10.2: Reversing Multiplication/Division

Suppose $2+5$ pounds of grain are loaded on the left scale of a balance. At the same time $7$ pounds are put on the right side.

The balance is in equilibrium.

If the number of pounds on the left is doubled $2(2+5)$ and simultaneously the number on the right side is doubled $2(7)$, the balance will remain in equilibrium.

Similarly we can divide the left side by 7 if we divide the right side by 7.

$\displaystyle \frac{2+5}{7}=\frac{7}{7}$

Given an equation $ax=b$ Multiply both sides by $c$ to get $acx=bc$ Divide both sides by $a\not= 0$ to get $\displaystyle \frac{ax}{a}=x=\frac{b}{a}$

$$\begin{picture}(0,4)(8,0) \put(0,0){\huge The Grand Canyon } \put(-7,-2){separates addition/subtraction from multiplication/division.} \end{picture}$$

Given an equation $x=a$ Add $b$ to both sides (members) $x+b=a+b$ Subtract $b$ from both sides $x-b=a-b$

Being able to multiply or divide by the same number on both sides of an equation is straightforward. A multiplication can be used to undo (cancel, void, reverse) the effect of a division and vice-versa.

If $3x=15$ then

$$\fdrac{3x}{3}=\dfrac{15}{3}$$

or

$$x=5.$$

Could we have subtracted $3$ from both sides of the original equation $3x=15$?

Yes, $3x-3=15-3$ is correct but counter productive. We need to get to a stage where only unknowns are on one side and constants on the other side of an equation.

Note that addition and subtraction undo each other. Multiplication and division undo each other also. Do not try to undo a multiplication by addition or subtraction.
A member of an equation consists of more than one term. Multiply/divide all the terms of the member by the same number.

If $3x+6=21$ (you should first subtract $6$ from both sides, then divide by $3$)
but if you insist on dividing then \(\begin{array}{rcl lll} \displaystyle \frac{3x+6}{3}&=&\displaystyle \frac{21}{3}&\hbox{Divide both\)3x$and$6$by$3\(}\\[10pt] \displaystyle \frac{3x}{3}+\frac{6}{3}&=&7\\[10pt] x+2&=&7&\hbox{subtract\)-2$from both sides}\\[10pt] x&=&5\\ \end{array}$

Example $\PageIndex{1}$

Solve $\displaystyle \frac{4}{5}x=-8$.

Solution:

$\begin{array}{rcl lll} \displaystyle \frac{4}{5}x&=&-8\\[10pt] \displaystyle \frac{5}{4}\cdot \frac{4}{5}x&=&\displaystyle -8\cdot \frac{5}{4}&\hbox{Ddivision by a fraction is equivalent to }\\[1pt] &&&\hbox{multiplication by the reciprocal of the fraction.}\\[10pt] x&=&\displaystyle -2\cdot \frac{5}{1}&\hbox{Reduce.}\\[10pt] x&=&-10 \end{array}$

Example $\PageIndex{2}$

Solve $2(y+6)=5(y-12)$?

Solution:

\(\begin{array}{rcl lll} 2(y+6)&=&5(y-12)\ \ \ \hbox{Distribute multiplication over addition}\\[5pt] &&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{and subtraction to ``free" the variables.}\\[5pt] 2y+2(6)&=&5y-5(12)\\[5pt] 2y+12&=&5y-60\ \ \ \ \ \ \hbox{Multiply.}\\[5pt] 2y+12-12&=&5y-60-12\ \ \ \hbox{Subtract\)12\(from both sides.}\\[5pt] 2y&=&5y-72\ \ \ \ \ \ \ %\hbox{Subtract $12$ from both sides.} \\[5pt] -3y&=&-72\ \ \ \ \ \ \ \ \ \ \ \ \hbox{Subtract\)5y\(from both sides.}\\[10pt] \displaystyle \frac{-3}{-3}y&=&\displaystyle \frac{-72}{-3}\ \ \ \ \ \ \ \ \ \ \ \ \hbox{Divide both sides by\)-3$.}\\[10pt] y&=&24 \end{array}$

Example $\PageIndex{3}$

Solve $3(x-0.8)+2.5x=2.7+5x$.

