How do you evaluate #(2+ 3) ^ { 2} - 4\times 5\div 3#?

2 Answers
Mar 15, 2018

Answer:

Using PEMDAS, you can calculate the solution to be #18 1/3#

Explanation:

Remember that PEMDAS defines the order of operations for all arithmetic. PEMDAS stands for:

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

So starting from the top of the acronym, we evaluate the terms inside of the parentheses:

#2+3=5#

Making our expression:

#5^2-4xx5-:3#

Now, we work on exponents. The only exponent here is the #5^2# term, and that evaluates to: #5^2=25#. Let's put that in the expression:

#25-4xx5-:3#

We're now at multiplication and division. we have one multiplication and one division exercise to do, let's combine them! Since the 4, 5, and 3 are all together, you can re-write that expression like so:

#4xx5-:3 rArr (4xx5)/3 rArr 20/3#

What I did here was that instead of dividing by 3, I multiplied by its inverse, #1/3#. This way, I was able to do both multiplications at the same time!

Finally, We'll skip addition (there's no addition in this expression!) and go straight to subtraction:

#25-20/3#

Let's raise the 25 so it's a function of thirds, making the arithmetic slightly easier. We'll then put that modified fraction into the expression:

#25*3/3=75/3 rArr 75/3-20/3=55/3#

Now we have a solution as an improper fraction, #55/3#. Let's make it a mixed fraction to finish things off:

#55/3=color(red)(18 1/3)#

Mar 16, 2018

Answer:

#18 1/3#

Explanation:

In any expression with multiple operations, identify the individual terms first:

#color(blue)((2+3)^2)color(green)( -4xx5 div3)" "larr# there are two terms.
#color(white)(xx)darrcolor(white)(xxxx)darr#
#=color(blue)((5)^2)color(green)(" "-20 div3)#
#color(white)(xx)darrcolor(white)(xxxxxx)darr#
#=color(blue)(25)color(green)(" "-" "20/3)#

#=25-6 2/3#

#= 19- 2/3#

#=18 1/3#