How do you simplify #3+ ( 4- ( 6- ( 1- 6- - 16\div 4) ) ) \cdot - 5#?

2 Answers
Sep 22, 2016

Answer:

18

Explanation:

There is one rule in order of operations, follow BODMAS.
Brackets (do the deepest level of brackets first), then orders (for example powers), followed by division, multiplication, addition and finally subtraction.

Deepest level of bracket first;

# (1-6--16/4) to (1-6--4) to (-5+4) to -1#

Note that even with that bracket, we follow BODMAS and do the division before subtraction

Then the next level of bracket;

#(6-(1-6--16/4)) to (6-(-1)) to 7#

Then the final, outermost set of brackets;

#(4- (6-(1-6--16/4)) to (4-(6-(-1)) to (4-(7)) to -3#

Then we multiply by #-5#

#-3 x -5 = 15#

and finally add #3#

#15 + 3 = 18#

Sep 22, 2016

Answer:

#18#

Explanation:

Count the number of terms first.
There are 2 terms - keep them separate until the last step.
Start by removing the innermost brackets each time.

#color(red)(3)+ ( 4- ( 6- ( 1- 6- (color(blue)(- 16 div 4)))) xx - 5#

#color(red)(3)+ ( 4- ( 6- ( 1- 6 color(lime)(-(-4))))xx - 5#

#color(red)(3)+ ( 4- ( 6- ( 1- 6 color(lime)(+4))xx - 5#

#color(red)(3)+ ( 4- ( 6- (color(magenta)( 1- 6 +4))))xx - 5#

#color(red)(3)+ ( 4- ( 6- (color(magenta)( 1+4-6))))xx - 5" "larr# re-arrange terms

#color(red)(3)+ ( 4- ( 6- (color(magenta)( -1))))xx - 5#

#color(red)(3)+ ( 4- (color(tomato)(6+1))xx - 5#

#color(red)(3)+ (color(sienna)( 4- 7)))xx - 5#

#color(red)(3)+ color(darkorange)(( -3xx - 5)#

#color(red)(3)+ color(darkorange)15#

#18#