Linear Expressions

Section 3.1 Linear Expressions

This section addresses the following topics.

Interpret data in various formats and analyze mathematical models

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Read and use mathematical models in a technical document

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This section covers the following mathematical concepts.

Solve linear, rational, quadratic, and exponential equations and formulas (skill)

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Read and interpret models (critical thinking)

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Use models including linear, quadratic, exponential/logarithmic, and trigonometric (skill)

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Section 1.3 presented models in general. This section presents linear models. First, we look at some examples and learn how the pieces of a linear model work. Next, we learn to write linear models given a description of a problem. After that we practice solving for different parts of a linear equation. Section 3.3 will introduce a more in depth look at identifying linear models.

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Subsection 3.1.1 Linear Models

This section presents examples of linear models and provides an explanation for the two parts of a linear model.

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A linear model (equation) can be written in the following, equivalent forms.

$\displaystyle y = mx+b$

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$\displaystyle Ay+Bx+C = 0$

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The second form can be solved for y which will make it look like the first form. The models in this section will be in the first form which you may know as slope intercept. The $m$ is commonly described as the slope which is a measure of how steep the line is. In these models we will generalize slope to a rate. The $b$ is commonly called the intercept which indicates a shift of the line up or down. This shift will have a meaning in each application.

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Model 3.1.1. Temperature Change with Altitude.

As a result of atmospheric physics, temperature decreases as the distance above the ground increases. For lower altitudes this can be modeled as

\begin{equation*} T_A = T_G - \left(\frac{3.5^\circ \text{ F}}{1000 \text{ ft}}\right)A. \end{equation*}

where

$T_A$ is the expected temperature in degrees Fahrenheit at the specified altitude.

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$T_G$ is the temperature at ground level in degrees Fahrenheit.

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$A$ is the specified altitude in number of feet above ground level.

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$\frac{3.5^\circ}{1000 \text{ ft}}$ is the rate of temperature decrease.

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All temperatures are in Fahrenheit.

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Before we use this model, we need to figure out what each part is. $T_G$ is a parameter that we need to find before we use the model. A parameter is not a variable, rather it is a value (number) that we obtain from the circumstances and write into the model (equation) before we do any work.

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In contrast the ratio $-\frac{3.5^\circ}{1000 \text{ ft}}$ is a constant (not a parameter), because it is a result of atmospheric physics. This ratio is always in the equation, because it is not dependent on the location for this simplified model.

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Temperature ($T_A$) and altitude ($A$) are variables which implies that the model shows a relationship between these two properties.

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The model subtracting from the starting temperature results in a decrease of temperature from $T_G\text{.}$ This implies that temperature decreases with altitude.

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Every linear model (equation) has a rate. In this case $m=-\frac{3.5}{1000}\text{.}$

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Every linear model has a shift. In this case $b=T_G\text{.}$

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Now that we have analyzed the model, we are prepared to use it.

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Example 3.1.2.

If the temperature at ground level is 43° what is the temperature 1000 ft above ground level (AGL)? 2000 ft AGL, 3000 ft AGL, 3500 ft AGL?

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Because fractions of a degree are not useful in making decisions like what clothes to wear, we will round to units.

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Note $T_G = 43^\circ\text{.}$ We need to calculate $T_A$ for $A=1000, 2000, 3000, 3500\text{.}$

\begin{align*} T_{1000} & = 43^\circ - \frac{3.5^\circ}{1000 \text{ ft}}(1000 \text{ ft})\\ & = 39.5\\ & = 43^\circ - 3.5^\circ\\ & = 39.5\\ & \approx 40.\\ T_{2000} & = 43^\circ - \frac{3.5^\circ}{1000 \text{ ft}}(2000 \text{ ft})\\ & = 43^\circ - 7.0^\circ\\ & = 36.\\ T_{3000} & = 43^\circ - \frac{3.5^\circ}{1000 \text{ ft}}(3000 \text{ ft})\\ & = 43^\circ - 10.5^\circ\\ & = 32.5\\ & \approx 33.\\ T_{3500} & = 43^\circ - \frac{3.5^\circ}{1000 \text{ ft}}(3500 \text{ ft})\\ & = 43^\circ - 12.25^\circ\\ & = 30.75\\ & \approx 31. \end{align*}

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Notice that we now know that it will be below freezing just above 3000 ft.

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An Equation for Time to Altitude.

