Int-Alg Chapter 1 Linear Models

Section A.1 Chapter 1 Linear Models

Subsection A.1.1 Creating a Linear Model

Subsubsection A.1.1.1 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*}

Example A.1.1.

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 A.1.2.

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 A.1.3.

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*}

Checkpoint A.1.4.

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 A.1.5.

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 A.1.6.

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 A.1.7.

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 A.1.8.

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 A.1.9.

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.

$F=-750+50m$

Subsubsection A.1.1.2 Plot points and graph a line

The simplest way to graph a line is to make a table and plot points. We will learn more efficient methods shortly.

Example A.1.10.

Make a table of values, plot the points, and graph the equation $y=-2x+6\text{.}$

Solution.

Choose both positive and negative values for $x\text{.}$ Calculate the $y$-value for each $x$-value by substituting the $x$-value into the equation.

$x$	$y$	$\hphantom{0000}$
$-3$	$12$	$y=-2(\alert{-3})+6=12$
$-2$	$10$	$y=-2(\alert{-2})+6=10$
$-1$	$8$	$y=-2(\alert{-1})+6=8$
$0$	$6$	$y=-2(\alert{0})+6=6$
$1$	$4$	$y=-2(\alert{1})+6=4$
$2$	$2$	$y=-2(\alert{2})+6=2$
$3$	$0$	$y=-2(\alert{3})+6=0$
$4$	$-2$	$y=-2(\alert{4})+6=-2$

Next, sketch a Cartesian coordinate system with appropriate scales on the $x$- and $y$-axes. Plot each of the points in the table of values and connect them with a straight line. The completed graph is shown at right.

$line$

Example A.1.11.

Byron borrowed $6000 from his uncle to help pay for his college education. Now that he has graduated and has a job, he is paying back the loan at $100 per month.

Write an equation showing the amount of money, $y\text{,}$ that Byron still owes his uncle after $x$ months.
Graph your equation.

Solution.

$\displaystyle y=6000-100x$
$x$ $y$ $\hphantom{0000}$

$0$ $6000$ $y=6000-100(\alert{0})=6000$

$10$ $5000$ $y=6000-100(\alert{10})=5000$

$20$ $4000$ $y=6000-100(\alert{20})=4000$

Now choose appropriate scales for the axes. A good choice would be to scale the $x$-axis by 10’s and the $y$-axis by 1000’s.

$line$

Checkpoint A.1.12.

Graph $~y=\dfrac{3}{4}x-3\text{.}$

$x$	$-8$	$-4$	$0$	$4$	$6$
$y$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$

Answer.

$x$	$-8$	$-4$	$0$	$4$	$6$
$y$	$-9$	$-6$	$-3$	$0$	$1.5$

$line$

Checkpoint A.1.13.

Graph $~y=-3x+2\text{.}$

$x$	$-2$	$-1$	$0$	$2$	$4$
$y$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$	$\hphantom{0000}$

Answer.

$x$	$-2$	$-1$	$0$	$2$	$4$
$y$	$8$	$5$	$2$	$-4$	$-10$

$line$

Checkpoint A.1.14.

Stuart invested $800 in a computer and now makes $5 a page typing research papers. Let $x$ represent the number of pages Stuart has typed, and let $y$ represent his profit.

Write an equation for $y$ in terms of $x\text{.}$
Complete the table and graph your equation.

$x$	$y$
$0$	$\hphantom{0000}$
$50$	$\hphantom{0000}$
$100$	$\hphantom{0000}$
$200$	$\hphantom{0000}$

$grid$

Answer.

$\displaystyle y=5x-800$
$x$ $y$

$0$ $-800$

$50$ $-550$

$100$ $-300$

$200$ $200$

$line$

Checkpoint A.1.15.

Ludmilla earns a commission of 5% of her real estate sales. Let $x$ represent her sales in thousands of dollars, and let $y$ represent the commission she earns from her sales, in thousands of dollars.

Write an equation for $y$ in terms of $x\text{.}$
Complete the table and graph the equation.

$x$	$y$
$250$	$\hphantom{0000}$
$600$	$\hphantom{0000}$
$800$	$\hphantom{0000}$
$1000$	$\hphantom{0000}$

$grid$

Answer.

