How do you solve #(5\cdot 3\cdot 2) \div ( 6+ 3\cdot - 1+ 2)#?

Question

Mila Anderson · Accepted Answer

6

We deal with the mixed operations in this calculation in the order set out in the acronym PEMDAS. That is by evaluating brackets first. #(5xx3xx2)÷(6+color(red)(3xx(-1))+2)# #=30÷(6color(red)(-3)+2)larr"multiplication before addition"# #=30÷5=6#

Roman Alvarez · Answer

#6#

In any expression involving different operations, count the number of terms first.

Each term will simplify to a single answer and these are added or subtracted in the last step

Within each term, the normal order of operation applies.

Brackets first, then work from the strongest to weakest operations : powers and roots then multiply and divide

add and subtract LAST

#(color(red)(5xx 3xx 2))div ( 6+ color(blue)(3xx (- 1)) + 2)" "larr# there is only 1 term

=#(color(red)(30))div ( 6 color(blue)(-3) + 2)#

= #30 div 5#

=#" "6#

Roman Alvarez · Answer

undefined To solve $ (5 \cdot 3 \cdot 2) \div (6 + 3 \cdot -1 + 2) $, follow the order of operations (PEMDAS/BODMAS):

1. Perform operations within parentheses first:
$$ 6 + 3 \cdot -1 + 2 = 6 - 3 + 2 = 5 $$

2. Substitute the result back into the original expression:
$$ (5 \cdot 3 \cdot 2) \div 5 $$

3. Perform multiplication:
$$ 5 \cdot 3 \cdot 2 = 30 $$

4. Perform division:
$$ \frac{30}{5} = 6 $$

So, $ (5 \cdot 3 \cdot 2) \div (6 + 3 \cdot -1 + 2) = 6 $.