A fixed wing aircraft flown optimally climbs from a starting altitude at a fixed climb rate.

\begin{equation*} A_t = A_G+C \cdot t. \end{equation*}

$A_t$ is the altitude after $t$ minutes.

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$A_G$ is the starting altitude (likely ground level) in feet mean sea level (MSL).

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$C$ is the rate of climb in feet per minute.

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$t$ is the time since the climb began in minutes.

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Before we can use this equation, let’s analyze it. We will need to know the parameters $A_G$ and $C\text{.}$ A parameter is not a variable, rather it is a value (number) that we obtain from the circumstances and write into the model (equation) before we do any work. $A_G$ varies by airport, because they are at different altitudes. The rate $C$ must be obtained for each plane and is often available in the aircraft’s Pilot’s Operating Handbook (POH).

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Final altitude ($A_t$) and time ($t$) are the variables which implies that the model shows a relationship between time climbing and how high the plane is.

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In this model everything is added which matches the increase of elevation over time (adding makes the altitude bigger).

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Every linear model (equation) has a rate (the slope $m$ in the equation). In this case $m=\frac{C \text{ ft}}{1 \text{ min}}\text{.}$ Every linear model has a shift (the $b$ in the equation), which may be zero. In this case $b=A_G\text{.}$ $b=A_G=0$ is possible because an aircraft can take off from sea level (e.g., float planes). This shift makes sense, because the climb starts at the altitude of the ground: the plane was already shifted up by being at that airport.

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Now that we have analyzed this equation, we are prepared to use it.

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Example 3.1.3.

If a plane begins at 160 ft MSL and is climbing at 700 ft/min, how high will it be after 5 minutes? 10 minutes? 15 minutes?

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These calculations are made as part of safety planning. The data is sufficiently accurate that rounding is not necessary. Rather we make conservative estimates of the parameters, so that there is always a safety buffer. In this case a conservative estimate for $A_G$ is to round down: this will give us a lower altitude. If that lower altitude is safe, then altitude 5 feet higher will be safe as well. For the climb rate a conservative estimate is to round down as well. If we can reach an altitude climbing at 700 ft/min, then if we climb at 720 ft/min we will reach that safe altitude a little sooner.

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Note

\begin{equation*} A_t = 160 \text{ ft}+\frac{700 \text{ ft}}{\text{min}} \cdot t. \end{equation*}

The expected altitudes are

\begin{align*} A_5 & = 160 \text{ ft}+\frac{700 \text{ ft}}{\text{min}} \cdot 5 \text{ min}\\ & = 160 \text{ ft}+3500 \text{ ft}\\ & = 3660.\\ A_{10} & = 160 \text{ ft}+\frac{700 \text{ ft}}{\text{min}} \cdot 10 \text{ min}\\ & = 160 \text{ ft}+7000 \text{ ft}\\ & = 7160.\\ A_{15} & = 160 \text{ ft}+\frac{700 \text{ ft}}{\text{min}} \cdot 15 \text{ min}\\ & = 160 \text{ ft}+10500 \text{ ft}\\ & = 10660. \end{align*}

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Based on these calculations the plane will climb above 4500 ft MSL in between 5 and 10 minutes (closer to 5).

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Equation for Fuel Remaining.

When operated at a fixed power setting a vehicle burns the same amount of gas per hour (or other time unit). This leads to the linear model

\begin{equation*} F_t = F_I - r \cdot t. \end{equation*}

$F_t$ is the amount of fuel remaining after $t$ minutes.

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$F_I$ is the amount of fuel at the beginning.

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$r$ is the rate (volume per time) at which fuel is being consumed.

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$t$ is the time the vehicle has been operated.

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Fuel amounts will be measured in units of volume like gallons or liters. Time will be measured in minutes or hours. The rate $r$ is then in units such as gallons/hour or liters/min.

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Before we use this model, let’s analyze it. We will need to know the parameters $F_I$ and $r\text{.}$ These parameters are not variables (they remains the same the whole time the model is in use), rather they are values (numbers) that we obtain from the circumstances and write into the model (equation) before we do any work.

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The initial fuel $F_I$ is obtained by checking the fuel tanks or fuel gauges. The rate $r$ is often is not shown during operation (fuel gauges show how much is remaining rather than how fast it is used). The rate can sometimes be obtained from vehicle documentation.

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Final fuel ($F_t$) and time ($t$) are the variables which implies that the model shows a relationship between time flown and fuel available (left in the tanks).