Solution:

\(\begin{array}{rcl lll} 3(x-0.8)+2.5x&=&2.7+5x\\[5pt] 3x-3(0.8)+2.5x&=&2.7+5x\ \ \ \hbox{Distribute multip. over subtraction.}\\[5pt] 3x-2.4+2.5x&=&2.7+5x\ \ \ \hbox{Multiply.}\\[5pt] 5.5x-2.4&=&2.7+5x\ \ \ \hbox{Combine like-terms.}\\[5pt] 5.5x-5x-2.4&=&\!2.7\!+\!5x\!- \!5x\ \ \hbox{Subtract\)5x\(from both sides.}\\[5pt] 0.5x-2.4&=&2.7\ \ \ \ \ \ \ \ \ \ \ \hbox{Subtract.}\\[5pt] 0.5x-2.4+2.4&=&2.7+2.4\ \ \hbox{Add\)2.4\(to both sides.}\\[5pt] 0.5x&=&5.1\ \ \ \ \ \ \ \ \ \ \hbox{Add. Variables are segregated on }\\[5pt] &&\ \ \ \ \ \ \ \ \ \ \ \hbox{the left, constants on the right side.}\\[15pt] \displaystyle \frac{0.5}{0.5}x&=&\displaystyle \frac{5.1}{0.5}\ \ \ \ \ \ \ \ \ \ \hbox{Divide both sides by\)0.5\(.}\\[15pt] x&=&2(5.1)\hspace{0.24in} \hbox{ Division by\)0.5$is equivalent to multiplying by$2$.}\\[5pt] x&=&10.2& \end{array}$

Example $\PageIndex{4}$

Solve $\displaystyle \frac{3}{10}x+\frac{7}{15}x-\frac{11}{30}(1-x)=\displaystyle \frac{1}{5}x$.

Solution:

\(\begin{array}{rcl lll} \displaystyle \frac{3}{10}x+\frac{7}{15}x-\frac{11}{30}(1-x)&=&\displaystyle \frac{1}{5}x\\[15pt] \!\!\!\!\!\!\displaystyle \frac{3\!\cdot \!30}{10}x\!+\!\frac{7\cdot 30}{15}x\!-\!\frac{11\!\cdot \!30}{30}(1\!-\!x)\!\!&=&\!\displaystyle \frac{1\!\cdot \!30}{5}x\!\ \ \hbox{Multiply by LCD\)=30.\(}\\[15pt] \!\!\!\!\displaystyle \frac{3\!\cdot \!3}{1}x\!+\!\frac{7\!\cdot \!2}{1}x\!-\!\frac{11\!\cdot \!1}{1}(1\!-\!x)&=&\displaystyle \frac{1\cdot 6}{1}x\ \ \hbox{Reduce.}\\[13pt] 9x+14x-11(1-x)&=&\displaystyle 6x\ \ \ \ \ \hbox{Simplify.}\\[9pt] 9x+14x-11-(-11)x&=&\!\!\displaystyle 6x\!\!\ \ \hbox{Distribute multiplication over subtraction.}\\[9pt] 9x+14x-11+11x&=&\displaystyle 6x\ \ \ \ \ \hbox{Rewrite subtraction.}\\[9pt] 34x-11&=&\!\!\displaystyle 6x\ \ \ \ \ \ \hbox{Compute.}\\[9pt] 34x-34x-11&=&\!6x\!-\!34x\!\!\ \ \ \hbox{Subtract\) -34x$from$2\(sides.}\\[9pt] -11&=&-28x\ \ \hbox{Compute.}\\[9pt] -28x&=&-11\ \ \hbox{Flip both sides. (Not required))}\\[9pt] \displaystyle \frac{-28}{-28}x&=&\displaystyle \frac{-11}{-28}\ \ \hbox{Divide both sides by\)-28$}\\[14pt] x&=&\displaystyle \frac{11}{28} \end{array}$

Example $\PageIndex{5}$

Solve $2(t-7)+t=3(t+1)$.

