We have $$a-c/2b$$ $$\text{minus}$$ $$aY-cY/4b^2$$
This simplifies to $$2ba - 2bc + cY - aY/4b^2$$
But, I don't understand how to do this (new to this kind of maths). I'm confused specifically because the first fraction has $2b$ on the denominator whereas the second has $4b^2$ on the denominator. Since denominators have to be the same to add/subtract, I don't know how to get the 2b to be the same as $4b^2.$
Can anyone walk me through this?
Thanks!
On
You wrote: $$a-c/2b$$ $$\text{minus}$$ $$aY-cY/4b^2$$
But it appears that what you actually meant was this: $$(a-c)/(2b)$$ $$\text{minus}$$ $$(aY-cY)/(4b^2)$$
In standard usage, $a-c/2b$ means $a$ minus a fraction, and the numerator of that fraction is $c.$
So you have $$ \frac{a-c}{2b} - \frac{aY-cY}{4b^2}. $$ The way to do this is first to multiply both the numerator and the denominator of that first fraction by $2b,$ so that its denominator becomes $4b^2.$ Then you have $$ \frac{2b(a-c)}{4b^2} - \frac{aY-cY}{4b^2} $$ which becomes $$ \frac{2ab - 2bc}{4b^2} - \frac{aY - cY}{4b^2}. $$ Then subtract the numerators: $$ \frac{2ab-2bc - (aY-cY)}{4b^2} $$ and then distribute the minus sign: $$ \frac{2ab-2bc-aY+cY}{4b^2}. $$
it is $$\frac{(a-c)2b}{4b^2}-\frac{aY-bY}{4b^2}=\frac{2ab-2bc-aY+cY}{4b^2}$$