$\displaystyle y=0.05x$
$x$ $y$

$250$ $12.5$

$600$ $30$

$800$ $40$

$1000$ $50$

$line$

Subsubsection A.1.1.3 Solve a linear equation

Recall that to solve an equation we want to "isolate" the variable on one side of the equals sign. We "undo" each operation performed on the variable by performing the opposite operation on both sides of the equation.

Example A.1.16.

Solve the equation $~~\dfrac{2}{3}x-5=7$

Solution.

\begin{align*} \dfrac{2}{3}x-5 \amp = 7 \amp \amp \blert{\text{Add 5 to both sides.}}\\ \dfrac{2}{3}x \amp = 12 \amp \amp \blert{\text{To divide both sides by } \dfrac{2}{3}, \text{we:}}\\ \dfrac{3}{2}\left(\dfrac{2}{3}x\right) \amp =\dfrac{3}{2}(12) \amp \amp \blert{\text{Multiply by the reciprocal of }\dfrac{2}{3}.}\\ x \amp = 18 \amp \amp \blert{\text{The solution is 18.}} \end{align*}

Example A.1.17.

Solve the equation $~~2x+7=4x-3$

Solution.

To begin, we must get both variable terms on the same side of the equation.

\begin{align*} 2x+7 \amp = 4x-3 \amp \amp \blert{\text{Subtract}~2x~\text{from both sides.}}\\ 7 \amp = 2x-3 \amp \amp \blert{\text{Add 3 to both sides.}}\\ 10 \amp =2x \amp \amp \blert{\text{Divide both sides by 2.}}\\ 5 \amp = x \amp \amp \blert{\text{The solution is 5.}} \end{align*}

Checkpoint A.1.18.

Solve the equation $~~10=1-\dfrac{3x}{7}$

Answer.

$-21$

Checkpoint A.1.19.

Solve the equation $~~6p-8=-3p-26$

Answer.

$-2$

Checkpoint A.1.20.

Solve the equation $~~12=\dfrac{7u+4}{5}$

Hint: Start by clearing the fraction: multiply both sides by 5.

Answer.

$8$

Checkpoint A.1.21.

Solve the equation $~~0=13q+25-17q+7$

Hint: Start by combining like terms.

Answer.

$8$

Subsubsection A.1.1.4 Solve a linear inequality

The rules for solving an inequality are the same as those for solving an equation, with one important difference:

Solving a Linear Inequality.

If we multiply or divide both sides of a linear inequality by a negative number, we must reverse the direction of the inequality symbol.

Example A.1.22.

Solve $~~-3x+1 \gt 7~~$ and graph the solutions on a number line.

Solution.

\begin{align*} -3x+1 \amp \gt 7 \amp \amp ~\blert{\text{ Subtract 1 from both sides.}}\\ -3x \amp \gt 6 \amp \amp\ \begin{array}{l} \blert{\text{Divide both sides by }{-3}\text{, and reverse }}\\ \blert{\text{the direction of the inequality.}} \end{array}\\ x \amp \lt -2 \end{align*}

The solutions are all the numbers less than $-2\text{.}$ The graph of the solutions is shown below.

Example A.1.23.

Solve $~~-3 \lt 2x-5 \le 6~~$ and graph the solutions on a number line.

Solution.

\begin{align*} -3 \amp \lt 2x-5 \le 6 \amp \amp \blert{\text{Add 5 on all three sides of the inequality.}}\\ 2 \amp \lt 2x \le 11 \amp \amp \blert{\text{Divide each side by 2.}}\\ 1 \amp \lt x \le \dfrac{11}{2} \amp \amp \blert{\text{Notice that we did not reverse the inequality.}} \end{align*}

The solutions are all the numbers greater than 1 but less than 5.5. The graph of the solutions is shown below.

$number line$

Recall that a solid dot on a number line indicates that the number is part of the solution; an open dot means that the number is not part of the solution.

Checkpoint A.1.24.

Solve the inequality $~~8-4x \gt -2~~$ and graph the solutions on a number line.

Answer.

$x \lt \dfrac{5}{2}$

$number line$

Checkpoint A.1.25.

Solve the inequality $~~-6 \le \dfrac{4-x}{3} \lt 2~~$ and graph the solutions on a number line.

Answer.