Solution:

\(\begin{array}{rcl lll} 2(t-7)+t&=&3(t+1)&\hbox{Distribute multiplication over}\\ &&&\hbox{addition and subtraction.}\\ 3t-14&=&3t+3&\hbox{Combine like-terms.}\\ 3t-3t-14&=&3t-3t+3&\hbox{Subtract\)3t$from both sides.}\\ -14&=&3 \end{array}$
This is false. No matter what you replace $t$ by in the original equation,the false statement cannot be remedied. There are no $x$s in the final step. The equation has no solution.

Example $\PageIndex{6}$

Solve $3x-\displaystyle \frac{2}{9}=4x-\displaystyle \frac{1}{18}+\frac{1}{6}-x-\frac{1}{3}$.

Solution:

\(\begin{array}{rcl lll} 3x-\displaystyle \frac{2}{9}&=&4x-\displaystyle \frac{1}{18}+\frac{1}{6}-x-\frac{1}{3}\ \ \hbox{Multiply\)2$sides by LCD$=18\(}\\[10pt] \!\!\!18\cdot 3x\!-\!\displaystyle \frac{2\!\cdot \!18}{9}&=&\!18\!\cdot \!4x\!-\displaystyle \frac{1\cdot 18}{18}+\frac{1\cdot \!18}{6}\!-\!18\!\cdot \!x\!-\!\frac{1\!\cdot \!18}{3}\ \ \hbox{Reduce.}\\[10pt] 54x-\displaystyle \frac{2\cdot 2}{1}&=&72x-\displaystyle \frac{1\cdot 1}{1}+\frac{1\cdot 3}{1}-18x-\frac{6}{1}\ \ \hbox{Simplify.}\\[10pt] 54x-4&=&72x-\displaystyle 1+3-18x-6\ \ \hbox{Simplify.}\\[10pt] 54x-4&=&54x+2-6\ \ \hbox{Combine like-terms.}\\[10pt] 54x-4&=&54x-4\ \ \hbox{Subtract.}\\[10pt] 54x-54x-4&=&54x-54x-4\ \ \hbox{Subtract\)54x\(from both sides.}\\[10pt] -4&=&-4\ \ \hbox{Add 4 to both sides.}\\[10pt] 0&=&0\ \ \hbox{\hspace{0.1in}This is true. No matter what you replace\)x\(by in}\\ &&\ \ \hbox{\hspace{-1.0in} the original equation, the true statement cannot be falsified. }\\ &&\ \ \hbox{\hspace{-1.0in} There are no\)x$s in the final step.}\\ \end{array}$
There are infinitely many solutions in the original equation. Every real number is a solution, even $\pi$ with its infinite number of digits.

Multiplying both sides of the original equation by the LCD is usually a better way of solving an equation with fractions.

Example $\PageIndex{7}$

Solve $3x-\displaystyle \frac{2}{9}=2x+\displaystyle \frac{5}{12}$.

Solution:

\(\begin{array}{rcl lll} 3x-\displaystyle \frac{2}{9}&=&2x+\displaystyle \frac{5}{12}\ \ \hbox{Multiply\)2$sides by LCD$=36\(.}\\[10pt] 36\cdot 3x-\displaystyle \frac{2\cdot 36}{9}&=&2\cdot 36 x+\displaystyle \frac{5\cdot 36}{12}\ \ \hbox{Reduce.}\\[10pt] 108x-\displaystyle \frac{2\cdot 4}{1}&=&72x+\displaystyle \frac{5\cdot 3}{1}\ \ \ \ \ \ \ \ \hbox{Simplify}\\[10pt] 108x-8&=&72x+15\ \ \ \ \ \ \ \ \ \ \ \hbox{Simplify.}\\[10pt] 108x-8+8&=&72x+15+8\ \ \ \ \ \hbox{Add\)8\(to both sides.}\\[10pt] 108x&=&72x+23\ \ \ \ \ \ \ \ \ \ \ \hbox{Add.}\\[10pt] 108x-72x&=&72x-72x+23\ \ \hbox{Subtract\)72x\(from both sides.}\\[10pt] 36x&=&23\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{Simplify.}\\[10pt] \displaystyle \frac{36}{36}x&=&\displaystyle \frac{23}{36}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \hbox{Divide both sides by\)36$.}\\[10pt] x&=&\displaystyle \frac{23}{36}\\[10pt] \end{array}$

Example $\PageIndex{8}$

A liter (1,000 mL) of sodium chloride solution is hooked up to a patient.The solution is injected at the rate of $\displaystyle \frac{75\hbox{ mL}}{\hbox{hour}}$. At midnight $700$ mL are already in the patient’s system. How much longer will it take the whole liter to be consumed?

