7.2: Exercises 7

Basics

You are paid $\$2+3\cdot 4$ per hour for a job that takes $10$ hours to complete. Your employer wants to pay you $\$14$ per hour but you feel cheated. You think you earned $\$20$ per hour. Who is right?

$$\$2+3\cdot 4=2+12=14$$

or

$$\$(2+3)\cdot 4=5\cdot 4=20?$$

In order to avoid confusion like this, smart people have agreed on the following order of operations:

1. If there is an innermost group (like Parentheses), perform the following in that group. The whole expression is group P if there is no other inermost group.
2. Carry out Exponentiation (if applicable) in that group.
3. Perform Multiplication and/or Division (in the order of occurrence from left to right), if any, in that group.
4. Perform Addition and/or Subtraction (in the order of occurrence from left to right), if any, in that group.

You are now down to one number (in arithmetic). Repeat steps $1-4$ in the next innermost group. If there is no innermost group, the entire expression is the group.

The term “Parentheses" above is misleading. It could refer to an expression inside parentheses, brackets, braces, under a square root sign, between absolute values, a numerator, a denominator, ….

Example $\PageIndex{1}$

Simplify $4+5[3^2(4^2\div 2^3\cdot 3-5-1)+7]$

Solution

(0,30)(0,-15) (0,0)$\begin{array}{cl lll lll} &4+5[3^2(\underbrace{4^2\div 2^3\cdot 3-5-1})+7]\ \ \hbox{Copy every unmodified symbol.}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{{\bf \large P}EMDAS -- innermost parentheses.}\\[10pt] =&4+5[3^2(16\div 8\cdot 3-5-1)+7]\ \ \hbox{Exponentiation P{\bf \large E}MDAS}\\[10pt] =&4+5[3^2(2\cdot 3-5-1)+7]\ \ \hbox{Division precedes multiplication.PEM{\bf \large D}AS}\\[10pt] =&4+5[3^2(6-5-1)+7]\ \ \hbox{Multiply PE{\bf \large M}DAS}\\[10pt] =&4+5[3^2(1-1)+7]\ \ \hbox{Subtract. PEMD{\bf \large A}S}\\[10pt] =&4+5[\underbrace{3^2(0)+7}\ \ \hbox{New group. {\bf \large PE}MDAS. Exponentiation.}\\[10pt] =&4+5[9(0)+7]\ \ \hbox{Multiplication. PE{\bf \large M}DAS}\\[10pt] =&4+5[0+7]\ \ \hbox{Addition. PEMD{\bf \large A}S}\\[10pt] =&\underbrace{4+5[7]}\ \ \hbox{New Group -- the entire expression -- Multiplication. PE{\bf \large M}DAS}\\[10pt] =&4+35=39\ \ \hbox{Addition. PEMD{\bf \large A}S} \end{array}$

Example $\PageIndex{1}$

Simplify $\displaystyle \frac{15+50 \div 5 \cdot 3}{50+15\div (3\cdot 5)-6}$

Solution

$\begin{array}{cl lll} &\displaystyle \frac{15+\underbrace{50 \div 5} \cdot 3}{50+15\div (\underbrace{3\cdot 5})-6}&\displaystyle \frac{pem{\bf \large D}as}{pe{\bf \large M}das}\\[20pt] =&\displaystyle \frac{15+10 \cdot 3}{50+15\div (15)-6}&\displaystyle \frac{pe{\bf \large M}das}{pem{\bf \large D}as}\\[20pt] =&\displaystyle \frac{15+30}{50+1-6}&\displaystyle \frac{pemd{\bf \large A}s}{pemd{\bf \large A}s}\\[20pt] =&\displaystyle \frac{45}{51-6}=1&\displaystyle \frac{45}{45}=\displaystyle \frac{\cdots}{pemda{\bf \large S}} %\\[20pt] %=&\displaystyle \frac{15+30}{50+1-6}\\[10pt] %=&\\[20pt] %=&1 \end{array}$

If you get the correct answer for the next example the first time, without help from any source, you are exceptionally good. Most students make at least one of three common mistakes. Copy all unmodified symbols vertically. Do scratch work on the right.

