Adding and subtracting fractions

[Probability]

So I've realised where I was previously confused with making numbers up when cancelling fractions (See other thread, cancelling fractions) When I thought I could make a number up that was different for the numerator and denominator I was wrong, but I knew I'd got the idea from somewhere, and now I think I've remembered where that idea came from.

Suppose you have a fraction;
[MATH]{4}/{9}+{5}/{6}[/MATH]The first part of the understanding now appears to be, how do I make the denominators the same! Mathematically I use the LCM of both (9) and (6). When you understand the subject, it is then easy to just multiply (9) by (2) and (6) by (3). I'll end up with (18) as a denominator for both fractions. You can see that I have just made up two numbers to find the LCM of (6) and (9), which is (18). (I thought I could do this previously cancelling fractions) but I know now that I can't, and although the idea was there to work out how to make the denominators the same, I'd never realised until now that the LCM is the correct method to determine how to make the denominators the same value. (another valued lesson learned)

The next problem now for me is what to do with the numerators!
[MATH]{4}[/MATH] and [MATH]{9}[/MATH]I can't just add them as this would be mathematically incorrect, so can I cross multiply the original fractions before I found the LCM?
[MATH]{4}/{9}+{5}/{6}[/MATH][MATH]{4}X{6}={24}[/MATH] and [MATH]{5}X{9}={45}[/MATH]Now I can say;
[MATH]{4}/{9}+{5}/{6}={24}/{18}+{45}/{18}={69}/{18}[/MATH]Now I know about cancelling fractions, but I'm now learning again because the LCM (18) could be cancelled down to (6) but for an unknown reason this is not carried out, but (69) divided by (3) = [MATH]{23}/{18}[/MATH]
Now alternatively I could have calculated the numerators by multiplying [MATH]{4}/{9}[/MATH]by[MATH]{2}[/MATH] and [MATH]{5}/{6}[/MATH] by [MATH]{3}[/MATH] but them numbers I've just made up!

I can't even say in this latter method that I've multiplied one fraction numerator with the same integer as the second numerator, as I've used different integers.

Somewhere here there is another way of thinking to work out fractions!

By the way [MATH]{23}/{18}[/MATH] can be converted to [MATH]{1}[/MATH] and [MATH]{5}/{18}[/MATH]
Any views on this subject would be greatly appreciated.

 

[Deleted member 4993]

So I've realised where I was previously confused with making numbers up when cancelling fractions (See other thread, cancelling fractions) When I thought I could make a number up that was different for the numerator and denominator I was wrong, but I knew I'd got the idea from somewhere, and now I think I've remembered where that idea came from.

Suppose you have a fraction;
[MATH]{4}/{9}+{5}/{6}[/MATH]The first part of the understanding now appears to be, how do I make the denominators the same! Mathematically I use the LCM of both (9) and (6). When you understand the subject, it is then easy to just multiply (9) by (2) and (6) by (3). I'll end up with (18) as a denominator for both fractions. You can see that I have just made up two numbers to find the LCM of (6) and (9), which is (18). (I thought I could do this previously cancelling fractions) but I know now that I can't, and although the idea was there to work out how to make the denominators the same, I'd never realised until now that the LCM is the correct method to determine how to make the denominators the same value. (another valued lesson learned)

The next problem now for me is what to do with the numerators!
[MATH]{4}[/MATH] and [MATH]{9}[/MATH]I can't just add them as this would be mathematically incorrect, so can I cross multiply the original fractions before I found the LCM?
[MATH]{4}/{9}+{5}/{6}[/MATH][MATH]{4}X{6}={24}[/MATH] and [MATH]{5}X{9}={45}[/MATH]Now I can say;
[MATH]{4}/{9}+{5}/{6}={24}/{18}+{45}/{18}={69}/{18}[/MATH]Now I know about cancelling fractions, but I'm now learning again because the LCM (18) could be cancelled down to (6) but for an unknown reason this is not carried out, but (69) divided by (3) = [MATH]{23}/{18}[/MATH]
Now alternatively I could have calculated the numerators by multiplying [MATH]{4}/{9}[/MATH]by[MATH]{2}[/MATH] and [MATH]{5}/{6}[/MATH] by [MATH]{3}[/MATH] but them numbers I've just made up!