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Because fuel decreases the $r \cdot t$ term is subtracted decreasing the amount from $F_I\text{.}$

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Every linear model (equation) has a rate. In this case $m=\frac{r \text{ gal}}{1 \text{ hr}}$ (or similar units).

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Every linear model has a shift. In this case $b=F_I\text{.}$ We can think of this shift in terms of the needle on the gas gauge moving up to represent the amount of fuel present.

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Now that we have analyzed this equation, we are prepared to use it.

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Example 3.1.4.

If a car begins with 20 gallons of fuel and burns 1.55 gallons per hour, how much fuel will it have after 1 hour, 2 hours, 3 hours, 36 minutes?

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A gallon is a large amount so we will maintain one decimal place precision. For safety we should always assume a larger fuel burn, so we will round fuel remaining down.

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Solution.

The model is

\begin{equation*} F_t = 20 \text{ gal} - \frac{1.55 \text{ gal}}{\text{hr}} \cdot t \text{ hr}\text{.} \end{equation*}

Thus

\begin{align*} F_1 & = 20 \text{ gal} - \frac{1.55 \text{ gal}}{\text{hr}} \cdot 1 \text{ hr}\\ & = 20 \text{ gal} - 1.55 \text{ gal}\\ & = 18.45\\ & \approx 18.4.\\ F_2 & = 20 \text{ gal} - \frac{1.55 \text{ gal}}{\text{hr}} \cdot 2 \text{ hrs}\\ & = 20 \text{ gal} - 3.10 \text{ gal}\\ & = 16.9.\\ F_3 & = 20 \text{ gal} - \frac{1.55 \text{ gal}}{\text{hr}} \cdot 3 \text{ hrs}\\ & = 20 \text{ gal} - 4.65 \text{ gal}\\ & = 15.35\\ & \approx 15.3.\\ F_{0.6} & = 20 \text{ gal} - \frac{1.55 \text{ gal}}{\text{hr}} \cdot 0.6 \text{ hrs}\\ & = 20 \text{ gal} - 0.93 \text{ gal}\\ & = 19.07\\ & \approx 19.0. \end{align*}

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Use this Checkpoint to practice using a linear model.

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Checkpoint 3.1.5.

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Subsection 3.1.2 Building Linear Models

The previous section presented linear models, and illustrated using the models provided. This section presents problems that can be modeled as linear equations, and illustrates writing the model (equation) before using it.

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A linear model has a starting point (shift, $b$) and rate (ratio, $m$). We need to identify these and then write the linear model

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

with these values. We should also label units and explain any parameters.

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Aside

Example 3.1.6.

Consider rope that costs $0.93 per foot with a shipping charge of $7.64. We will produce a model for the cost of each purchase.

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We can start by calculating the cost of some specific orders (experiment). Suppose we are purchasing 20 feet of this rope. The units (dollars/foot) suggest that we can multiply to calculate the cost. $20 \text{ ft} \cdot \frac{\$0.93}{\text{ft}}\text{.}$

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Notice this multiplication is also the same as using a ratio (proportion). We could setup $\frac{\$0.93}{1 \text{ft}} = \frac{C}{20 \text{ ft}}$ When we solve this we end up with the same multiplication $20 \text{ ft} \cdot \frac{\$0.93}{\text{ft}}\text{.}$

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Next we must add the shipping charge. Thus the final cost is $20 \text{ ft} \cdot \frac{\$0.93}{\text{ft}}+\$7.64 = \$26.24\text{.}$ Note there is no rounding because all numbers are exact and no fractions of a cent occurred.

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Suppose we are purchasing 100 feet of this rope. The cost for the 100 feet will be

\begin{equation*} 100 \text{ ft} \cdot \frac{\$0.93}{\text{ft}}\text{.} \end{equation*}

Then we must add the shipping charge. Thus the final cost is $100 \text{ ft} \cdot \frac{\$0.93}{\text{ft}}+\$7.64 = \$100.64\text{.}$

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We can now notice that this calculation is the same with any number of feet (unless the shipping charge increases for larger orders). So in general we can write the cost as

\begin{equation*} s \text{ ft} \cdot \frac{\$0.93}{\text{ft}}+\$7.64 = C\text{.} \end{equation*}

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This equation has a ratio ($0.93/1 ft), which is the cost per foot, and also has a shift (+$7.64), which is the fixed shipping cost. Thus this is another linear equation.

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When cost is set per linear foot, or per square yard, or similar per unit pricing we often end up with a linear model.

