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Please could someone explain what is wrong with my workings?

$$\left(\frac19+\frac12\right)+\left(\frac13-\frac12\right)=?$$

My workings are:

$$\left(18\left(\frac19+\frac12\right)\right)+\left(6\left(\frac13-\frac12\right)\right) = (2+6)+(2-3)=8-1=7$$

WHERE DID I GO WRONG ?

user8143349
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4 Answers4

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we get $$\frac{1}{9}+\frac{1}{2}=\frac{2+9}{18}=$$ and $$\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=$$ or $$\frac{1}{9}+\frac{1}{3}=\frac{1+3}{9}=\frac{4}{9}$$

Dr. Sonnhard Graubner
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Suppose you had $2+3$. Then answer would be $5$. If you decided to multiply by $6$, your answer would change to $6(2+3)=30.$ Same thing happens in case of fractions. If you multiply the equation with some number, you have to divide by it too so that the value of the equation doesn't change. This is because $\frac mm=1.$ So multiplying and dividing by a number is equivalent to multiplying it by $1$ which doesn't change the value.

$$\left(\frac19+\frac12\right)+\left(\frac13-\frac12\right)\\ =\frac{18}{18}\left(\frac19+\frac12\right)+\frac66\left(\frac13-\frac12\right)\\=\frac{1}{18}(2+9)+\frac16(2-3)\\=\frac{11}{18}-\frac16=\frac{11-3}{18}\\=\frac8{18}=\frac49$$

Sahiba Arora
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  • Thank you but why did you multiply by 18/18 on one part of the expression and 6/6 on thr other part ? – user8143349 Jun 17 '17 at 12:11
  • Well, the same reason you multiplied by $18$ on one part and $6$ on the other. – Sahiba Arora Jun 17 '17 at 12:12
  • But my workings were wrong. – user8143349 Jun 17 '17 at 12:12
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    Because you multiplied by $18$ and not $\frac{18}{18}=1$. The LCM of $2$ and $9$ is $18$ and the LCM of $3$ and $2$ is $6$. Thus I multiplied with $18/18$ on one part and $6/6$ on another. – Sahiba Arora Jun 17 '17 at 12:15
  • @user8143349 You were right to pick $18$ and $6$, but you can't multiply equations by numbers like that or you will change the value of the equation. If you want to use the $18$ and $6$, because it simplifies the fractions, but don't want to change the value of the equation (which you should try never to do), then you need to divide by $18$ and $6$ as well. Because if you multiply by $18$ and then divide by $18$, that's like multiplying by $18/18$, which is $1$! Multiplying by $1$ doesn't change the value, because any number multiplied by $1$ is that same number again. – Ducky Jun 17 '17 at 12:21
  • Thank you. I've never learn't that at secondary school neither at college ! – user8143349 Jun 17 '17 at 12:23
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    @user8143349 There is a similar principle when adding. If you for some reason had an equation and wanted to add $3$ to one side of it to simplify it, then you can't just stop there because you've changed the value of your equation and it may no longer be true. In this case you would need to subtract $3$ as well, so that in total you've added $3-3=0$, and adding $0$ does not change the value of any number. – Ducky Jun 17 '17 at 12:24
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If you multiply some part by any number you also must divide by that number!

So you get: $$\frac{\left(18\left(\frac19+\frac12\right)\right)}{18}+\frac{\left(6\left(\frac13-\frac12\right)\right)}{6} = \frac{(2+9)}{18}+\frac{(2-3)}{6}=11/18-1/6=\frac{11-3}{18}=\frac{4}{9}$$

Also you made a miscalculation: $18(1/9+1/2)=2+9 $ and not $2+6$.

Mesmerized student
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$(\frac19+\frac12)+(\frac13-\frac12)=\frac19+\frac13=\frac19+\frac39=\frac49$

In your proposed solution, you multiplied by various numbers that changed the value of the expression.

paw88789
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  • So am I not allowed to multiply different parts of an equation with different numbers? – user8143349 Jun 17 '17 at 12:02
  • @user8143349: No, you aren't. If you were allowed to do that, you could get from $1+1=2$ to $3\cdot 1+5\cdot 1=7\cdot 2$, that is $8=14$, which is clearly wrong. – celtschk Jun 17 '17 at 12:52