Skills Refresher:  Working with Formulas (2.4, 2.6, 2.7) — Intermediate Algebra

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13 Skills Refresher: Working with Formulas (2.4, 2.6, 2.7) — Intermediate Algebra

2.4 Fractional Linear Equations — Intermediate Algebra

When working with fractions built into linear equations, it is often easiest to remove the fraction in the very first step. This generally means finding the LCD of the fraction and then multiplying every term in the entire equation by the LCD.

Example 2.4.1

Solve for $x$ in the equation $\dfrac{3}{4}x - \dfrac{7}{2} = \dfrac{5}{6}.$

For this equation, the LCD is 12, so every term in this equation will be multiplied by 12.

$\dfrac{3}{4}x(12) - \dfrac{7}{2}(12) = \dfrac{5}{6}(12)$

Cancelling out the denominator yields:

$3x(3) - 7(6) = 5(2)$

Multiplying results in:

$\begin{array}{rrrrr} \\ \\ \\ 9x&-&42&=&10 \\ &+&42&&+42 \\ \midrule &&\dfrac{9x}{9}&=&\dfrac{52}{9} \\ \\ &&x&=&\dfrac{52}{9} \end{array}$

Example 2.4.2

Solve for $x$ in the equation $\dfrac{3\left(\dfrac{5}{9}x+\dfrac{4}{27}\right)}{2}=3.$

First, remove the outside denominator 2 by multiplying both sides by 2:

$\left(2\right)\dfrac{3\left(\dfrac{5}{9}x+\dfrac{4}{27}\right)}{2}=3(2)$

$3\left(\dfrac{5}{9}x+\dfrac{4}{27}\right)=6$

Now divide both sides by 3, which leaves:

$\dfrac{5}{9}x + \dfrac{4}{27} = 2$

To remove the 9 and 27, multiply both sides by the LCD, 27:

$\dfrac{5}{9}x\left(27\right) + \dfrac{4}{27}\left(27\right) = 2(27)$

This leaves:

$\begin{array}{rrrrl} 5x(3)&+&4&=&54 \\ &-&4&&-4 \\ \midrule &&15x&=&50 \\ \\ &&x&=&\dfrac{50}{15}\text{ or }\dfrac{10}{3} \end{array}$

2.6 Working With Formulas — Intermediate Algebra

In algebra, expressions often need to be simplified to make them easier to use. There are three basic forms of simplifying, which will be reviewed here. The first form of simplifying expressions is used when the value of each variable in an expression is known. In this case, each variable can be replaced with the equivalent number, and the rest of the expression can be simplified using the order of operations.

Example 2.6.1

Evaluate $p(q + 6)$ when $p = 3$ and $q = 5.$

$\begin{array}{rl} (3)((5)+(6))&\text{Replace }p\text{ with 3 and }q\text{ with 5 and evaluate parentheses} \\ (3)(11)&\text{Multiply} \\ 33&\text{Solution} \end{array}$

Whenever a variable is replaced with something, the new number is written inside a set of parentheses. Notice the values of 3 and 5 in the previous example are in parentheses. This is to preserve operations that are sometimes lost in a simple substitution. Sometimes, the parentheses won’t make a difference, but it is a good habit to always use them to prevent problems later.

Example 2.6.2

Evaluate $x + zx(3 - z)\left(\dfrac{x}{3}\right)$ when $x = -6$ and $z = -2.$

$\begin{array}{rl} (-6)+(-2)(-6)\left[(3)-(-2)\right]\left(\dfrac{-6}{3}\right)&\text{Evaluate parentheses} \\ \\ -6+(-2)(-6)(5)(-2)&\text{Multiply left to right} \\ -6+12(5)(-2)&\text{Multiply left to right} \\ -6+60(-2) &\text{Multiply} \\ -6-120 & \text{Subtract} \\ -126& \text{Solution}\end{array}$

Isolating variables in formulas is similar to solving general linear equations. The only difference is, with a formula, there will be several variables in the problem, and the goal is to solve for one specific variable. For example, consider solving a formula such as $A = \pi r^2+ \pi rs$ (the formula for the surface area of a right circular cone) for the variable $s.$ This means isolating the $s$ so the equation has $s$ on one side. So a solution might look like $s = \dfrac{A - \pi r^2}{\pi r}.$ This second equation gives the same information as the first; they are algebraically equivalent. However, one is solved for the area $A,$ while the other is solved for the slant height of the cone $s.$

When solving a formula for a variable, focus on the one variable that is being solved for; all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown: the first is a normal one-step equation, and the second is a formula that you are solving for $x.$

Example 2.6.3

Isolate the variable $x$ in the following equations.