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Example 3.1.7.

At lower altitudes the barometric pressure typically drops 1 inHg for every 1000 feet of elevation gained (the air is less dense higher up). We will produce a model for pressure at altitude.

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As in the previous example, we will start by calculating some specific pressures (experiment). If the pressure on the ground is 29.76 inHg, what do we expect the pressure to be flying at 4500 ft above ground level?

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The pressure drop is a ratio $\frac{1 \text{ inHg}}{1000 \text{ ft}}\text{.}$ The units suggest we can multiply $4500 \text{ ft} \cdot \frac{1 \text{ inHg}}{1000 \text{ ft}} = 4.5 \text{ inHg}\text{.}$ This is the drop in pressure. To calculate the resulting pressure we need $29.76 \text{ inHg} - 4.5 \text{ inHg} = 25.26 \text{ inHg}\text{.}$ We retain 2 decimal places because that is the traditional amount for reporting by meteorologists. Written as one calculation this is $T = 29.76 \text{ inHg}-\left(4500 \text{ ft} \cdot \frac{1 \text{ inHg}}{1000 \text{ ft}}\right). $

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If the pressure on the ground is 30.02 inHg what do we expect the pressure to be flying at 6000 ft above ground level?

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The pressure drop is a ratio $\frac{1 \text{ inHg}}{1000 \text{ ft}}\text{.}$ The units suggest we can multiply $6000 \text{ ft} \cdot \frac{1 \text{ inHg}}{1000 \text{ ft}} = 6 \text{ inHg}\text{.}$ This is the drop in pressure. To calculate the resulting pressure we need $30.02 \text{ inHg} - 6 \text{ inHg} = 24.02 \text{ inHg}\text{.}$ Written as one calculation this is $P = 30.02 \text{ inHg}-\left(6000 \text{ ft} \cdot \frac{1 \text{ inHg}}{1000 \text{ ft}}\right). $

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Notice we could do this same calculation for any altitude. So in general we can write

\begin{equation*} P_A = P_G-\left(A \text{ ft} \cdot \frac{1 \text{ inHg}}{1000 \text{ ft}}\right). \end{equation*}

$P_A$ is the pressure at the specified altitude. $P_G$ is the pressure at ground level. $A$ is the altitude above ground level. This is a linear equation with a ratio of (-1/1000) which is the drop in pressure with altitude, and a shift of $P_G\text{,}$ which is the pressure on the ground.

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Example 3.1.8.

We will construct a model (equation) that converts temperature in Fahrenheit to temperature in Celsius. Every 9 degrees Fahrenheit is only 5 degrees Celsius, so to convert between them, we must scale the degrees. Also they use different values for the starting point (which is the freezing point of water). Fahrenheit starts at 32° and Celsius starts at 0°. Notice we have a ratio and a shift, so this already looks like a linear model.

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To begin with we will convert 52° F to Celsius. We will round to units, because this is just an example (no one will be injured in the demonstration of this model).

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The first step is to determine how many degrees above freezing. Because 32° F is the freezing temperature of water, 52° F is $52^\circ-32^\circ=20^\circ$ above freezing.

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The next step is to scale the degrees. The conversion ratio is $\frac{9^\circ \text{ F}}{5^\circ \text{ C}}\text{.}$ Converting the 20° above freezing is now a unit conversion. The units suggest we can multiply $20^\circ \text{ F} \cdot \frac{5^\circ \text{ C}}{9^\circ \text{ F}} \approx 11^\circ \text{ C}\text{.}$ Notice we flipped the conversion ratio so the Fahrenheit degrees would divide to one. The result is 11° C above the freezing point of water in Celsius.

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Finally we can add the degrees Celsius to the starting point (freezing temperature of water). Because that is 0°, we have $0^\circ+11^\circ=11^\circ$ C.

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If we write all of that as one step we obtain

\begin{equation*} (52^\circ \text{ F}-32^\circ \text{ F})\frac{5^\circ \text{ C}}{9^\circ \text{ F}}+0^\circ \text{ C} = 11^\circ \text{ C}\text{.} \end{equation*}

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Notice we could do this with any temperature. So if we call the temperature to convert $T$ we have

\begin{equation*} C = (T-32)\frac{5}{9}+0\text{.} \end{equation*}

This may not look like a starting point plus a ratio scaled. If we expand the expression we obtain

\begin{equation*} C = (T-32)\frac{5}{9}+0 = \frac{5}{9}T-\frac{160}{9}. \end{equation*}

So this is a linear model. We prefer the first form of the equation because the numbers have meaning (e.g., 32° is the freezing point of water as opposed to $160/9$ which has no useful meaning.

