How do I do this? #(3/8)^2 + 6/7cdot2/9 + (-5)#? #(3/8)^2 -: 6/7cdot2/9 + (-5)#? #(3/8)^2 + 6/7cdot2/9 -: (-5)#? #(3/8)^2 -: 6/7cdot2/9 -: (-5)#?

1 Answer
Truong-Son N.
Dec 20, 2017

It's hard to tell if some of the operators are #+# or #-:#. So there are four possibilities...

The answers for all four possibilities are #-6275/1344#, #-953/192#, #689/6720#, and #-7/960#, respectively. They couldn't be more different!

Recall the acronym PEMDAS for order of operations:

  • Parentheses or Exponents
  • Multiplication or Division
  • Addition or Subtraction

And our four options are...

  1. #(3/8)^2 + 6/7 cdot 2/9 + (-5)# #" "bb((1))#
  2. #(3/8)^2 -: 6/7 cdot 2/9 + (-5)# #" "bb((2))#
  3. #(3/8)^2 + 6/7 cdot 2/9 -: (-5)# #" "bb((3))#
  4. #(3/8)^2 -: 6/7 cdot 2/9 -: (-5)# #" "bb((4))#

OPTION 1

#(3/8)^2 + 6/7 cdot 2/9 + (-5)#

#= 9/64 + 6/7 cdot 2/9 + (-5)#
[apply the square exponent]

#= 9/64 + 12/63 + (-5)#
[multiply the middle terms]

#= 9/64 + 12/63 - 5#
[distribute the plus sign]

Now we require common denominators. You should of course use a calculator to manage this.

#= 9/64 cdot 63/63 + 12/63 cdot 64/64 - 5 cdot (63 cdot 64)/(63 cdot 64)#

Now apply the multiplication first.

#= 567/4032 + 768/4032 - 20160/4032#

Now add these together.

#= -18825/4032#

Both numbers are divisible by #3# because the sum of their digits is as well. This is as reduced as it gets:

#= color(blue)(-6275/1344)#

OPTION 2

Now that we have done this much, the others should be easier. But be sure to do multiplication/division from left to right! In this case, the new thing to us is that dividing is the same as multiplying by the reciprocal (the upside-down term).

#(3/8)^2 -: 6/7 cdot 2/9 + (-5)#

#= 9/64 cdot 7/6 cdot 2/9 + (-5)#

#= 9/64 cdot 7/27 + (-5)#

#= 63/1728 cdot (1//3)/(1//3) - 5#
[do the multiplication and apply the plus operation]

#= 21/576 - 5 cdot 576/576#
[get common denominators]

#= 21/576 - 2880/576#
[multiply through]

#= -2859/576#

This is as reduced as it gets:

#= color(blue)(-953/192)#

OPTION 3

See what happens when you have blurry images? :)

#(3/8)^2 + 6/7 cdot 2/9 -: (-5)#

#= 9/64 + 12/63 -: (-5)#

The first bit is almost the same as in #(1)#:

#= 567/4032 + 768/4032 -: (-5)#

This time, we do the division first. Careful, the #5# is negative.

#= 567/4032 + 768/4032 cdot -1/5#

#= 567/4032 cdot 5/5 - 768/20160#
[common denominators]

#= 2835/20160 - 768/20160#
[apply multiplication]

#= 2067/20160 cdot (1//3)/(1//3)#
[divide through]

#= color(blue)(689/6720)#

OPTION 4

Last one... A tip is to make all the similar operations (#xx//-:#, #+//-#) identical first before going from left to right.

#(3/8)^2 -: 6/7 cdot 2/9 -: (-5)#

#= 9/64 cdot 7/6 cdot 2/9 -: (-5)#

#= 126/3456 cdot -1/5#

#= -126/17280 cdot (1//3)/(1//3)#

#= -42/5760 cdot (1//3)/(1//3)#

#= -14/1920 cdot (1//2)/(1//2)#

#= color(blue)(-7/960)#

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