$\begin{array}{ll} \begin{array}{rrl} 3x&=&12 \\ \\ \dfrac{3x}{3}&=&\dfrac{12}{3} \\ \\ x&=&4 \end{array} & \hspace{0.5in} \begin{array}{rrl} wx&=&z \\ \\ \dfrac{wx}{w}&=&\dfrac{z}{w} \\ \\ x&=&\dfrac{z}{w} \end{array} \end{array}$

The same process is used to isolate $x$ in $3x = 12$ as in $wx = z.$ Because $x$ is being solved for, treat all other variables as numbers. For these two equations, both sides were divided by 3 and $w,$ respectively. A similar idea is seen in the following example.

Example 2.6.4

Isolate the variable $n$ in the equation $m+n=p.$

To isolate $n,$ the variable $m$ must be removed, which is done by subtracting $m$ from both sides:

$\begin{array}{rrrrl} m&+&n&=&p \\ -m&&&&\phantom{p}-m \\ \midrule &&n&=&p-m \end{array}$

Since $p$ and $m$ are not like terms, they cannot be combined. For this reason, leave the expression as $p - m.$

Example 2.6.5

Isolate the variable $a$ in the equation $a(x - y) = b.$

This means that $(x-y)$ must be isolated from the variable $a.$

$\dfrac{a(x-y)}{(x-y)}=\dfrac{b}{(x-y)}\hspace{0.25in}\Rightarrow\hspace{0.25in}a=\dfrac{b}{(x-y)}$

If no individual term inside parentheses is being solved for, keep the terms inside them together and divide by them as a unit. However, if an individual term inside parentheses is being solved for, it is necessary to distribute. The following example is the same formula as in Example 2.6.5, but this time, $x$ is being solved for.

Example 2.6.6

Isolate the variable $x$ in the equation $a(x - y) = b.$

First, distribute $a$ throughout $(x-y)$ :

$\begin{array}{rrrrr} a(x&-&y)&=&b \\ ax&-&ay&=&b \end{array}$

Remove the term $ay$ from both sides:

$\begin{array}{rrrrl} ax&-&ay&=&b \\ &+&ay&&\phantom{b}+ay \\ \midrule &&ax&=&b+ay \end{array}$

$ax$ is then divided by $a$ :

$\dfrac{ax}{a}=\dfrac{b+ay}{a}$

The solution is $x=\dfrac{b+ay}{a},$ which can also be shown as $x=\dfrac{b}{a}+y.$

Be very careful when isolating $x$ not to try and cancel the $a$ on the top and the bottom of the fraction. This is not allowed if there is any adding or subtracting in the fraction. There is no reducing possible in this problem, so the final reduced answer remains $x = \dfrac{b + ay}{a}.$ The next example is another two-step problem.

Example 2.6.7

Isolate the variable $m$ in the equation $y=mx+b.$

First, subtract $b$ from both sides:

$\begin{array}{lrrrr} y&=&mx&+&b \\ \phantom{y}-b&&&-&b \\ \midrule y-b&=&mx&& \end{array}$

Now divide both sides by $x$ :

$\dfrac{y-b}{x}=\dfrac{mx}{x}$

Therefore, the solution is $m=\dfrac{y-b}{x}.$

It is important to note that a problem is complete when the variable being solved for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation.

The next example is also a two-step equation. It is a problem from earlier in the lesson.

