How do you evaluate #7- 4[ 8- 7( 2- 3) ]#?

2 Answers

Answer:

Start from the inside and work out following the order of operations.

Explanation:

Start from the inside and work out following the order of operations
When doing the order of operations, Follow the PEDMAS order:
Parenthenses, Exponents, Divide, Multiply, Add, Subtract.

One way to remember this is If you get hurt in PE call an MD Asap

P = parentheses E = Exponent PE is physical education class.
M = multiplication D= Division MD is a medical doctor, a doctor is one person not two, so multiplication and division must be done together working from left to right
A = addition S = subtraction Asap is as soon as possible is one immediate time right now. so addition and subtraction must be done together working from left to right

In doing the order of operation do parenthesis and exponents first and then multiplication and division together and addition and subtraction together.

The first parentheses is # 7xx(2-3)# Take what is inside the bracket first

# 7 xx (2-3)# This equals

# 7 xx -1 = -7 # This leaves

# 7 - 4{ 8 - (-7) }# #( - xx - = +)# so -(-7) = +7 The result is

# 7- 4{ 8 +7} #

Think about this way taking away (-) something bad or negative(-) is good or positive (+)

Now add 8+7 = 15 to simplify the second parenthesis This gives

# 7- 4 (15) # # 4 xx 15 = 60 # so

7 - 60 after removing the second parenthesis.

7- 60 = - 53

Mar 6, 2017

Answer:

#7 - 4[8-7(2-3)] = -53#

Explanation:

Rather than blindly following PEDMAS, BEMDAS, BODMAS, BOMDAS or any other acronym, I prefer the concept of counting terms.

Terms are separated by + and - signs.

Within each term, the order applies of Brackets, (parentheses) first, then powers and roots and lastly multiply and divide.

Once each term has been simplified to a single answer, then we add or subtract the terms, either by working from left to right, or by re-arranging the terms with the additions at the front. (Just adding first often leads to wrong answers.)
Work with the entire expression, doing whatever is possible in each term.

#color(limegreen)(7) - 4[8-7(color(red)(2-3))]#

#=color(limegreen)(7) - 4[8-7(color(red)(-1))]#

#=color(limegreen)(7) - 4[color(blue)(8+7)]" "larr (-7xx-1 =+7)#

#=color(limegreen)(7) - 4[color(purple)(15)]#

#=color(limegreen)(7) - color(purple)(60)#

#=-53#