Solution:

Let $x$ be the time (in minutes) needed to finish the sodium chloride solution.
The portion of the solution injected in $x$ minutes is

$\!\!\displaystyle \frac{x\hbox{ minutes}}{1}\!\cdot \!\frac{1 \hbox{ hour}}{60\hbox{ minutes}}\!\cdot \!\frac{75\hbox{ mL}}{\hbox{hour}}\!=\!\frac{x}{1}\!\cdot \!\frac{1}{4\cdot 15}\!\cdot \!\frac{5\cdot 15\hbox{ mL}}{1}\!=\!\frac{5}{4}x$ mL.

The units are so useful (factor out $\displaystyle \frac{15}{15}=1$, cancel units).

The patient is still scheduled for $1,000-700=300$ mL.

Equation:

$\begin{array}{rcl lll} \displaystyle \frac{5x\hbox{ mL}}{4}&=&300\hbox{ mL}\\[15pt] \displaystyle \frac{4}{5}\cdot \frac{5x}{4}&=&\displaystyle \frac{4}{5}\cdot 300\\[15pt] %x&=&\displaystyle \frac{4}{1}\cdot 60\\[10pt] x&=&4\cdot 60\hbox{ minutes}&=&4\hbox{ hours}\\[10pt] \end{array}$.
The patient will have to be patient for four more hours.

Note: This problem can be solved more quickly by straightforward reasoning (keep units). The purpose here is to demonstrate how algebra can be used (in more complicated problems.)

Example $\PageIndex{9}$

Milan, Italy, produced a certain number of micrograms of particulate matter per cubic meter (mcgpcm) in $2004$.
Dehli, India, generated five times that number.
Caracas, Venezuela, had $20$ mcgpcm less than Milan.
The sum of the numbers of particulates for all three cities equals $10$ less than five times the sum of the number of particulates for Milan and Caracas.
Find the number of particulates for Milan.

Solution:

Let $x$ be the number of particulates in Milan.
Dehli generates five times that number: $5x$
Caracas had $20$ mcgpcm less than Milan: $x-20$
Sum of particulates for all three cities is $x+5x+x-20=7x-20$.
Sum of particulates for Milan and Caracas: $x+x-20=2x-20$.
$10$ less than $5\times$ the sum for Milan and Caracas $5(2x\!\!-\!20)\!-\!10\!=\!10x\!-\!100\!-\!10\!=\!10x\!-\!110$

Equation:

$\begin{array}{rcl lll} 7x-20&=&10x-110\\ 20&&\ \ \ \ \ \ \ \ \ \ 20\\$ $\\[-8pt] \cline{1-3}$ $\\[-8pt] 7x&=& \ \ \ 10x-90\\ -10x&&-10x\\$ $\\[-8pt]\cline{1-3}$ $\\[-8pt] -3x&=&-90\\[8pt] \displaystyle \frac{-3}{-3}x&=&\displaystyle \frac{-90}{-3}\\[8pt] x&=&30 \end{array}$

Milan had $30$ mcgpcm of particulates in $2004$.

Example $\PageIndex{10}$

A telephone company used to charge a basic fee of $\$20$ per month in addition to $10$ cents per minute of use. Jimmy’s bill for July was $\$50$. For how many minutes did Jimmy use the phone in July?

Solution:

Let $x$ be the number of minutes used during July.
The cost is: basic fee $+$ $\displaystyle \frac{x\hbox{ calls}}{1}\cdot \frac{0.1\hbox{ dollars}}{\hbox{ call}}$

Keep units. They will help you decide whether to multiply or divide by a ratio.

