I,M,R = Did not inverse the operation from + to -

ReplyDeleteStudent

Student F = Correct

Student D = Forgot to divide P

Student J = The subject should only be on the left

By Sylvia and Esther

This comment has been removed by the author.

ReplyDeleteStudent A,B,L,Q,C,E,H,I,M,R = Did not inverse the operation from + to -

ReplyDeleteStudent

Student F = Correct

Student D = Forgot to divide P

Student J = The subject should only be on the left

By Sylvia and Esther

Student "I" has left the "ans)" there and did not find the solution for z-2z which should be -z.

ReplyDeleteStudent C:

ReplyDeleteDid not change the right hand side of the equation from multiplication to division. So in stead of 2z/p it became 2zp.

Student N- The 2z should be a fraction with the p instead, 2z the numerator while p the denominator .

ReplyDeleteStudent C did not divide the equation by p and he did not change the signs for both sides of the equation.

ReplyDeleteStudent C did not make the 2z divide by p

ReplyDeleteStudent N - The 2z should actually be the fraction with p.

ReplyDeletestudent D did not make the 2z divided by p, but multiplied it instead

ReplyDeleteOnly Student F got it correct, Student D is close but he/she did not divide p. Student J is correct except the fact that the subject is on the right side. The rest got the question wrong because they did not inverse the answer.

ReplyDeleteStudent G :

ReplyDeleteDid not change + to - when changing sides for 2z

pz is supposed to be a + not a -

Student C:

ReplyDeleteArithmetic error, when he transferred it over the equals sign, he forgot to change it to a divide sign.

Distributive error, did not distribute the - to the z

Student P has put way too many spaces. the Majority of the mistakes made by students is due to wrong conversions. They forgot about the - and + sometimes leading it to be confusing.

ReplyDeleteStudent G - Arithmetic error and did not bracket the numerator, confusion if [2z-(zp/p)] OR [(2z-zp)/p]

ReplyDeleteStudent D:

ReplyDeleteDid not divide the p

Student N

ReplyDeleteIt is supposed to be 2Z as the numerator of the fraction where P is the denominator and not the other way around with Z being the numerator.