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The temperature conversion example illustrates an idea about models. We describe a linear model as having a ratio and a (that is one) shift. However, in the temperature conversion example there is a shift a ratio and another shift, or two shifts. We showed these can be combined as one shift. That is always true. However, sometimes the version with multiple shifts is easier to understand. This will be true when we look at graphs of quadratics and exponentials (other forms of models).

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Subsection 3.1.3 Solving Linear Equations

The first section demonstrated using linear models to calculate values by plugging in the inputs. However, sometimes we know the result and want to calculate the input. This requires solving the linear equation. This section reviews solving linear equation starting with non-contextualized examples and then using some of the models presented above.

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Before reading farther solve the equation $5x-7=12\text{.}$ What steps did you use? Why do they work? Example 3.1.9 is an example of solving another linear equation.

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Example 3.1.9.

Solve $-8x-3=5\text{.}$

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\begin{align*} -8x-3 & = 5.\\ -8x-3+3 & =5+3.\\ -8x & = 8.\\ \frac{-8x}{-8} & = \frac{8}{-8}.\\ x & = -1. \end{align*}

We added three because it eliminates the -3 (undoes subtraction of 3). We divided by negative eight because it eliminates the -8 (undoes the multiplication by -8). These steps isolate the variable for which we are solving.

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Checkpoint 3.1.10.

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Some linear equations need one more technique. What would you need to solve $17-4y = 14-y\text{?}$ Below is an example of solving a similar linear equation.

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Example 3.1.11.

Solve $17-4y = 5+2y\text{.}$

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\begin{align*} 17-4y & = 5+2y.\\ -5+17-4y & = -5+5+2y. & & \text{Collect constants on one side.}\\ 12-4y & = 2y.\\ 12-4y+4y & = 2y+4y. & & \text{Collect the variable on the other side.}\\ 12 & = 2y+4y.\\ 12 & = (2+4)y. & & \text{Factoring reduces to one instance of the variable.}\\ 12 & = 6y.\\ \frac{12}{6} & = \frac{6y}{6}. & & \text{Arithmetic to isolate } y\\ 2 & = y. \end{align*}

Because there were multiple instances of the variable in the initial equation, we had to combine like terms (factor and add).

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Checkpoint 3.1.12.

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Another linear equation is $\frac{x}{3}+\frac{x}{4}=\frac{7}{12}\text{.}$ How would you solve it?

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We can solve this the same as in Example 3.1.11 but because there are factions as coefficients we will use an additional technique.

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Example 3.1.13.

Solve $\frac{x}{5}+\frac{2x}{7}=\frac{34}{35}$

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\begin{align*} \frac{x}{5}+\frac{2x}{7} & = \frac{34}{35}.\\ 5 \cdot \left(\frac{x}{5}+\frac{2x}{7}\right) & = 5 \cdot \frac{34}{35}. & & \text{Eliminate a denominator}\\ \frac{5x}{5}+\frac{10x}{7} & = 5 \cdot \frac{34}{35}. & & \text{Distribute}\\ x+\frac{10x}{7} & = \frac{34}{7}.\\ 7 \cdot \left(x+\frac{10x}{7}\right) & = 7 \cdot \frac{34}{7}. & & \text{Eliminate the other denominator}\\ 7x+\frac{7 \cdot 10x}{7} & = 7 \cdot \frac{34}{7}. & & \text{Distribute}\\ 7x+10x & = 34.\\ (7+10)x & = 34. & & \text{Factor}\\ 17x & = 34.\\ \frac{17x}{17} & = \frac{34}{17}.\\ x & = 2. \end{align*}

This is referred to as clearing denominators. We are once again eliminating division by multiplying. Always remember to distribute. Note, we could multiply once if we figured out the correct number (it would be 35 in this case), but there are no prizes for doing this fast, so you can do this either way.

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Now that we have practiced solving linear equations, we can use this skill with the models.

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Example 3.1.14.

Given the temperature model in Temperature Change with Altitude and supposing the temperature at ground level is 65, determine at what altitude we expect the temperature to be freezing. Round to the tens position; our measurements and sensors are not more precise.

