MFG-tool Linear Models

Section 1.2 Linear Models

Subsection 1. Write a linear model

When we say "Express $y$ in terms of $x\text{,}$" we mean to write an equation that looks like

\begin{equation*} y = \text{algebraic expression in}~ x \end{equation*}

We say that $x$ is the input variable, and $y$ is the output variable.

In particular, a linear model has the form

\begin{equation*} \blert{y = \text{starting value} + \text{rate} \times x} \end{equation*}

Subsubsection Examples

Example 1.25.

Steve bought a Blu-Ray player for $269 and a number of discs at $14 each. Write an expression for Steve's total bill, $B$ (before tax), in terms of the number of discs he bought, $d\text{.}$

Solution

We want an equation of the form

\begin{equation*} B = \text{starting value} + \text{rate} \times d \end{equation*}

where Steve's bill started with the Blu-Ray player or $269, and then increased by a number of discs at a rate of $14 each. Substituting those values, we have

\begin{equation*} B=269+14d \end{equation*}

Example 1.26.

At 6 am the temperature was 50$\degree\text{,}$ and it has been falling by 4$\degree$ every hour. Write an equation for the temperature, $T\text{,}$ after $h$ hours.

Solution

We want an equation of the form

\begin{equation*} T = \text{starting value} + \text{rate} \times h \end{equation*}

The temperature started at 50$\degree\text{,}$ and then decreased each hour at the rate of 4$\degree$ per hour, so we subtract $4h$ from 50 to get

\begin{equation*} T=50-4h \end{equation*}

Example 1.27.

Kyli's electricity company charges her $6 per month plus $0.10 per kilowatt hour (kWh) of energy she uses. Write an equation for Kyli's electric bill, $E\text{,}$ if she uses $w$ kWh of electricity.

Solution

Kyli's bill starts a6 $6 and increases by $0.10 for each kWh, $w\text{.}$ Thus,

\begin{equation*} E=6+0.10w \end{equation*}

Subsubsection Exercises

Checkpoint 1.28.

Salewa saved $5000 to go to school full time. She spends $200 per week on living expenses. Write an equation for Salewa's savings, $S\text{,}$ after $w$ weeks.

Answer

$S=5000-200w$

Checkpoint 1.29.

As a student at City College, Delbert pays a $50 registration fee plus $15 for each unit he takes. Write an equation that gives Delbert's tuition, $T\text{,}$ if he takes $u$ units.

Answer

$T=50+15u$

Checkpoint 1.30.

Greta's math notebook has 100 pages, and she uses on average 6 pages per day for notes and homework. How many pages, $P\text{,}$ will she have left after $d$ days?

Answer

$P=100-6d$

Checkpoint 1.31.

Asa has typed 220 words of his term paper, and is still typing at a rate of 20 words per minute. How many words, $W\text{,}$ will Asa have typed after $m$ more minutes?

Answer

$W=220+20m$

Checkpoint 1.32.

The temperature in Nome was $-12 \degree$ F at noon. It has been rising at a rate of 2 $\degree$ F per hour all day. Write an equation for the temperature, $T\text{,}$ after $h$ hours.

Answer

$T=-12+2h$

Checkpoint 1.33.

Francine borrowed money from her mother, and she owes her $750 right now. She has been paying off the debt at a rate of $50 per month. Write an equation for Francine's financial status, $F\text{,}$ in terms of $m\text{,}$ the number of months from now.

Answer

$F=-750+50m$

Subsection 2. Graph a linear equation by the intercept method

To graph a line by the intercept method, we find the $x$- and $y$-intercepts of the line and plot those points.

Example 1.34.

Graph the equation $3x+2y=7$ by the intercept method.

Solution

First, we find the $x$- and $y$-intercepts of the graph. To find the $y$-intercept, we substitute $0$ for $x$ and solve for $y\text{:}$

\begin{align*} 3(\alert{0})+2y \amp =7 \amp \amp \blert{\text{Simpify the left side.}}\\ 2y \amp =7 \amp \amp \blert{\text{Divide both sides by 2.}}\\ y \amp =\dfrac{7}{2}=3\dfrac{1}{2} \end{align*}

The $y$-intercept is the point $\left(0, 3\dfrac{1}{2}\right)\text{.}$ To find the $x$-intercept, we substitute $0$ for $y$ and solve for $x\text{:}$

\begin{align*} 3x+2(\alert{0}) \amp =7 \amp \amp \blert{\text{Simpify the left side.}}\\ 3x \amp =7 \amp \amp \blert{\text{Divide both sides by 3.}}\\ x \amp =\dfrac{7}{3}=2\dfrac{1}{3} \end{align*}

The $x$-intercept is the point $\left(2\dfrac{1}{3}, 0\right)\text{.}$

A table with the two intercepts is shown below. We plot the intercepts and connect them with a straight line.

$x$	$y$
$0$	$3\dfrac{1}{2}$
$2\dfrac{1}{3}$	$0$

Subsubsection Exercises

Checkpoint 1.35.

Graph the line $y=\dfrac{-4}{3}x+8$ by the intercept method.

$x$	$y$
$~~~~~~~~$	$~~~~~~~~$
$~~~~~~~~$	$~~~~~~~~$

Answer

Checkpoint 1.36.

Graph the line $\dfrac{x}{6}+\dfrac{y}{8}=-1$ by the intercept method.

$x$	$y$
$~~~~~~~~$	$~~~~~~~~$
$~~~~~~~~$	$~~~~~~~~$

Answer

Subsection 3. Interpret the intercepts

The values of the variables at the intercepts often tell us something important about a linear model

Example 1.37.