Equation:

$\begin{array}{rcl lll} 20+0.1x&=&50\\[10pt] 20-20+0.1x&=&50-20\\[10pt] 0.1x&=&30\\[10pt] \displaystyle \frac{0.1}{0.1}x&=&\displaystyle \frac{30}{0.1}\\[10pt] x&=&\displaystyle \frac{300}{1}\\[10pt] \end{array}$

Jimmy spent $300$ minutes ($5$ hours) on the phone in July. Jimmy is not the typical integer.

Exercises 9

1. Solve $\displaystyle \frac{7}{5}x=-35$.
2. Solve $5(y-7)=2(y+13)$?
3. Solve $6(x-0.9)+2.7x=2.8+0.5x$.
4. Solve $\displaystyle \frac{7}{20}x-\frac{4}{25}x-\frac{13}{50}(2-x)=\displaystyle \frac{1}{5}x-\frac{1}{100}x$.
5. Solve $5(t-2)+2t=7(t+3)$.
6. Solve $10y-\displaystyle \frac{3}{8}=9y-\displaystyle \frac{5}{24}-\frac{1}{4}+y+\frac{1}{12}$.
7. Solve $3y+\displaystyle \frac{2}{7}=5y+\displaystyle \frac{5}{21}$.
8. A liter (1,000 mL) of sodium chloride solution is hooked up to a patient. The solution is injected at the rate of $\displaystyle \frac{50\hbox{ mL}}{\hbox{hour}}$. At midnight $400$ mL are already in the patient’s system. How much longer will it take the whole liter to be consumed?
9. According to the World Almanac, $2010$ edition, National league statistics for $2009$, the Arizona Diamodbackers batter’s averages at bat were as follows:
   Gerardo Parra averaged a certain number at bat.
   Justin Upton’s average was $0.01$ more.
   MaxReynold’s average was $0.3$ less than Parra’s average.
   The sum of the averages for all three batters was $0.33$ less than twice the average for Upton and Parra.
   Find the average for Gerardo Parra.
10. A car company charges a basic fee of $\$20$ per day in addition to $0.05$ dollars per mile. Jimmy’s budget for driving is $\$45$. What is the maximum number of miles he can drive?