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In this case the model is $T_A = 65^\circ-\frac{3.5^\circ}{1000 \text{ ft}}A\text{.}$ We know $T_A=32^\circ$ and we want to calculate $A\text{,}$ the altitude in feet.

\begin{align*} 32^\circ & = 65^\circ-\frac{3.5^\circ}{1000 \text{ ft}}A\\ -65^\circ + 32^\circ & = -65^\circ + 65^\circ-\frac{3.5^\circ}{1000 \text{ ft}}A\\ -33^\circ & = -\frac{3.5^\circ}{1000 \text{ ft}}A\\ -\frac{1000 \text{ ft}}{3.5^\circ} ( -33^\circ) & = -\frac{1000 \text{ ft}}{3.5^\circ} \left( -\frac{3.5^\circ}{1000 \text{ ft}}A \right)\\ 9428.571428 \text{ ft} & = A.\\ 9430 \text{ ft} & \approx A. \end{align*}

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Example 3.1.15.

Given the time to altitude model in An Equation for Time to Altitude and supposing that we are climbing from 80 ft MSL to 5000 ft MSL with a climb rate of 700 ft/min, how long will it take to complete the climb?

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In this case the model is $A_t = 80 \text{ ft} + \frac{700 \text{ ft}}{\text{min}}t\text{.}$ We know $A_t=5000$ ft, and we want to know the time $t\text{.}$

\begin{align*} 5000 \text{ ft} & = 80 \text{ ft} + \frac{700 \text{ ft}}{\text{min}}t.\\ -80 \text{ ft} + 5000 \text{ ft} & = -80 \text{ ft} + 80 \text{ ft} + \frac{700 \text{ ft}}{\text{min}}t.\\ 4920 \text{ ft} & = \frac{700 \text{ ft}}{\text{min}}t.\\ \frac{\text{min}}{700 \text{ ft}} \cdot 4920 \text{ ft} & = \frac{\text{min}}{700 \text{ ft}} \cdot \frac{700 \text{ ft}}{\text{min}}t.\\ 7.028571429 \text{ min} & = t.\\ 8 \text{ min} & = t. \end{align*}

We round up as a safety margin: it is better to assume we need 8 minutes and be higher, than to hope we can recognize 0.03 of a minute (not quite 2 seconds).

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Use this Checkpoint to try solving a linear model.

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Checkpoint 3.1.16.

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Subsection 3.1.4 Identifying Linear Expressions

All of the equations in this section are linear. What can we use to identify linear expressions or linear equations? Table 3.1.17 shows examples of linear expressions and non-linear expressions.

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Table 3.1.17. Linear and Non-linear

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Linear	Non-linear
$5x+3$	$5x^2-x+3$
$y=11-\frac{7}{13}x$	$y=\frac{17}{x}$
$7x-9y=8$	$3-2xy=12$

Some equations that may not appear to be linear can be solved using the same methods.

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Example 3.1.18.

Solve $\frac{11}{x}+2 = \frac{18}{x}-5\text{.}$

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\begin{align*} \frac{11}{x}+2 & = \frac{18}{x}-5.\\ x \cdot \left(\frac{11}{x}+2\right) & = x \cdot \left(\frac{18}{x}-5\right). & & \text{Clearing the denominators}\\ \frac{11x}{x}+2x & = \frac{18x}{x}-5x. & & \text{Distribute. }\\ 11+2x & = 18-5x. & & \text{Now it appears linear.}\\ -11+11+2x & = -11+18-5x.\\ 2x & = 7-5x.\\ 2x+5x & = 7-5x+5x.\\ 7x & = 7.\\ \frac{7x}{7} & = \frac{7}{7}.\\ x & = 1. \end{align*}

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In Section 3.3 we will learn to identify linear models from data.

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Exercises 3.1.5 Exercises

Identify Shift and Ratio.

For each model identify the shift (y-intercept) and ratio (slope).

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1.

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2.

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Linear expressions in applications.

Answer these questions using the linear model described.

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Solving in Models.

Solve each equation for the variable indicated. Treat all other variables as parameters.

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10.

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11.

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12.

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13.

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14.

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15.

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16.

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Contextless Linear Equations.

Solve each of these equations.

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17.

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18.

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Identifying Linear Models.

For each of the following equations whether in context or not, determine whether it is linear.

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19.

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20.

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21.

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Prev Top Next

Linear	Non-linear
\(5x+3\)	\(5x^2-x+3\)
\(y=11-\frac{7}{13}x\)	\(y=\frac{17}{x}\)
\(7x-9y=8\)	\(3-2xy=12\)