$22 \ge x \gt -2$

$number line$

Checkpoint A.1.26.

Solve the inequality $~~3x-5 \lt -6x+7~~$ and graph the solutions on a number line.

Answer.

$x \lt \dfrac{4}{3}$

$number line$

Checkpoint A.1.27.

Solve the inequality $~~-6 \gt 4-5b \gt -21~~$ and graph the solutions on a number line.

Answer.

$2 \lt b \lt 5$

$number line$

Subsection A.1.2 Graphs and Equations

Subsubsection A.1.2.1 Verify a solution

We can always check a solution to an equation by verifying that it makes the equation true.

Example A.1.28.

Verify that $x=-5$ is a solution of the equation

\begin{equation*} ~~x^2+2x-15=0 \end{equation*}

Solution.

We show that substituting $-5$ for $x$ makes the equation true. When we substitute a negative number for a variable, we should enclose the number in parentheses.

\begin{equation*} \begin{aligned}[t] x^2+2x-15 \amp = (\alert{-5})^2+2(\alert{-5})-15\\ \amp = 25-10-15=0 \end{aligned} \end{equation*}

Because the expression does equal $0\text{,}$ we see that $x=-5$ is a solution.

Example A.1.29.

Verify that $x=-3$ is not a solution of the equation

\begin{equation*} ~~\sqrt{2x+10}-3x=8 \end{equation*}

Solution.

We show that substituting $-3$ for $x$ does not make the equation true.

\begin{equation*} \begin{aligned}[t] \sqrt{2x+10}-3x \amp = \sqrt{2(\alert{-3})+10}-3(\alert{-3})\\ \amp = \sqrt{4}+9=2+9=11 \not= 8 \end{aligned} \end{equation*}

The left side of the equation does not equal 8 when $x=-3\text{,}$ so $x=-3$ is not a solution.

Checkpoint A.1.30.

Decide whether the given value is a solution of the equation.

\begin{equation*} x^3-3x^2-4x+2=10;~~~~x=-2 \end{equation*}

Answer.

Yes

Checkpoint A.1.31.

Decide whether the given value is a solution of the equation.

\begin{equation*} \sqrt{3x+5}=10+\sqrt{x+7};~~~~x=9 \end{equation*}

Answer.

Checkpoint A.1.32.

Decide whether the given value is a solution of the equation.

\begin{equation*} \dfrac{2x-1}{x+1}+2=\dfrac{x+1}{x-1};~~~~x=2 \end{equation*}

Answer.

Yes

Checkpoint A.1.33.

Decide whether the given value is a solution of the equation.

\begin{equation*} 9-4x=5\sqrt{x+3};~~~~x=6 \end{equation*}

Answer.

Subsubsection A.1.2.2 Solve linear equations and inequalities with parentheses

Strategy for solving linear equations.

Simplify each side of the equation: apply the distributive law, combine like terms.
Use addition and subtraction to get all the variable terms on one side of the equation, and all constsnt terms on the other side.
Divide both sides by the coefficient of the variable.

Example A.1.34.

Solve $~~3(2a-4) \ge 4-(1-3a)$

Solution.

First, we remove parentheses by applying the distributive law. Then we can combine like terms on each side of the equation.

Note that the minus sign in front of the parentheses on the right side of the equation applies to both terms inside the parentheses.

\begin{align*} 3(2a-4) \amp \ge 4-(1-3a) \amp \amp \blert{\text{Apply the distributive law.}}\\ 6a-12 \amp \ge 4\alert{-}1\alert{+}3a \amp \amp \blert{\text{Simplify the right side.}}\\ 6a-12 \amp \ge 3+3a \amp \amp \blert{\text{Subtract}~ 3a~ \text{from both sides.}}\\ 3a-12 \amp \ge 3 \amp \amp \blert{\text{Add 12 to both sides.}}\\ 3a \amp \ge 15 \amp \amp \blert{\text{Divide both sides by 3.}}\\ a \amp \ge 5 \end{align*}

Example A.1.35.

Solve the inequality $~~25-6x \gt 3x-2(4-x)$

Solution.

