# (Pre-calc) How do I simplify this expression?

How would I go about simplifying $4(a-2(b-c)-(a-(b-2)))$. Show working out and steps please.

I'd show my working out but I'm not really sure where to start. Firstly, I would want to get rid of the 4 so I'd times everything else by 4 right? No idea.

Consider re-writing the equation in different brackets. Mathematics has three different type of parentheses for a reason - to distinguish between each pair of brackets. \begin{align} 4(a-2(b-c)-(a-(b-2)))&=4\left\{a-2[b-c]-[a-(b-2)]\right\}\\ &=4\left\{a-2[b-c]-[a-b+2]\right\}\\ &=4\left\{a-2[b-c]-a+b-2\right\}\\ &=4\left\{a-2b+2c-a+b-2\right\}\\ &=4\left\{a-a-2b+b+2c-2\right\}\\ &=4\left\{-b+2c-2\right\}\\ &=-4b+8c-8\\ \end{align} As an exercise, figure out what I did step by step. This is very long-winded but I hope you see what happens as I remove brackets.
You can start by separating $4(a-2(b-c)-(a-(b-2)))$ into $4(a+?-(a-(b-2)))$ and $? = -2(b-c)$. Then solve $? = -2(a-c) = -2a+2c$ and put it back as $4(a-2a+2c-(a-(b-2)))$.
Also, from the other side, you could start by separating $4(a-2(b-c)-(a-(b-2)))$ into $? = 2(b-c)$, $! = a-(b-2)$ and $4(a-?-!)$. Then solve $4(a-?-!) = 4a - 4\cdot ? - 4\cdot!$ and put the extracted bits back as $4a - 4\cdot 2(b-c) - 4\cdot(a-(b-2))$.