1. Solve $\displaystyle \frac{7}{5}x=-35$.
   Solution:
   $\begin{array}{rcl lll} \displaystyle \frac{7}{5}x&=&-35\\[17pt] \displaystyle \frac{5}{7} \cdot \frac{7}{5}x&=&\displaystyle -(35)\frac{5}{7}\\[17pt] x&=&\displaystyle -(5)\frac{5}{1}\\[17pt] x&=&-25 \end{array}$
2. Solve $5(y-7)=2(y+13)$?
   Solution:
   $\begin{array}{rcl lll} 5(y-7)&=&2(y+13)\\[7pt] 5y-5(7)&=&2y+2(13)\\[7pt] 5y-35&=&2y+26\\[7pt] 5y-35+35&=&2y+26+35\\[7pt] 5y&=&2y+61\\[7pt] 5y-2y&=&2y-2y+61\\[7pt] 3y&=&61\\[15pt] \displaystyle \frac{3}{3}y&=&\displaystyle \frac{61}{3}\\[15pt] y&=&\displaystyle \frac{61}{3}\\[10pt] \end{array}$
3. Solve $6(x-0.9)+2.7x=2.8+0.5x$.
   Solution:
   $\begin{array}{rcl lll} 6(x-0.9)+2.7x&=&2.8+0.5x\\[5pt] 6x-6(0.9)+2.7x&=&2.8+0.5x\\[5pt] 6x-5.4+2.7x&=&2.8+0.5x\\[5pt] 8.7x-5.4&=&2.8+0.5x\\[5pt] 8.7x-5.4+5.4&=&2.8+5.4+0.5x\\[5pt] 8.7x&=&8.2+0.5x\\[5pt] 8.7x-0.5x&=&8.2+0.5x-0.5x\\[5pt] 8.2x&=&8.2\\[10pt] \displaystyle \frac{8.2}{8.2}x&=&\displaystyle \frac{8.2}{8.2}\\[10pt] x&=&1 \end{array}$
4. Solve $\displaystyle \frac{7}{20}x-\frac{4}{25}x-\frac{13}{50}(2-x)=\displaystyle \frac{1}{5}x-\frac{1}{100}x$.
   Solution:
   $\begin{array}{rcl lll} \displaystyle \frac{7}{20}x-\frac{4}{25}x-\frac{13}{50}(2-x)&=&\displaystyle \frac{1}{5}x-\frac{1}{100}x\\[10pt] \!\!\!\!\!\!\!\!\!\!\displaystyle \frac{7\cdot 100}{20}x-\frac{4\cdot 100}{25}x-\frac{13\cdot 100}{50}(2-x)&=&\displaystyle \frac{1\cdot 100}{5}x-\frac{1\cdot 100}{100}x\\[10pt] \displaystyle \frac{7\cdot 5}{1}x-\frac{4\cdot 4}{1}x-\frac{13\cdot 2}{1}(2-x)&=&\displaystyle \frac{1\cdot 20}{1}x-\frac{1}{1}x\\[10pt] 35x-16x-26(2-x)&=&20x-x\\[10pt] 35x-16x-52+26x&=&20x-x\\[10pt] 35x+10x-52&=&20x-x\\[10pt] 45x-52&=&19x\\[10pt] 45x-45x-52&=&19x-45x\\[10pt] -52&=&-26x\\[10pt] \displaystyle \frac{-52}{-26}&=&\displaystyle \frac{-26}{-26}x\\[10pt] 2&=&x\\[10pt] x&=&2 \end{array}$
5. Solve $5(t-2)+2t=7(t+3)$.
   Solution:
   $\begin{array}{rcl lll} 5(t-2)+2t&=&7(t+3)\\[5pt] 5t-5(2)+2t&=&7t+7(3)\\[5pt] 5t-10+2t&=&7t+21\\[5pt] 7t-10&=&7t+21\\[5pt] 7t-7t-10&=&7t-7t+21\\[5pt] -10&=&21&\hbox{\hspace{-0.5in}False statement. Can't be corrected. }\\[5pt] &&&\hbox{\hspace{-0.5in}The original equation has no solution.} \end{array}$
6. Solve $10y-\displaystyle \frac{3}{8}=9y-\displaystyle \frac{5}{24}-\frac{1}{4}+y+\frac{1}{12}$.
   Solution:
   \(\begin{array}{rcl lll} 10y-\displaystyle \frac{3}{8}&=&9y-\displaystyle \frac{5}{24}-\frac{1}{4}+y+\frac{1}{12}\\[10pt] \!\!\!\!\!\!\!10\cdot 24 y-\displaystyle \frac{3\cdot 24}{8}&=&\!9\cdot 24 y\!-\!\displaystyle \frac{5\cdot 24}{24}\!-\!\frac{1\cdot 24}{4}\!+\!24\cdot y\!+\!\frac{1\cdot 24}{12}\\[10pt] 240 y-\displaystyle \frac{3\cdot 3}{1}&=& 216y-\displaystyle \frac{5\cdot 1}{1}-\frac{1\cdot 6}{1}+24y+\frac{1\cdot 2}{1}\\[10pt] 240 y-9&=& 216y-5-6+24y+2\\[10pt] 240 y-9&=& 240y-9\\[10pt] \!\!\!\!\!\!\!240 y-240y-9&=& 240y-240y-9\\[10pt] -9&=& -9\\[10pt] 0&=&0&\hbox{\hspace{-2.5in}True statement. Can't be falsified by}\\ &&&\hbox{\hspace{-2.5in}using any value of\)x$.} \end{array}$
   The original equation has infinitely many solutions. Any real number is a solution.
7. Solve $3y+\displaystyle \frac{2}{7}=5y+\displaystyle \frac{5}{21}$.
   Solution:
   $\begin{array}{rcl lll} 3y+\displaystyle \frac{2}{7}&=&5y+\displaystyle \frac{5}{21}\\[10pt] 3\cdot 21 y+\displaystyle \frac{2\cdot 21}{7}&=&5\cdot 21 y+\displaystyle \frac{5\cdot 21}{21}\\[9pt] 63 y+\displaystyle \frac{2\cdot 3}{1}&=&105y+\displaystyle \frac{5\cdot 1}{1}\\[9pt] 63y+6&=&105y+5\\[9pt] 63y-63y+6&=&105y-63y+5\\[9pt] 6&=&42y+5\\[9pt] 6-5&=&42y+5-5\\[9pt] 1&=&42y\\[9pt] 42y&=&1\\[10pt] \displaystyle \frac{42}{42}y&=&\displaystyle \frac{1}{42}\\[10pt] y&=&\displaystyle \frac{1}{42} \end{array}$
8. A liter (1,000 mL) of sodium chloride solution is hooked up to a patient. The solution is injected at the rate of $\displaystyle \frac{50\hbox{ mL}}{\hbox{hour}}$. At midnight $400$ mL are already in the patient’s system. How much longer will it take the whole liter to be consumed?