We begin by the same way we solve an equation. For this example, we start by removing the parentheses.

\begin{align*} 25-6x \amp \gt 3x-2(4-x) \amp \amp \blert{\text{Apply the distributive law.}}\\ 25-6x \amp \gt 3x-8+2x \amp \amp \blert{\text{Combine like terms.}}\\ 25-6x \amp \gt 5x-8 \amp \amp \blert{\text{Subtract}~ 5x~ \text{from both sides.}}\\ 25-11x \amp \gt -8 \amp \amp \blert{\text{Subtract 25 from both sides.}}\\ -11x \amp \gt -33 \amp \amp \blert{\text{Divide both sides by -11.}}\\ x \amp \alert{\lt} 3 \amp \amp \blert{\text{Don't forget to reverse the inequality symbol.}} \end{align*}

Recall that if we multiply or divide both sides of an inequality by a negative number, we must reverse the direction of the inequality symbol.

Checkpoint A.1.36.

Solve the inequality $~~-4(x+2)+3(x-2) \ge -2$

Answer.

$x \le -12$

Checkpoint A.1.37.

Solve the equation $~~4(2-3w)=9-3(2w-1)$

Answer.

$\dfrac{-1}{2}$

Checkpoint A.1.38.

Solve the inequality $~~2(3h-6) \lt 5-(h-4)$

Answer.

$h \lt 3$

Checkpoint A.1.39.

Solve the equation $~~0.25(x+3)-0.45(x-3)=0.30$

Answer.

$9$

Subsubsection A.1.2.3 Solutions of an equation in two variables

A solution of an equation in two variables $x$ and $y$ is written as an ordered pair, $(x,y)\text{.}$ For example, the solution $(-2,5)$ means that $x=-2$ and $y=5\text{.}$

Example A.1.40.

Is $(-3,2)$ a solution of the equation $~~x^2+4y^2=25$ ?

Solution.

We substitute $x=\alert{-3}$ and $y=\alert{2}$ into the equation.

\begin{equation*} (\alert{-3})^2+4(\alert{2})^2=9+4(4)=9+16=25 \end{equation*}

The ordered pair $(-3,2)$ satisfies the equation, so it is a solution.

Example A.1.41.

Which of the following ordered pairs are solutions of the equation whose graph is shown?

$\displaystyle (-3,-2)$
$\displaystyle (-5,0)$
$\displaystyle (1,-4)$
$\displaystyle (-1,-6)$

$parabola$

Solution.

The graph of an equation is just a picture of its solutions, so points that lie on the graph are solutions of the equation.

The points $(-3,-2)$ and $(-1,-6)$ lie on the graph, so they represent solutions of the equation. The points $(-5,0)$ and $(1,-4)$ do not lie on the graph, so they are not solutions of the equation.

Checkpoint A.1.42.

Find a solution of the equation with the given coordinate.

\begin{equation*} 6x-5y=-3,~~~~(2,?) \end{equation*}

Answer.

$(2,3)$

Checkpoint A.1.43.

Find a solution of the equation with the given coordinate.

\begin{equation*} y=\frac{3}{4}x+8,~~~~(?,-1) \end{equation*}

Answer.

$(-12,-1)$

Checkpoint A.1.44.

Find a solution of the equation with the given coordinate.

Answer.

$(16,4)$

Checkpoint A.1.45.

Find a solution of the equation with the given coordinate.

Answer.

$(2,-5)$

Subsection A.1.3 Intercepts

Subsubsection A.1.3.1 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 A.1.46.

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.

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

$line$

Checkpoint A.1.47.

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

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

$8 by 8 grid$

Answer.

$line$

Checkpoint A.1.48.

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

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

Answer.

$line$

Subsubsection A.1.3.2 Interpret the intercepts

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

Example A.1.49.

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*}

$Temperature in Nome$

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.

Checkpoint A.1.50.

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.

$grid$

Answer.

$grid$

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

Checkpoint A.1.51.

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.

$grid$

Answer.

$grid$

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

Subsubsection A.1.3.3 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.

Example A.1.52.

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

Solution.

\begin{align*} 2x-3y \amp = 8 \amp \amp \blert{\text{Subtract }2x \text{ 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 A.1.53.

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 }c \text{ from both sides.}}\\ \dfrac{2A}{h}-c \amp = b \end{align*}

Checkpoint A.1.54.

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

Answer.

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

Checkpoint A.1.55.

