How do you evaluate #15- 2+ 7- 12\div 4#?

3 Answers
May 29, 2018

#17#

Explanation:

Using BODMAS

First Divide:
#15 − 2 + 7 − 12 ÷ 4#
#=15 − 2 + 7 − 3#

As there is no multiply you can move on.

Now, Add:
#15 + (-2) + 7 - 3#
#=15 + 5 - 3#

Here, to complete the addition be sure to take the '-' symbol with the number 2.

Add once again:
#15 + 5 - 3#
#=20 - 3#

Now Subtract:
#20 - 3#
#= 17#

May 29, 2018

#17#

Explanation:

#15-2+7-12 divide 4#

Here we use the order of operations PEMDAS/PEDMAS, whichever acronym it is you've been taught.

What many students misunderstand is that PEMDAS is just a way to memorize the order of operations, it's not the literal order of operations.

Let me break down the real order of operations for you:
(1) P - parenthesis
(2) E - exponent
(3) M or D - multiplication or division
(4) A or S - addition or subtraction

The (1) to (4) are the levels of priority. So how the order of operations works is that first you look at the level of priority and THEN you do the first operation that comes up from left to right.

For example, #2 * 5 divide 10#, multiplication and division are in the same level of priority so which one do you do first? The answer is you do the multiplication #2*5# first.

What happens if you get a #3 + 10 divide 2 * 4#, which operation do you do first? The answer is the division #10 divide 2#, multiplication and division have a higher priority than addition, and division came up before the multiplication, so you do the division. The same concept goes for the addition and subtraction level as well.

So back to our problem, #15-2+7-12 divide 4#.

We do the division first.

#=15-2+7 - 3#

#=13 + 7 - 3#

#=20 - 3#

#=17#

May 29, 2018

#17#

Explanation:

When you are working with an expression which has different operations, it is important to count the number of terms first.

Each term will simplify to a single answer and they can be added or subtracted in the last step.

Within each term, the order is:

  • brackets
  • powers and roots
  • multiply and divide

#color(purple)(15)" " color(lime)(-2)" " color(blue)(+7)" " color(red)(-12div4)" "larr # there are four terms.

Only the last one requires a calculation.

#color(purple)(15)" " color(lime)(-2)" " color(blue)(+7)" " color(red)(-12div4)#
#color(white)(xxxxxxxxxxxxxx)color(red)(darr)#
#=color(purple)(15)" " color(lime)(-2)" " color(blue)(+7)" " color(red)(-3)" "larr# each term is a single value.

To reduce errors, re-arrange the expression with the additions at the beginning, then add left to right.

#=color(purple)(15)" " color(blue)(+7)" " color(lime)(-2)" " color(red)(-3)#

#=22" "color(lime)(-2)" " color(red)(-3)#

#=20" " color(red)(-3)#

#=17#