I can't even say in this latter method that I've multiplied one fraction numerator with the same integer as the second numerator, as I've used different integers.

Somewhere here there is another way of thinking to work out fractions!

By the way [MATH]{23}/{18}[/MATH] can be converted to [MATH]{1}[/MATH] and [MATH]{5}/{18}[/MATH]
Any views on this subject would be greatly appreciated.

You may want to read:

https://search.freefind.com/find.ht...&_charset_=UTF-8&bcd=÷&query=Adding+fractions

 

[Steven G]

Whenever you want to change what a fractions looks like, (if you want to change the denominator from 6 to 8) you multiply by 1. Multiplying by 1 does not change the value. You just have to understand that 1 is not the only way to write one. 7/7, .9/.9, (-2/3)/(-2/3),... all equal 1.

If you 2/3 to have an 18 in the denominator you multiply 2/3 by 6/6. Do you see that?

 

[Probability]

Whenever you want to change what a fractions looks like, (if you want to change the denominator from 6 to 8) you multiply by 1. Multiplying by 1 does not change the value. You just have to understand that 1 is not the only way to write one. 7/7, .9/.9, (-2/3)/(-2/3),... all equal 1.

If you 2/3 to have an 18 in the denominator you multiply 2/3 by 6/6. Do you see that?

I need to come back to that when I've resolved this misunderstanding that I'm trying to solve!

I have a fraction;

2 (1/7) + 4 (2/7) = (2+4) + (1/7) + (2/7) = 6 + (7/7) + (14/7) = (21/7) = 6 (3/7)

I reasonable confident its correct although this is the long version to a solution.

 

[Steven G]

I have a number of comments to make.

First, 2(1/7) is a multiplication problem. That is, 2 (1/7) = 2*(1/7) = 2/7.

If you want to write 2 and 1/7, then please write 2 1/7 NOT 2 (1/7)

2 1/7 + 4 2/7 = (2+4) + (1/7) + (2/7) = 6 + (7/7) + (14/7) = (21/7) = 6 (3/7)

I agree with the 1st equal sign---good!

Then you replaced 1/7 with 7/7 and 2/7 with 14/7. Those are not equal. 1/7 < 1 while 7/7 =1 !! 2/7<1 while 14/7=2 !!

I disagree with the 3rd equal sign. I agree that 7/7 + 14/7 = 21/7 which is what you got BUT you were supposed to add 6 + 7/7 + 14/7!! What happened to the 6.

The last equal sign is wrong. You should know and I bet do know that 21/7 = 3!!!! But you wrote that 21/7 = 6 3/7. What happened???

 

[Probability]

I have a number of comments to make.

First, 2(1/7) is a multiplication problem. That is, 2 (1/7) = 2*(1/7) = 2/7.

If you want to write 2 and 1/7, then please write 2 1/7 NOT 2 (1/7)

2 1/7 + 4 2/7 = (2+4) + (1/7) + (2/7) = 6 + (7/7) + (14/7) = (21/7) = 6 (3/7)

I agree with the 1st equal sign---good!

Then you replaced 1/7 with 7/7 and 2/7 with 14/7. Those are not equal. 1/7 < 1 while 7/7 =1 !! 2/7<1 while 14/7=2 !!

I disagree with the 3rd equal sign. I agree that 7/7 + 14/7 = 21/7 which is what you got BUT you were supposed to add 6 + 7/7 + 14/7!! What happened to the 6.

The last equal sign is wrong. You should know and I bet do know that 21/7 = 3!!!! But you wrote that 21/7 = 6 3/7. What happened???

Would you or somebody good with latex give me a fast course in writing this;

2 1/7 + 4 2/7 = (2+4) + (1/7) + (2/7) = 6 + (7/7) + (14/7) = (21/7) = 6 (3/7) in latex format.

I can see loads of confusion occurring. Another member commented on me not using brackets and now using them is also confusing. I need the crash course in latex to solve this confusion on the forum.

