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$6÷2(1+2)=$?

$6÷2\times3=9$

We need to solve first the parentheses or we can distribute this way:

$6÷(2\times1+2\times2)=$?

$6÷(2+4)=$?

$6÷6=1$

What is the correct? $9$ or $1$?

Chris Godsil
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deldev
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2 Answers2

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Which is the correct answer depends on whether the intended interpretation was $$6÷(2(1+2))$$ or $$(6÷2)(1+2)$$ There would not be the slightest amount of confusion if it were written $$\frac{6}{2(1+2)}$$ or $$\frac{6}{2}(1+2)$$

Wouter
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  • so, the first equation is wrote from wrong way? we can say that is an invalid equation? – deldev Feb 21 '18 at 13:04
  • The first equation is written in a way that is not optimized for clarity. It may even be a deliberate trick question. Even if careful application of operator precedence allows you to unambiguously assign a value, how confident are you that whoever wrote it was as careful as you are? – Wouter Feb 21 '18 at 13:09
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Go here to answer your question. Because answers that just provide links are discouraged — and for good reason too — I will give you some insight on how it approaches the problem. The link takes you to a YouTube video which is entirely dedicated to explaining the correct answer to the equation, $$6\div 2(1 +2) = \, ?$$ It uses an order-of-operations technique called PEDMAS / BODMAS such that, $$\begin{array}{cc} \mathbf{PEDMAS} & \mathbf{BODMAS} \\ \hline \\ \ \ \ \ \ \ \ \ \ \mathrm{Parentheses} \quad \ \ \diagup &\mathrm{Brackets} \\ \ \ \ \ \ \ \ \ \ \ \mathrm{Exponents} \quad \ \ \ \diagup &\mathrm{Orders} \\ \ \ \ \ \ \ \ \mathrm{Multiplication} \quad — &\mathrm{Division} \\ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{Addition} \quad \ \ \ \ — &\mathrm{Subtraction} \\ \end{array}$$

Mr Pie
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