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

Answer.

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

Checkpoint A.1.56.

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

Answer.

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

Checkpoint A.1.57.

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

Answer.

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

Subsection A.1.4 Slope

Subsubsection A.1.4.1 Use ratios for comparison

Slope is a type of ratio that compares vertical distance per unit of horizontal distance. We use ratios for comparison in other situations, for example, when shopping we might compute price per unit.

Example A.1.58.

You are choosing between two brands of iced tea. Which is a better bargain: a 28-ounce bottle of Teatime for $1.82, or a 36-ounce bottle of Leafdream for $2.25?

Solution.

Compute the ratio price per ounce for each brand.

\begin{equation*} \text{Teatime:}~~~\dfrac{182 \text{ cents}}{28 \text{ ounces}}= 6.5 \text{ cents per ounce} \end{equation*}

\begin{equation*} \text{Leafdream:}~~~\dfrac{225 \text{ cents}}{36 \text{ ounces}}= 6.25 \text{ cents per ounce} \end{equation*}

Leafdream is the better bargain.

Example A.1.59.

The trail to Lookout Point gains 780 feet in elevation over a distance of 1.3 miles. The trail to Knife Edge gains 950 feet in elevation over a distance of 1.6 miles. Which trail is steeper?

Solution.

Compute the ratio of elevation gain to horizontal distance traveled for each trail.

\begin{equation*} \text{Lookout Point:}~~~\dfrac{780 \text{ feet}}{1.3 \text{ miles}}= 600 \text{ feet per mile} \end{equation*}

\begin{equation*} \text{Knife Edge:}~~~\dfrac{950 \text{ feet}}{1.6 \text{ miles}}= 593.75 \text{ feet per mile} \end{equation*}

The Lookout Point trail is steeper.

Checkpoint A.1.60.

Rachel drove 292.4 miles on 8.6 gallons of gasoline. Reuben drove 390 miles on 12 gallons of gasoline. Who got the better gas mileage?

Hint: Compute the ratio miles per gallon.

Answer.

Rachel: 34 miles per gallon; Reuben: 32.5 miles per gallon

Checkpoint A.1.61.

Leslie drove 168 miles in 2.8 hours, and Mark drove 224 miles in 3.5 hours. Who drove at the greater average speed?

Hint: Compute the ratio miles per hour.

Answer.

Mark: 64 miles per hour; Leslie: 60 miles per hour

Subsubsection A.1.4.2 Calculate slope from a graph

We often think of slope as measuring the "steepness" of a graph, but the appearance of steepness is also affected by the scales on the axes.

Example A.1.62.

Calculate the slope of the line.

$line on grid$

Solution.

Choose two points on the line, and calculate the ratio of vertical change to horizontal change. Use the grid lines on the graph, but don’t forget to note the scales on the axes.

$line marked with horizontal and vertical change between two points$

The slope is the ratio $\dfrac{\Delta h}{\Delta t}\text{.}$ The variable on the horizontal axis increases by 4 units, from 2 to 6, so $\Delta t=4\text{.}$ The variable on the vertical axis increases by 8 grid lines, but each grid line represents 2 units, so $\Delta h=16\text{.}$ Thus, the slope is $\dfrac{\Delta h}{\Delta t} = \dfrac{16}{4}=4\text{.}$

Example A.1.63.

Calculate the slope of the line.

$line on grid$

Solution.

Choose two points on the line, and calculate the ratio of vertical change to horizontal change. Use the grid lines on the graph, but don’t forget to note the scales on the axes.

$line marked with horizontal and vertical change between two points$

The slope is the ratio $\dfrac{\Delta V}{\Delta t}\text{.}$ The horizontal variable, $t\text{,}$ increases by 6 grid lines, but each grid line represents 2 units, so $\Delta t=12\text{.}$ The vertical variable, $V\text{,}$ decreases by 3 grid lines, or 6 units, so $\Delta V=-6\text{.}$ Thus, $\dfrac{\Delta V}{\Delta t} = \dfrac{-6}{12}=\dfrac{-1}{2}\text{.}$

Checkpoint A.1.64.

Calculate the slope of the line.

Hint: Find two points that lie on the intersubsection of grid lines, so that it’s easy to read their coordinates. For example, you could use $(2, 300)$ and $(8, 600)\text{.}$

$line on grid$

Answer.

