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 ?
On
$(\frac19+\frac12)+(\frac13-\frac12)=\frac19+\frac13=\frac19+\frac39=\frac49$
In your proposed solution, you multiplied by various numbers that changed the value of the expression.
On
we get $$\frac{1}{9}+\frac{1}{2}=\frac{2+9}{18}=$$ and $$\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=$$ or $$\frac{1}{9}+\frac{1}{3}=\frac{1+3}{9}=\frac{4}{9}$$
On
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.
$$\left(\frac19+\frac12\right)+\left(\frac13-\frac12\right)\\ =\frac{18}{18}\left(\frac19+\frac12\right)+\frac66\left(\frac13-\frac12\right)\\=\frac{1}{18}(2+9)+\frac16(2-3)\\=\frac{11}{18}-\frac16=\frac{11-3}{18}\\=\frac8{18}=\frac49$$
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 $2+6$.