Fractions: adding and subtracting, numerical

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Ben Brawn

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From users who are members of pre-algebra Numeracy and Arithmetic :

Ben Brawn

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Ready to use

3 years, 5 months ago

From users who are not members of pre-algebra Numeracy and Arithmetic :

Paul Jainta

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Should not be used

3 years, 5 months ago

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Ben Brawn commented 3 years, 5 months ago

Fractions are in reduced form by design now and advice involves dealing with cases where the lowest common denominator is one of the denominators.

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Ben Brawn 3 years, 5 months ago

Gave some feedback: Ready to use

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Paul Jainta 3 years, 5 months ago

Gave some feedback: Should not be used

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Ben Brawn 3 years, 5 months ago

Published this.

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Ben Brawn 3 years, 5 months ago

Gave some feedback: Ready to use

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Ben Brawn 3 years, 5 months ago

Gave some feedback: Has some problems

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Ben Brawn 3 years, 5 months ago

Created this as a copy of Fractions: adding and subtracting, numerical, already with common denominator.

Name	Status	Author	Last Modified
Fractions: adding and subtracting, numerical, already with common denominator	Ready to use	Ben Brawn	11/02/2025 23:17
Fractions: adding and subtracting, algebraic, already with common denominator	Ready to use	Ben Brawn	26/03/2025 02:38
Fractions: adding and subtracting, numerical	Ready to use	Ben Brawn	11/02/2025 23:20
Fractions: adding and subtracting, algebraic	Ready to use	Ben Brawn	26/03/2025 02:38
Fractions: multiplying, algebraic	Ready to use	Ben Brawn	26/03/2025 04:18
Merryn's copy of Fractions: adding and subtracting, numerical	Ready to use	Merryn Horrocks	07/03/2024 01:07
Merryn's copy of Fractions: adding and subtracting, numerical, already with common denominator	Ready to use	Merryn Horrocks	07/03/2024 01:07

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Question statement

Express the following as a single fraction (using / as the fraction bar).

Give any introductory information the student needs.

			Name	Type	Generated Value

numbers	list	[ 19, 17 ]
num1	integer	19
num2	integer	17
addSub	integer	1
cf	integer	2
extrafactor	list	[ 5, 1 ]
den1	integer	10
den2	integer	2
ansden	integer	10
ansnum	integer	104
finalcf	number	2

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
shuffle([2,3,5,7,11,13,17,19,23,29])[0..2]

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `list`

[ 19, 17 ]

→ Used by:

num1
num2

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1. Gap Number entry
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$=$ Unnamed gap

Advice

We need to get the denominators to be the same, preferably the lowest common denominator, at which point, we can simply addsubtract the new numerators and put the result over the common denominator.

Since $\var{den1}=\var{cf}\times\var{extrafactor[0]}$ and $\var{den2}=\var{cf}\times \var{extrafactor[1]}$ the lowest common denominator will be the product $\var{cf}\times \var{extrafactor[0]}\times \var{extrafactor[1]}$ , that is, $\var{ansden}$ . Therefore, we will multiply the first fraction by $\frac{\var{extrafactor[1]}}{\var{extrafactor[1]}}$ and the second fraction by $\frac{\var{extrafactor[0]}}{\var{extrafactor[0]}}$ as follows.

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}+\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}\times \var{extrafactor[1]}}{\var{den1}\times \var{extrafactor[1]}}+\dfrac{\var{num2}\times \var{extrafactor[0]}}{\var{den2}\times \var{extrafactor[0]}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}}{\var{ansden}}+\dfrac{\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}+\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}-\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}\times \var{extrafactor[1]}}{\var{den1}\times \var{extrafactor[1]}}-\dfrac{\var{num2}\times \var{extrafactor[0]}}{\var{den2}\times \var{extrafactor[0]}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}}{\var{ansden}}-\dfrac{\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}-\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

Since $\var{den2}=\var{cf}\times \var{extrafactor[1]}$ and the other denominator is $\var{den1}$ , the lowest common denominator will be the product $\var{cf}\times \var{extrafactor[1]}$ , that is, $\var{ansden}$ . Therefore, we will multiply the first fraction by $\frac{\var{extrafactor[1]}}{\var{extrafactor[1]}}$ as follows.

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}+\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}\times \var{extrafactor[1]}}{\var{den1}\times \var{extrafactor[1]}}+\dfrac{\var{num2}}{\var{den2}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}}{\var{ansden}}+\dfrac{\var{num2}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}+\var{num2}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}-\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}\times \var{extrafactor[1]}}{\var{den1}\times \var{extrafactor[1]}}-\dfrac{\var{num2}}{\var{den2}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}}{\var{ansden}}-\dfrac{\var{num2}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1*extrafactor[1]}-\var{num2}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

Since $\var{den1}=\var{cf}\times\var{extrafactor[0]}$ and the other denominator is $\var{den2}$ , the lowest common denominator will be the product $\var{cf}\times \var{extrafactor[0]}$ , that is, $\var{ansden}$ . Therefore, we will multiply the second fraction by $\frac{\var{extrafactor[0]}}{\var{extrafactor[0]}}$ as follows.

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}+\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}}{\var{den1}}+\dfrac{\var{num2}\times \var{extrafactor[0]}}{\var{den2}\times \var{extrafactor[0]}}\\[3pt]&=\dfrac{\var{num1}}{\var{ansden}}+\dfrac{\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1}+\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

$\begin{align*}\dfrac{\var{num1}}{\var{den1}}-\dfrac{\var{num2}}{\var{den2}}&=\dfrac{\var{num1}}{\var{den1}}-\dfrac{\var{num2}\times \var{extrafactor[0]}}{\var{den2}\times \var{extrafactor[0]}}\\[3pt]&=\dfrac{\var{num1}}{\var{ansden}}-\dfrac{\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{num1}-\var{num2*extrafactor[0]}}{\var{ansden}}\\[3pt]&=\dfrac{\var{ansNum}}{\var{ansden}}\end{align*}$

Since there are no common factors between the two denominators (other than $1$ ) the lowest common denominator will just be the product of the two denominators. Therefore, we will multiply the first fraction by $\frac{\var{extrafactor[1]}}{\var{extrafactor[1]}}$ and the second fraction by $\frac{\var{extrafactor[0]}}{\var{extrafactor[0]}}$ as follows.

This is our final answer since there are no common factors to cancel.

At this point, we can simplify the fraction by noticing a common factor of $\var{finalcf}$ :

$\begin{align}\dfrac{\var{ansNum}}{\var{ansden}}&=\dfrac{\var{ansNum}\div\var{finalcf}}{\var{ansden}\div\var{finalcf}}\\&=\var[fractionnumbers]{{ansnum/ansden}}\end{align}$

Which in this case was actually the same as just performing the division $\dfrac{\var{ansNum}}{\var{ansden}}$ .

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This question is used in the following exams:

Pre-Algebra by Chris Templet in Chris's workspace.
Fractions: adding and subtracting, numeric and algebraic by Ben Brawn in Algebra.
addition by Michelle Kwok in Michelle's workspace.
Week 1 by P Oak in P's workspace.

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Fractions: adding and subtracting, numerical

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Ben Brawn commented 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Paul Jainta 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

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Generated value: `list`

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Fractions: adding and subtracting, numerical

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Ben Brawn commented 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Paul Jainta 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Ben Brawn 3 years, 5 months ago

Error in variable testing condition

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Generated value: list

Parts

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Generated value: `list`