Example 2.6.8

Isolate the variable $s$ in the equation $A= \pi r^2+\pi rs.$

Subtract $\pi r^2$ from both sides:

$\begin{array}{rrrrr} A\phantom{- \pi r^2}&=&\pi r^2&+&\pi rs \\ \phantom{A}-\pi r^2&&-\pi r^2&& \\ \midrule A- \pi r^2&=&\pi rs&& \end{array}$

Divide both sides by $\pi r$ :

$\dfrac{A-\pi r^2}{\pi r}=\dfrac{\pi rs}{\pi r}$

The solution is:

$s=\dfrac{A-\pi r^2}{\pi r}$

Formulas often have fractions in them and can be solved in much the same way as any fraction. First, identify the LCD, and then multiply each term by the LCD. After reducing, there will be no more fractions in the problem.

Example 2.6.9

Isolate the variable $m$ in the equation $h=\dfrac{2m}{n}.$

To clear the fraction, multiply both sides by $n$ :

$(n)h=\dfrac{2m}{n}(n)$

This leaves:

$nh=2m$

Divide both sides by 2:

$\dfrac{nh}{2}=\dfrac{2m}{2}$

Which reduces to:

$m=\dfrac{nh}{2}$

Example 2.6.10

Isolate the variable $b$ in the equation $A=\dfrac{a}{2-b}.$

To clear the fraction, multiply both sides by $(2-b)$ :

$(2-b)A=\dfrac{a}{2-b}(2-b)$

Which reduces to:

$A(2-b)=a$

Distribute $A$ throughout $(2-b),$ then isolate:

$\begin{array}{rrrrl} 2A&-&Ab&=&a \\ -2A&&&&\phantom{a}-2A \\ \midrule &&-Ab&=&a-2A \end{array}$

Finally, divide both sides by $-A$ :

$\dfrac{-Ab}{-A}=\dfrac{a-2A}{-A}$

Solution:

$b=\dfrac{a-2A}{-A}\text{ or }b=\dfrac{2A-a}{A}$

2.7 Variation Word Problems — Intermediate Algebra

Direct Variation Problems

There are many mathematical relations that occur in life. For instance, a flat commission salaried salesperson earns a percentage of their sales, where the more they sell equates to the wage they earn. An example of this would be an employee whose wage is 5% of the sales they make. This is a direct or a linear variation, which, in an equation, would look like:

$\begin{array}{c} \text{Wage }(x)=5\%\text{ Commission }(k)\text{ of Sales Completed }(y) \\ \\ \text{or} \\ \\ x=ky \\ \\ \text{(The constant }k\text{ comes from the German word for constant, which is }\emph{konstant}) \end{array}$

A historical example of direct variation can be found in the changing measurement of pi, which has been symbolized using the Greek letter π since the mid 18th century. Variations of historical π calculations are Babylonian $\left(\dfrac{25}{8}\right),$ Egyptian $\left(\dfrac{16}{9}\right)^2,$ and Indian $\left(\dfrac{339}{108}\text{ and }10^{\frac{1}{2}}\right).$ In the 5th century, Chinese mathematician Zu Chongzhi calculated the value of π to seven decimal places (3.1415926), representing the most accurate value of π for over 1000 years.

Pi is found by taking any circle and dividing the circumference of the circle by the diameter, which will always give the same value: 3.14159265358979323846264338327950288419716… (42 decimal places). Using an infinite-series exact equation has allowed computers to calculate π to 10¹³ decimals.

$\begin{array}{c} \text{Circumference }(c)=\pi \text{ times the diameter }(d) \\ \\ \text{or} \\ \\ c=\pi d \end{array}$

All direct variation relationships are verbalized in written problems as a direct variation or as directly proportional and take the form of straight line relationships. Examples of direct variation or directly proportional equations are:

- $x$ varies directly as $y$
- $x$ varies as $y$
- $x$ varies directly proportional to $y$
- $x$ is proportional to $y$
- $x$ varies directly as the square of $y$
- $x$ varies as $y$ squared
- $x$ is proportional to the square of $y$
- $x$ varies directly as the cube of $y$
- $x$ varies as $y$ cubed
- $x$ is proportional to the cube of $y$
- $x$ varies directly as the square root of $y$
- $x$ varies as the root of $y$
- $x$ is proportional to the square root of $y$

Example 2.7.1

Find the variation equation described as follows:

The surface area of a square surface $(A)$ is directly proportional to the square of either side $(x).$

Solution:

$\begin{array}{c} \text{Area }(A) =\text{ constant }(k)\text{ times side}^2\text{ } (x^2) \\ \\ \text{or} \\ \\ A=kx^2 \end{array}$

Example 2.7.2

When looking at two buildings at the same time, the length of the buildings’ shadows $(s)$ varies directly as their height $(h).$ If a 5-story building has a 20 m long shadow, how many stories high would a building that has a 32 m long shadow be?