   Solution:
   Let $x$ be the time (in hours) needed to finish the sodium chloride solution.

   The portion of the solution injected in $x$ hours is

   $\displaystyle \frac{x\hbox{ hours}}{1}\cdot \frac{50\hbox{ mL}}{\hbox{hour}}=50x$ mL.

   The units are so useful.

   The patient is still getting $1,000-400=600$ mL.

   Equation:
   $\begin{array}{rcl lll} 50x\hbox{ mL}&=&600\hbox{ mL}\\[10pt] \displaystyle \frac{50}{50}x&=&\displaystyle \frac{600}{50}\\[10pt] x&=&12 \end{array}$.
   The patient will have to be patient for $12$ more hours.

   Warning: The temptation is to solve this problem by inspection. You will not be able to find the answer by inspection on a test, because the numbers will not be this obvious. Learn to follow algebraic steps.

9. According to the World Almanac, $2010$ edition, National league statistics for $2009$, the Arizona Diamodbackers batters averages at bat were as follows:
   Gerardo Parra averaged a certain number at bat.
   Justin Upton’s average was $0.01$ more.
   MaxReynold’s average was $0.3$ less than Parra’s average.
   The sum of the averages for all three batters was $0.33$ less than twice the average for Upton and Parra.
   Find the average for Gerardo Parra.
   Solution:
   Let $x$ be the average for Parra.
   Upton’s average was $0.01$ more: $x+0.01$
   Reynold’s was $0.3$ less than Parra’s: $x-0.3$
   The sum of the averages for all three batters was
   $x+x+0.01+x-0.3=3x-0.29$.
   The sum of the averages for Upton and Parra was
   $x+x+0.01=2x+0.01$.
   The sum of the averages for all three batters $3x-0.29$ was $0.33$ less than twice the average for Upton and Parra $2x+0.01$.
   Equation:
   $\begin{array}{rcl lll} 3x-0.29&=&2(2x+0.01)-0.33\\ 3x-0.29&=&4x+0.02-0.33\\ 3x-0.29&=&4x-0.31\\ 3x-3x-0.29&=&4x-3x-0.31\\ -0.29&=&x-0.31\\ -0.29+0.31&=&x-0.31+0.31\\ 0.02&=&x\\ \end{array}$

   Gerardo Parra’s average at bat was $0.020$.

10. A car company charges a basic fee of $\$20$ per day in addition to $0.05$ dollars per mile. Jimmy’s budget for driving is $\$45$. What is the maximum number of miles he can drive?
    Solution:
    Let $x$ be the maximum number of miles to be driven.
    The cost is: basic fee $+$ $\displaystyle \frac{x\hbox{ miles}}{1}\cdot \frac{0.05\hbox{ dollars}}{\hbox{ mile}}$
    Equation:
    $\begin{array}{rcl lll} 20+0.05x&=&45\\[10pt] 20-20+0.05x&=&45-20\\[10pt] 0.05x&=&25\\[15pt] \displaystyle \frac{0.05}{0.05}x&=&\displaystyle \frac{25}{0.05}\\[15pt] x&=&\displaystyle \frac{2500}{5}=500\\[15pt] \end{array}$

    Jimmy can drive a maximum of $500$ miles.