$50$

Checkpoint A.1.65.

Calculate the slope of the line.

Hint: Find two points that lie on the intersubsection of grid lines. For example, you could use $(0, 60)$ and $(3, -12)\text{.}$

$line on grid$

Answer.

$-24$

Subsubsection A.1.4.3 Interpret the slope

Example A.1.66.

Audrey can drive 150 miles on 6 gallons of gas, and 225 miles on 9 gallons of gas.

$graph$

Compute the slope of the graph, including units.
Interpret the slope as a rate; what does it tell you about the problem?

Solution.

$\displaystyle \dfrac{\Delta d}{\Delta g} = \dfrac{225-150}{9-6} = \dfrac{25\text{ miles}}{1\text{ gallon}} $
The car gets 25 miles per gallon in gas mileage

Checkpoint A.1.67.

The sales tax on a $15 purchase is 60 cents, and 80 cents on a $20 purchase.

$line on grid$

Compute the slope of the graph, including units.
Interpret the slope as a rate; what does it tell you about the problem?

Answer.

$\displaystyle \dfrac{4~ \text{cents}}{\text{dollar}}$
The tax rate is 4 cents per dollar, or 4%

Checkpoint A.1.68.

Lynette is saving money for the down payment on a new car. The figure below shows the amount $A$ she has saved, in dollars, $w$ weeks after the first of the year.

$graph of line$

Compute the slope of the graph, including units.
What does the slope tell you about the problem?

Answer.

$~\dfrac{\Delta A}{\Delta w}=50~$ dollars/week
Lynette is saving $50 per week.

Checkpoint A.1.69.

Jason is raising a rabbit for the county fair. The figure below shows the rabbit’s weight $W$ when it was $t$ weeks old.

$graph of line$

Compute the rabbit’s rate of growth, including units.
What does the slope tell you about the problem?

Answer.

$~\dfrac{\Delta W}{\Delta t}=1~$ pound/week
The rabbit gained 1 pound per week in weight.

Subsection A.1.5 Equations of Lines

Subsubsection A.1.5.1 Slope-Intercept Form

Because the $y$-intercept $(0,b)$ is the "starting value" of a linear model, and its rate of change is measured by its slope,$m\text{,}$ the equation for a linear model

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

can be expressed in symbols as

\begin{equation*} y = b + mx \end{equation*}

Slope-Intercept Form.

If we write the equation of a linear function in the form,

\begin{equation*} f (x) = b + mx \end{equation*}

then $m$ is the slope of the line, and $b$ is the $y$-intercept.

Example A.1.70.

The temperature inside a pottery drying oven starts at 70 degrees and is rising at a rate of 0.5 degrees per minute. Write a function for the temperature, $H\text{,}$ inside the oven after $t$ minutes.

Solution.

At $t=0\text{,}$ the temperature is 70 degrees, so $b=70\text{.}$

The slope is given by the rate of increase of $H\text{,}$ so $m=0.5\text{.}$

Thus, the function is

\begin{equation*} H=70+0.5t \end{equation*}

Example A.1.71.

A perfect score on a driving test is 120 points, and you lose 4 points for each wrong answer. Write a function for your score, $S\text{,}$ if you give $n$ wrong answers.

Solution.

If $n=0\text{,}$ your score is 120, so $b=120\text{.}$

Your score decreases by 4 points per wrong answer, so $m=\dfrac{\Delta S}{\Delta n} = 4\text{.}$

The function is

\begin{equation*} S=120-4n \end{equation*}

Checkpoint A.1.72.

Monica has saved $7800 to live on while she attends college. She spends $600 a month. Write a function for the amount, $S\text{,}$ in Monica’s savings account after $t$ months.

Answer.

$b=7800~~$ and $~~m=-400~~\text{,}$ so $~~S=7800-600t$

Checkpoint A.1.73.

Jesse opened a new doughnut shop in an old store-front. He invested $2400 in remodeling and set-up, and he makes about $400 per week from the business. Write a function giving the shop’s financial standing, $F\text{,}$ after $w$ weeks.

Answer.