The equation that describes this variation is:

$h=kx$

Breaking the data up into the first and second parts gives:

$\begin{array}{ll} \begin{array}{rrl} \\ &&\textbf{1st Data} \\ s&=&20\text{ m} \\ h&=&5\text{ stories} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ h&=&kx \\ 5\text{ stories}&=&k\text{ (20 m)} \\ k&=&5\text{ stories/20 m}\\ k&=&0.25\text{ story/m} \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ s&=&\text{32 m} \\ h&=&\text{find 2nd} \\ k&=&0.25\text{ story/m} \\ \\ &&\text{Find }h\text{:} \\ h&=&kx \\ h&=&(0.25\text{ story/m})(32\text{ m}) \\ h&=&8\text{ stories} \end{array} \end{array}$

Inverse Variation Problems

Inverse variation problems are reciprocal relationships. In these types of problems, the product of two or more variables is equal to a constant. An example of this comes from the relationship of the pressure $(P)$ and the volume $(V)$ of a gas, called Boyle’s Law (1662). This law is written as:

$\begin{array}{c} \text{Pressure }(P)\text{ times Volume }(V)=\text{ constant} \\ \\ \text{ or } \\ \\ PV=k \end{array}$

Written as an inverse variation problem, it can be said that the pressure of an ideal gas varies as the inverse of the volume or varies inversely as the volume. Expressed this way, the equation can be written as:

$P=\dfrac{k}{V}$

Another example is the historically famous inverse square laws. Examples of this are the force of gravity $(F_{\text{g}}),$ electrostatic force $(F_{\text{el}}),$ and the intensity of light $(I).$ In all of these measures of force and light intensity, as you move away from the source, the intensity or strength decreases as the square of the distance.

In equation form, these look like:

$F_{\text{g}}=\dfrac{k}{d^2}\hspace{0.25in} F_{\text{el}}=\dfrac{k}{d^2}\hspace{0.25in} I=\dfrac{k}{d^2}$

These equations would be verbalized as:

The force of gravity $(F_{\text{g}})$ varies inversely as the square of the distance.
Electrostatic force $(F_{\text{el}})$ varies inversely as the square of the distance.
The intensity of a light source $(I)$ varies inversely as the square of the distance.

All inverse variation relationship are verbalized in written problems as inverse variations or as inversely proportional. Examples of inverse variation or inversely proportional equations are:

- $x$ varies inversely as $y$
- $x$ varies as the inverse of $y$
- $x$ varies inversely proportional to $y$
- $x$ is inversely proportional to $y$
- $x$ varies inversely as the square of $y$
- $x$ varies inversely as $y$ squared
- $x$ is inversely proportional to the square of $y$
- $x$ varies inversely as the cube of $y$
- $x$ varies inversely as $y$ cubed
- $x$ is inversely proportional to the cube of $y$
- $x$ varies inversely as the square root of $y$
- $x$ varies as the inverse root of $y$
- $x$ is inversely proportional to the square root of $y$

Example 2.7.3

Find the variation equation described as follows:

The force experienced by a magnetic field $(F_{\text{b}})$ is inversely proportional to the square of the distance from the source $(d_{\text{s}}).$

Solution:

$F_{\text{b}} = \dfrac{k}{{d_{\text{s}}}^2}$

Example 2.7.4

The time $(t)$ it takes to travel from North Vancouver to Hope varies inversely as the speed $(v)$ at which one travels. If it takes 1.5 hours to travel this distance at an average speed of 120 km/h, find the constant $k$ and the amount of time it would take to drive back if you were only able to travel at 60 km/h due to an engine problem.