The temperature, $T\text{,}$ in Nome was $-12 \degree$ at noon and has been rising at a rate of $2 \degree$ per hour all day.

Write and graph an equaton for $T$ in terms of $h\text{,}$ the number of hours after noon.
Find the intercepts of the graph and interpret their meaning in the context of the problem situation.

Solution

An equation for $T$ at time $h$ is

\begin{equation*} T=-12 +2h \end{equation*}

To find the $T$-intercept, we set $h=0$ and solve for $T\text{.}$

\begin{equation*} T=-12+2(\alert{0})=-12 \end{equation*}

The $T$-intercept is $(0,-12)\text{.}$ This point tells us that when $h=0, T=-12\text{,}$ or the temperature at noon was $-12 \degree\text{.}$ To find the $h$-intercept, we set $T=0$ and solve for $h\text{.}$

\begin{align*} \alert{0} \amp = -12+2h \amp \amp \blert{\text{Add 12 to both sides.}}\\ 12 \amp = 2h \amp \amp \blert{\text{Divide both sides by 2.}}\\ 6 \amp =h \end{align*}

The $h$-intercept is the point $(6,0)\text{.}$ This point tells us that when $h=6, T=0\text{,}$ or the temperature will reach zero degrees at six hours after noon, or 6 pm.

Subsubsection Exercises

Checkpoint 1.38.

Sheri bought a bottle of multivitamins for her family. The number of vitamins lt in the bottle after $d$ days is given by

\begin{equation*} N=300-5d \end{equation*}

Find the intercepts and use them to make a graph of the equation.

$d$ $N$

$~~~~~~~~$ $~~~~~~~~$

$~~~~~~~~$ $~~~~~~~~$
Explain what each intercept tells us about the vitamins.

Answer

$(0,300)$ There were 300 vitamins to start.
$(60,0)$ The vitamin bottle is empty after 60 days.

Checkpoint 1.39.

Delbert bought some equipment and went into the dog-grooming business. His profit is increasing according to the equation

\begin{equation*} P=-600+40d \end{equation*}

where $d$ is the number of dogs he has groomed.

Find the intercepts and use them to make a graph of the equation.

$d$ $P$

$~~~~~~~~$ $~~~~~~~~$

$~~~~~~~~$ $~~~~~~~~$
Explain what each intercept tells us about Delbert's dog-grooming business.

Answer

$(0,-600)$ To start, Delbert's profit is $-$600\text{.}$ (He is $600 in debt.)
$(15,0)$ Delbert breaks even after grooming 15 dogs.

Subsection 4. Solve an equation for one of the variables

It is usually easier to study a model and draw its graph if it is in the form

\begin{equation*} y = \text{starting value} + \text{rate} \times x \end{equation*}

To put an equation into this form, we want to "isolate" the output variable on one side of the equation.

Subsubsection Examples

Example 1.40.

Solve the equation $2x-3y=8$ for $y\text{.}$

Solution

\begin{align*} 2x-3y \amp = 8 \amp \amp \blert{\text{Subtract from both sides.}} \\ -3y \amp = 8-2x \amp \amp \blert{\text{Divide both sides by} -3.}\\ y \amp =\dfrac{8-2x}{-3} \amp \amp \blert{\text{Divide each term of the numerator by} -3.}\\ y \amp =\dfrac{8}{3}-\dfrac{2}{3}x \end{align*}

Example 1.41.

Solve the equation $A=\dfrac{h}{2}(b+c)$ for $b\text{.}$

Solution

It is nearly always best to clear fractions from an equation first, so we begin by multiplying both sides by 2.

\begin{align*} \alert{2}A \amp = \alert{\cancel{2}}\left(\dfrac{h}{\cancel{2}}(b+c)\right) \amp \amp \blert{\text{Multiply both sides by}~ \alert{2}.}\\ 2A \amp = h(b+c) \amp \amp \blert{\text{Divide both sides by}~h.}\\ \dfrac{2A}{h} \amp = b+c \amp \amp \blert{\text{Subtract from both sides.}}\\ \dfrac{2A}{h}-c \amp = b \end{align*}

Subsubsection Exercises

Checkpoint 1.42.

Solve $f=s+at$ for $t$

Answer

$t=\dfrac{f-a}{s}$

Checkpoint 1.43.

Solve $2x-4y=k$ for $y$

Answer

$y=\dfrac{-1}{4}k+\dfrac{1}{2}x$

Checkpoint 1.44.

Solve $P=2l+2w$ for $l$

Answer

$l=\dfrac{P}{2}-w$

Checkpoint 1.45.

Solve $\dfrac{x}{a}+\dfrac{y}{b}=1$ for $x$

Answer

$x=a-\dfrac{ay}{b}$

\(x\)	\(y\)
\(0\)	\(3\dfrac{1}{2}\)
\(2\dfrac{1}{3}\)	\(0\)

\(x\)	\(y\)
\(~~~~~~~~\)	\(~~~~~~~~\)
\(~~~~~~~~\)	\(~~~~~~~~\)

\(x\)	\(y\)
\(~~~~~~~~\)	\(~~~~~~~~\)
\(~~~~~~~~\)	\(~~~~~~~~\)

\(d\)	\(N\)
\(~~~~~~~~\)	\(~~~~~~~~\)
\(~~~~~~~~\)	\(~~~~~~~~\)

\(d\)	\(P\)
\(~~~~~~~~\)	\(~~~~~~~~\)
\(~~~~~~~~\)	\(~~~~~~~~\)