$b=-2400~~$ and $~~m=400~~\text{,}$ so $~~F=-2400+400w$

Subsubsection A.1.5.2 Calculate slope using a formula

Recall that the subscripts on the coordinates in $~P_1(x_1,y_1)~$ and $~P_2(x_2,y_2)~$ just mean "first point" and "second point".

Two-Point Formula for Slope.

The slope of the line joining points $~P_1(x_1,y_1)~$ and $~P_2(x_2,y_2)~$ is

\begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{y_2-y_1}{x_2-x_1}~~~~~~\text{if}~~~~x_2 \not= x_1 \end{equation*}

Example A.1.74.

Compute the slope of the line joining $(-6,2)$ and $(3,-1)\text{.}$

Solution.

It doesn’t matter which point is $P_1$ and which is $P_2\text{,}$ so we choose $P_1$ to be $(-6,2)\text{.}$ Then $(x_1,y_1)=(-6,2)$ and $(x_2,y_2)=(3,-1)\text{.}$ Thus,

\begin{align*} m \amp = \dfrac{y_2-y_1}{x_2-x_1}\\ \amp = \dfrac{-1-2}{3-(-6)} = \dfrac{-3}{9} = \dfrac{-1}{3} \end{align*}

Caution A.1.75.

Make sure that you subtract both the $x$ and $y$ coordinates in the same order! That is, do NOT calculate

\begin{equation*} \dfrac{y_{\alert{2}}-y_{\alert{1}}}{x_{\alert{1}}-x_{\alert{2}}}~~~~~~~~~~\blert{\text{Incorrect!}} \end{equation*}

or your slope will have the wrong sign.

Checkpoint A.1.76.

Compute the slope of the line joining the points $(5,2)$ and $(8,7)\text{.}$

Answer.

$\dfrac{5}{3}$

Checkpoint A.1.77.

Compute the slope of the line joining the points $(-3,-4)$ and $(-7,1)\text{.}$

Answer.

$\dfrac{-5}{4}$

Subsubsection A.1.5.3 Point-Slope Form

If we don’t know the $y$-intercept of a line but we do know one other point and the slope, we can still find an equation for the line.

Point-Slope Formula.

To find an equation for the line of slope $m$ passing through the point $(x_1,y_1)\text{,}$ use the point-slope formula

\begin{equation*} \dfrac{y-y_1}{x-x_1} = m \end{equation*}

\begin{equation*} y-y_1=m(x-x_1) \end{equation*}

Example A.1.78.

Find an equation for the line that passes through $(1,3)$ and has slope $-2\text{.}$

Solution.

We substitute $x_1=\alert{1}\text{,}$ $y_1=\alert{3}\text{,}$ and $m=\alert{-2}$ into the point-slope formula.

\begin{align*} y-\alert{3} \amp = \alert{-2} (x-\alert{1}) \amp \amp \blert{\text{Apply the distributive law.}}\\ y-3 \amp = -2x + 2 \amp \amp \blert{\text{Add 3 to both sides.}}\\ y \amp = -2x + 5 \end{align*}

Example A.1.79.

Find an equation for the line of slope $\dfrac{-1}{2}$ that passes through $(-3,-2)\text{.}$

Solution.

We substitute $x_1=\alert{-3}\text{,}$ $y_1=\alert{-2}\text{,}$ and $m=\alert{\dfrac{-1}{2}}$ into the point-slope formula.

\begin{align*} \dfrac{y-(\alert{-2})}{x-(\alert{-3})} \amp = \alert{\dfrac{-1}{2}} \amp \amp \blert{\text{Simplify the left side.}}\\ \dfrac{y+2}{x+3} \amp = \dfrac{-1}{2} \amp \amp \blert{\text{Cross-multiply.}}\\ 2(y+2) \amp = -1(x+3) \amp \amp \blert{\text{Apply the distributive law.}}\\ 2y+4 \amp = -x-3 \amp \amp \blert{\text{Subtract 4 from both sides.}}\\ 2y \amp = -x-7 \amp \amp \blert{\text{Divide both sides by 2.}}\\ y \amp = \dfrac{-1}{2}x-\dfrac{7}{2} \end{align*}

Checkpoint A.1.80.

Find an equation for the line of slope $-4$ that passes through $(2,-5)\text{.}$

Answer.

$y=-4x+3$

Checkpoint A.1.81.