The equation that describes this variation is:

$t=\dfrac{k}{v}$

Breaking the data up into the first and second parts gives:

$\begin{array}{ll} \begin{array}{rrl} &&\textbf{1st Data} \\ v&=&120\text{ km/h} \\ t&=&1.5\text{ h} \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ k&=&tv \\ k&=&(1.5\text{ h})(120\text{ km/h}) \\ k&=&180\text{ km} \end{array} & \hspace{0.5in} \begin{array}{rrl} \\ \\ \\ &&\textbf{2nd Data} \\ v&=&60\text{ km/h} \\ t&=&\text{find 2nd} \\ k&=&180\text{ km} \\ \\ &&\text{Find }t\text{:} \\ t&=&\dfrac{k}{v} \\ \\ t&=&\dfrac{180\text{ km}}{60\text{ km/h}} \\ \\ t&=&3\text{ h} \end{array} \end{array}$

Joint or Combined Variation Problems

In real life, variation problems are not restricted to single variables. Instead, functions are generally a combination of multiple factors. For instance, the physics equation quantifying the gravitational force of attraction between two bodies is:

$F_{\text{g}}=\dfrac{Gm_1m_2}{d^2}$

where:

$F_{\text{g}}$ stands for the gravitational force of attraction
$G$ is Newton’s constant, which would be represented by $k$ in a standard variation problem
$m_1$ and $m_2$ are the masses of the two bodies
$d^2$ is the distance between the centres of both bodies

To write this out as a variation problem, first state that the force of gravitational attraction $(F_{\text{g}})$ between two bodies is directly proportional to the product of the two masses $(m_1, m_2)$ and inversely proportional to the square of the distance $(d)$ separating the two masses. From this information, the necessary equation can be derived. All joint variation relationships are verbalized in written problems as a combination of direct and inverse variation relationships, and care must be taken to correctly identify which variables are related in what relationship.

Example 2.7.5

Find the variation equation described as follows:

The force of electrical attraction $(F_{\text{el}})$ between two statically charged bodies is directly proportional to the product of the charges on each of the two objects $(q_1, q_2)$ and inversely proportional to the square of the distance $(d)$ separating these two charged bodies.

Solution:

$F_{\text{el}}=\dfrac{kq_1q_2}{d^2}$

Solving these combined or joint variation problems is the same as solving simpler variation problems.

First, decide what equation the variation represents. Second, break up the data into the first data given—which is used to find $k$ —and then the second data, which is used to solve the problem given. Consider the following joint variation problem.

Example 2.7.6

$y$ varies jointly with $m$ and $n$ and inversely with the square of $d.$ If $y = 12$ when $m = 3, n = 8,$ and $d = 2,$ find the constant $k,$ then use $k$ to find $y$ when $m = -3, n = 18,$ and $d = 3.$

The equation that describes this variation is:

$y=\dfrac{kmn}{d^2}$

Breaking the data up into the first and second parts gives:

$\begin{array}{ll} \begin{array}{rrl} \\ \\ \\ && \textbf{1st Data} \\ y&=&12 \\ m&=&3 \\ n&=&8 \\ d&=&2 \\ k&=&\text{find 1st} \\ \\ &&\text{Find }k\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ 12&=&\dfrac{k(3)(8)}{(2)^2} \\ \\ k&=&\dfrac{12(2)^2}{(3)(8)} \\ \\ k&=& 2 \end{array} & \hspace{0.5in} \begin{array}{rrl} &&\textbf{2nd Data} \\ y&=&\text{find 2nd} \\ m&=&-3 \\ n&=&18 \\ d&=&3 \\ k&=&2 \\ \\ &&\text{Find }y\text{:} \\ y&=&\dfrac{kmn}{d^2} \\ \\ y&=&\dfrac{(2)(-3)(18)}{(3)^2} \\ \\ y&=&12 \end{array} \end{array}$

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2.4 Fractional Linear Equations — Intermediate Algebra

2.6 Working With Formulas — Intermediate Algebra

2.7 Variation Word Problems — Intermediate Algebra

Direct Variation Problems

Inverse Variation Problems

Joint or Combined Variation Problems

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