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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.

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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.

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Basic algebra is all about term rewriting. You need to identify complete parts of the expression, extract them, solve them and place the results back where they were taken from.

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 $4a - 4\cdot 2(b-c) - 4\cdot(a-(b-2))$.

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4(a−2(b−c)−(a−(b−2))) =4(a-2b+2c-(a-b+2)) =4(a-2b+2c-a+b-2) =4(2c-b-2) =8c-4b-8

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