Find an equation for the line of slope $\dfrac{2}{3}$ that passes through $(-6,1)\text{.}$

Answer.

$y=\dfrac{2}{3}x+5$

Subsubsection A.1.5.4 Graphing a linear equation by the point-slope method

If we know one point on a line and its slope, we can sketch its graph without having to make a table of values.

Example A.1.82.

Graph the line $~y=\dfrac{3}{4}x-2$

Solution.

Step1: Begin by plotting the $y$-intercept, $(0,-2)\text{.}$

Step 2: We use the slope, $\dfrac{\Delta y}{\Delta x} = \dfrac{3}{4}\text{,}$ to find another point on the line, as follows. Start at the point $(0,-2)$ and move 3 units up and 4 units to the right. Plot a second point here, at $(4,1)\text{.}$

Step 3: Find a third point by writing the slope as $\dfrac{\Delta y}{\Delta x} = \dfrac{-3}{-4}\text{:}$ from $(0,-2)\text{,}$ move down 3 units and 4 units to the left. Plot a third point here, at $(-4,-5)\text{.}$

Finally, draw a line through the three points.

Example A.1.83.

Graph the line of slope $\dfrac{-1}{2}$ that passes through $(-3,-2)\text{.}$

Solution.

Step 1: Begin by plotting the point $(-3,-2)\text{.}$

Step 2: Use the slope, $\dfrac{\Delta y}{\Delta x} = \dfrac{-1}{2}\text{,}$ to find another point on the line, as follows. Start at the point $(-3,-2)$ and move 1 unit down and 2 units to the right. Plot a second point here, at $(-1,-3)\text{.}$

Step 3: Find a third point by writing the slope as $\dfrac{\Delta y}{\Delta x} = \dfrac{1}{-2}\text{:}$ from $(-3,-2)\text{,}$ move 1 unit up and 2 units to the left. Plot a third point here, at $(-5,-1)\text{.}$

Finally, draw a line through the three points.

Checkpoint A.1.84.

Graph the line $~y=\dfrac{-1}{3}x-3$

$10 by 10 grid$

Answer.

$y = -x/3-3$

Checkpoint A.1.85.

Graph the line with slope $m=\dfrac{3}{2}$ passing through $(-1,-2)\text{.}$

Answer.

$through (-1,-2) slope 3/2$

You have attempted of activities on this page.

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\(x\)	\(y\)	\(\hphantom{0000}\)
\(-3\)	\(12\)	\(y=-2(\alert{-3})+6=12\)
\(-2\)	\(10\)	\(y=-2(\alert{-2})+6=10\)
\(-1\)	\(8\)	\(y=-2(\alert{-1})+6=8\)
\(0\)	\(6\)	\(y=-2(\alert{0})+6=6\)
\(1\)	\(4\)	\(y=-2(\alert{1})+6=4\)
\(2\)	\(2\)	\(y=-2(\alert{2})+6=2\)
\(3\)	\(0\)	\(y=-2(\alert{3})+6=0\)
\(4\)	\(-2\)	\(y=-2(\alert{4})+6=-2\)

\(x\)	\(y\)	\(\hphantom{0000}\)
\(0\)	\(6000\)	\(y=6000-100(\alert{0})=6000\)
\(10\)	\(5000\)	\(y=6000-100(\alert{10})=5000\)
\(20\)	\(4000\)	\(y=6000-100(\alert{20})=4000\)

\(x\)	\(-8\)	\(-4\)	\(0\)	\(4\)	\(6\)
\(y\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)

\(x\)	\(-8\)	\(-4\)	\(0\)	\(4\)	\(6\)
\(y\)	\(-9\)	\(-6\)	\(-3\)	\(0\)	\(1.5\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(2\)	\(4\)
\(y\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)	\(\hphantom{0000}\)

\(x\)	\(y\)
\(250\)	\(\hphantom{0000}\)
\(600\)	\(\hphantom{0000}\)
\(800\)	\(\hphantom{0000}\)
\(1000\)	\(\hphantom{0000}\)

\(x\)	\(y\)
\(0\)	\(-800\)
\(50\)	\(-550\)
\(100\)	\(-300\)
\(200\)	\(200\)

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

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

\(d\)	\(N\)
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