 

[lev888]

Would you or somebody good with latex give me a fast course in writing this;

2 1/7 + 4 2/7 = (2+4) + (1/7) + (2/7) = 6 + (7/7) + (14/7) = (21/7) = 6 (3/7) in latex format.

I can see loads of confusion occurring. Another member commented on me not using brackets and now using them is also confusing. I need the crash course in latex to solve this confusion on the forum.

LaTeX - Basic Code

How to write basic code in math mode.

www.malinc.se

It's linked from "Read before posting" (not directly).

 

[Probability]

[MATH]2\frac{1}{7}[/MATH][MATH]+4\frac{2}{7}[/MATH][MATH]=(2+4)+\frac{1}{7}+\frac{2}{7}=6+\frac{7}{7}+\frac{14}{7}=\frac{21}{7}=6\frac{3}{7}[/MATH]
Using cross multiplication I believe the above to be correct.

Same fraction using mixed fraction form;

[MATH]2\frac{1}{7}+4\frac{2}{7}=\frac{15}{7}+\frac{30}{7}=\frac{45}{7}=6\frac{3}{7}[/MATH]
Jomo if I getting something wrong here please post back and advise.

Thanks lev888 for the latex link as you can see it has been most helpful.

 

[Steven G]

I have the same exact comments as before. You replaced 1/7 with 7/7 and they are not equal. You replaced 2/7 with 14/7 and they are not equal. You claim that 21/7 equals 6 3/7. 21/7 equals 3 and 3 does not equal 6 3/7!

You can not divide DIFFERENT numbers say by 9 and think they will be equal!

 

[Probability]

I'll get back to you soon Jomo.

 

[Deleted member 4993]

I have the same exact comments as before. You replaced 1/7 with 7/7 and they are not equal. You replaced 2/7 with 14/7 and they are not equal. You claim that 21/7 equals 6 3/7. 21/7 equals 3 and 3 does not equal 6 3/7!

You can not divide DIFFERENT numbers say by 9 and think they will be equal!

I just want to point out:

You can multiply 1/7 by 7/7 -the product will be equivalent fraction.

You can replace 1/7 by 49/7 - which is again equivalent fraction.

 

[Probability]

I think I have been seriously over thinking this and for some reason trying to make it much more complicated than it needs to be.

[MATH]2\frac{1}{7}+4\frac{2}{7}=(2+4)+\frac{1}{7}+\frac{2}{7}=6\frac{3}{7}=\frac{45}{7}[/MATH]
Why I was cross multiplying the fractions[MATH]\frac{1}{7} and \frac{2}{7}[/MATH]I've no idea now!!

 

[Deleted member 4993]

I think I have been seriously over thinking this and for some reason trying to make it much more complicated than it needs to be.

[MATH]2\frac{1}{7}+4\frac{2}{7}=(2+4)+\frac{1}{7}+\frac{2}{7}=6\frac{3}{7}=\frac{45}{7}[/MATH]
Why I was cross multiplying the fractions[MATH]\frac{1}{7} and \frac{2}{7}[/MATH]I've no idea now!!

Your process is correct but unnecessarily long!

[MATH]2\frac{1}{7}+4\frac{2}{7}=\frac{15}{7}+\frac{30}{7}=\frac{45}{7}[/MATH]

 

[Probability]

Your process is correct but unnecessarily long!

[MATH]2\frac{1}{7}+4\frac{2}{7}=\frac{15}{7}+\frac{30}{7}=\frac{45}{7}[/MATH]

Its the work book I'm following instructions from, they show different or alternative methods to solve fractions, which this one is a question in the book. Your method above was my first one in the book which I did not mention that method because I thought it was straight forward enough to complete it that way. The alternative method I got a bit too involved in it. Thank you for your advice.

 

[Steven G]

Your process is correct but unnecessarily long!

[MATH]2\frac{1}{7}+4\frac{2}{7}=\frac{15}{7}+\frac{30}{7}=\frac{45}{7}[/MATH]

To op: I agree with Subhotosh's work IF you want to get a single fraction. But if you want to end up with a mixed number then the way you did it is quicker. For the record, mixed numbers are better since you have a better feel for the final number.