How do you evaluate #\frac { 2} { 3} - \frac { 4} { 5} + ( 6\cdot \frac { 1} { 3} ) - \frac { 8} { 2} + \sqrt { 4/ 9} - ( \frac { -3} { 7} ) ^ { 2}#?

1 Answer
MathFact-orials.blogspot.com
Feb 6, 2018

#-1213/735#

Explanation:

We'll follow PEDMAS:

PEDMAS

  • #color(red)(P)# - Parentheses (also known as Brackets)
  • #color(blue)(E)# - Exponents
  • #color(green)(M)# - Multiplication
  • #color(green)(D)# - Division (this has the same weight as M and so I gave it the same colour)
  • #color(brown)(A)# - Addition
  • #color(brown)(S)# - Subtraction - again, same weight as A and so the same colour)

We start with:

#2/3-4/5+(6xx1/3)-8/2+sqrt(4/9)-((-3)/7)^2#

First let's see that there is a bracket that needs working:

#2/3-4/5+color(red)((6xx1/3))-8/2+sqrt(4/9)-((-3)/7)^2#

#2/3-4/5+color(red)(2)-8/2+sqrt(4/9)-((-3)/7)^2#

We now have the last two terms that have exponents (the square root is exponent value #1/2#):

#2/3-4/5+2-8/2+color(blue)(sqrt(4/9))-color(blue)(((-3)/7)^2)#

#2/3-4/5+2-8/2+color(blue)(2/3)-color(blue)(9/49)#

There's no multiplication, but we can work a division:

#2/3-4/5+2-color(green)(8/2)+2/3-9/49#

#2/3-4/5+2-color(green)(4)+2/3-9/49#

Let's simplify a little before adding all this up (the 49 in the denominator is going to make for big fractions. If the intent was to only have the #-3# be squared, this answer will need to be modified).

#4/3-4/5-2-9/49#

Now we have a list of addition/subtraction that requires a common denominator. That denominator is #3xx5xx7xx7=735#:

#4/3(245/245)-4/5(147/147)-2(735/735)-9/49(15/15)#

#980/735-588/735-1470/735-135/735=-1213/735#

#1213# is prime and so we can't simplify further.

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