Example $\PageIndex{1}$

Simplify $8-5(2^7\div 2^4\cdot 2^3+6-5\cdot 3-15\div 3-20-10-5 )$

Solution

$\begin{array}{cl|l lll} \!\!\!&\!8\!-\!5(2^7\!\div \!2^4\!\cdot\! 2^3\!+\!6\!-\!5\cdot 3\!-\!15\div 3\!-\!20\!-\!10\!-\!5)\!&\!2^3=8, \ 2^4\!=16,\\[10pt] \!\!\!=\!\!\!&\!\!\!8\!-\!5(128\!\div \!16\!\cdot \!8\!+\!6\!-\!5\cdot 3\!-\!15\div 3\!-\!20\!-\!10\!-\!5)\!\!&\!\!2^7\!\!=\!\!(2^3)\!(2^4)\!=\!(8)\!(16)\!=\!128.\\[10pt] \!\!\!=\!\!\!&8-5(8\cdot 8+6-5\cdot 3-5-20-10-5)&\hbox{Division first.}\\[10pt] \!\!\!=\!\!\!&8-5(64+6-15-5-20-10-5)&\hbox{Multiplication next.}\\[10pt] \!\!\!=\!\!\!&8-5(70-15-5-20-10-5)&\hbox{Addition.}\\[10pt] \!\!\!=\!\!\!&8-5(55-5-20-10-5)&\hbox{Subtraction.}\\[10pt] \!\!\!=\!\!\!&8-5(50-20-10-5)&\hbox{Subtraction.}\\[10pt] \!\!\!=\!\!\!&8-5(30-10-5)&\hbox{Subtraction.}\\[10pt] %\!\!\!=\!\!\!&8-5(20-5)&\hbox{Subtraction.}\\[10pt] \!\!\!=\!\!\!&8-5(15)&\hbox{Subtraction.}\\[10pt] \!\!\!=\!\!\!&8-75&\hbox{Multiplication.}\\[10pt] \!\!\!=\!\!\!&-67&\hbox{Subtraction.} \end{array}$

Note 1:

$$\begin{align*} 2^7\div 2^4\cdot 2^3 &= \displaystyle \frac{2^7}{2^4}\cdot 2^3 \\[4pt] &=\frac{2\cdot \not 2\cdot \not 2\cdot \not 2\cdot \not 2\cdot 2\cdot 2 } {\not 2\cdot \not 2\cdot \not 2\cdot \not 2}\cdot 2^3 \\[4pt] &=2^3\cdot 2^3=2^{3+3}=2^6 \end{align*}$$

Note 2:

• First common mistake: $$8-5=3$$ (not in this example!)
• Second common mistake: $$2^7\div 2^4\cdot 2^3=2^7\div (2^4\cdot 2^3)=2^7\div 2^7=1$$ WRONG. The offender belongs to the IPS (Invisible Parentheses Society) and needs to give up his/her membership immediately.
• Third common mistake: $6-5=1$ (not here) $3-15=-12$ (not here) $3-20=-17$ (not here) $20-10=10$ (not here) $15-5=10$.

Note 3:

You will be able to take shortcuts once you become proficient. If you have doubts, follow the order of operations as outlined here.

Exercises 7

1. Simplify $4-3\cdot 15$
2. Simplify $4-3\cdot 15\div 3$
3. Simplify $4-81\div 3\div 3$
4. Simplify $(4-15)\cdot 3\div 3$
5. Simplify $\displaystyle \frac{-18\cdot 2+9}{-(-18)-(2+16)}$
6. Simplify $5+3^2[2^9\div 2^7\cdot 2^4+2\cdot 3-2(5^2-2^5)-30-40]$

1. Simplify $4-3\cdot 15$
   Solution:
   $\begin{array}{rcl lll} 4-3\cdot 15&=&4-45\\ &=&-41 \end{array}$
2. Simplify $4-3\cdot 15\div 3$
   Solution:
   $\begin{array}{rcl lll} 4-3\cdot 15\div 3&=&4-45\div 3\\ &=&4-15\\ &=&-11 \end{array}$
3. Simplify $4-81\div 3\div 3$
   Solution:
   $\begin{array}{rcl lll} 4-81\div 3\div 3&=&4-27\div 3\\ &=&4-9\\ &=&-5 \end{array}$
4. Simplify $(4-15)\cdot 3\div 3$
   Solution:
   $\begin{array}{rcl lll} (4-15)\cdot 3\div 3&=&(-11)\cdot 3\div 3\\ &=&-33\div 3\\ &=&-11 \end{array}$
5. Simplify $\displaystyle \frac{-18\cdot 2+9}{-(-18)-(2+9)}$
   Solution:
   $\begin{array}{rcl lll} \displaystyle \frac{-18\cdot 2+9}{-(-18)-(2+16)}&=&\displaystyle \frac{-36+9}{-(-18)-(18)}\\[10pt] &=&\displaystyle \frac{-27}{18-18} &=&\displaystyle \frac{-27}{0} \end{array}$
   The operation of dividing by 0 is undefined. (Do not use $\emptyset$)
6. Simplify $5+3^2[2^9\div 2^7\cdot 2^4+2\cdot 3-2(5^2-2^5)-30-40]$

   Solution:

   $\begin{array}{cl lll} &5+3^2[2^9\div 2^7\cdot 2^4+2\cdot 3-2(5^2-2^5)-30-40]\\[5pt] =&5+3^2[2^9\div 2^7\cdot 2^4+2\cdot 3-2(25-32)-30-40]\\[5pt] =&5+3^2[2^9\div 2^7\cdot 2^4+2\cdot 3-2(-7)-30-40]\\[5pt] =&5+3^2[512\div 128\cdot 16+2\cdot 3-2(-7)-30-40]\\[5pt] =&5+3^2[4\cdot 16+6-2(-7)-30-40]\\[5pt] =&5+3^2[64+6-(-14)-30-40]\\[5pt] =&5+3^2[64+6+14-30-40]\\[5pt] =&5+3^2[70+14-30-40]\\[5pt] =&5+3^2[84-30-40]\\[5pt] =&5+3^2[54-40]\\[5pt] =&5+3^2[14]\\[5pt] =&5+9[14]\\[5pt] =&5+126\\[5pt] =&131 \end{array}$