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Section 3.5 Slope-Intercept Form

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permalinkIn this section, we will explore what is perhaps the most common way to write the equation of a line. It's known as slope-intercept form.

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Figure 3.5.1. Alternative Video Lesson
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Subsection 3.5.1 Slope-Intercept Definition

permalinkRecall Example 3.4.5, where Yara had $50 in her savings account when the year began, and decided to deposit $20 each week without withdrawing any money. In that example, we model using x to represent how many weeks have passed. After x weeks, Yara has added 20x dollars. And since she started with $50, she has

y=20x+50

permalinkin her account after x weeks. In this example, there is a constant rate of change of 20 dollars per week, so we call that the slope as discussed in Section 3.4. We also saw in Figure 3.4.7 that plotting Yara's balance over time gives us a straight-line graph.

permalinkThe graph of Yara's savings has some things in common with almost every straight-line graph. There is a slope, and there is a place where the line crosses the y-axis. Figure 3.5.3 illustrates this in the abstract.

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Figure 3.5.2. Yara's savings
permalinka coordinate plane with the line y=20x+50; the line extends to the left and the right
Figure 3.5.3. Generic line with slope and y-intercept

permalinkWe already have an accepted symbol, m, for the slope of a line. The y-intercept is a point on the y-axis where the line crosses. Since it's on the y-axis, the x-coordinate of this point is 0. It is standard to call the point (0,b) the y-intercept, and call the number b the y-coordinate of the y-intercept.

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Checkpoint 3.5.4.

Use Figure 3.4.7 to answer this question.

permalinkOne way to write the equation for Yara's savings was

y=20x+50

permalinkwhere both m=20 and b=50 are immediately visible in the equation. Now we are ready to generalize this.

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Definition 3.5.5. Slope-Intercept Form.

When x and y have a linear relationship where m is the slope and (0,b) is the y-intercept, one equation for this relationship is

(3.5.1)y=mx+b

and this equation is called the slope-intercept form of the line. It is called this because the slope and y-intercept are immediately discernible from the numbers in the equation.

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Checkpoint 3.5.6.
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Remark 3.5.7.

The number b is the y-value when x=0. Therefore it is common to refer to b as the initial value or starting value of a linear relationship.

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Example 3.5.8.

With a simple equation like y=2x+3, we can see that this is a line whose slope is 2 and which has initial value 3. So starting at y=3 on the y-axis, each time we increase the x-value by 1, the y-value increases by 2. With these basic observations, we can quickly produce a table and/or a graph.

x y
start on
y-axis ⟶
0 3 initial
⟵ value
increase
by 1⟶
1 5 increase
⟵ by 2
increase
by 1⟶
2 7 increase
⟵ by 2
increase
by 1⟶
3 9 increase
⟵ by 2
increase
by 1⟶
4 11 increase
⟵ by 2
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Example 3.5.9.

Decide whether data in the table has a linear relationship. If so, write the linear equation in slope-intercept form (3.5.1).

x-values y-values
0 āˆ’4
2 2
5 11
9 23
Explanation

To assess whether the relationship is linear, we have to recall from Section 3.3 that we should examine rates of change between data points. Note that the changes in \(y\)-values are not consistent. However, the rates of change are calculated as follows:

  • When \(x\) increases by \(2\text{,}\) \(y\) increases by \(6\text{.}\) The first rate of change is \(\frac{6}{2}=3\text{.}\)

  • When \(x\) increases by \(3\text{,}\) \(y\) increases by \(9\text{.}\) The second rate of change is \(\frac{9}{3}=3\text{.}\)

  • When \(x\) increases by \(4\text{,}\) \(y\) increases by \(12\text{.}\) The third rate of change is \(\frac{12}{4}=3\text{.}\)

Since the rates of change are all the same, \(3\text{,}\) the relationship is linear and the slope \(m\) is \(3\text{.}\) According to the table, when \(x=0\text{,}\) \(y=-4\text{.}\) So the starting value, \(b\text{,}\) is \(-4\text{.}\) So in slope-intercept form, the line's equation is \(y=3x-4\text{.}\)

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Checkpoint 3.5.10.
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Subsection 3.5.2 Graphing Slope-Intercept Equations

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Example 3.5.11.

The conversion formula for a Celsius temperature into Fahrenheit is F=95C+32. This appears to be in slope-intercept form, except that x and y are replaced with C and F. Suppose you are asked to graph this equation. How will you proceed? You could make a table of values as we do in Section 3.2 but that takes time and effort. Since the equation here is in slope-intercept form, there is a nicer way.

Since this equation is for starting with a Celsius temperature and obtaining a Fahrenheit temperature, it makes sense to let C be the horizontal axis variable and F be the vertical axis variable. Note the slope is 95 and the vertical intercept (here, the F-intercept) is (0,32).

  1. Set up the axes using an appropriate window and labels. Considering the freezing temperature of water (0∘ Celsius or 32∘ Fahrenheit), and the boiling temperature of water (100∘ Celsius or 212∘ Fahrenheit), it's reasonable to let C run through at least 0 to 100 and F run through at least 32 to 212.

  2. Plot the F-intercept, which is at (0,32).

  3. Starting at the F-intercept, use slope triangles to reach the next point. Since our slope is 95, that suggests a ā€œrunā€ of 5 and a ā€œriseā€ of 9 might work. But as Figure 3.5.12 indicates, such slope triangles are too tiny. You can actually use any fraction equivalent to 95 to plot using the slope, as in 1810, 9050, 90050, or 4525 which all reduce to 95. Given the size of our graph, we will use 9050 to plot points, where we will try a ā€œrunā€ of 50 and a ā€œriseā€ of 90.

  4. Connect your points with a straight line, use arrowheads, and label the equation.

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Figure 3.5.12. Graphing F=95C+32
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Example 3.5.13.

Graph y=āˆ’23x+10.

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permalinka coordinate plane with a scale of 2 on the x-axis and 5 on the y-axis; the y-intercept of (0,10) is plotted
(a) Setting up the axes in an appropriate window and making sure that the y-intercept will be visible, and that any ā€œrunā€ and ā€œriseā€ amounts we wish to use will not make triangles that are too big or too small.
permalinkthe previous graph with slope triangles drawn; to the right the run is 3 and the rise is -2; to the left the run is -3 and the rise is 2
(b) The slope is āˆ’23=āˆ’23=2āˆ’3. So we can try using a ā€œrunā€ of 3 and a ā€œriseā€ of āˆ’2 or a ā€œrunā€ of āˆ’3 and a ā€œriseā€ of 2.
permalinkthe previous graph with a smooth line drawn through all the points; there are arrowheads on the line and the line is labeled with the equation y=-2/3x+10
(c) Connecting the points with a straight line and adding labels.
Figure 3.5.14. Graphing y=āˆ’23x+10
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Example 3.5.15.

Graph y=3x+5.

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permalinka coordinate plane with a scale of 2 on the x-axis and 5 on the y-axis; the y-intercept of (0,5) is plotted
(a) Setting up the axes to make sure that the y-intercept will be visible, and that any ā€œrunā€ and ā€œriseā€ amounts we wish to use will not make triangles that are too big or too small.
permalinkthe previous graph with slope triangles drawn; to the right the run is 1 and the rise is 3; to the left the run is -1 and the rise is -3
(b) The slope is a whole number 3. Every 1 unit forward causes a change of positive 3 in the y-values.
permalinkthe previous graph with a smooth line drawn through all the points; there are arrowheads on the line and the line is labeled with the equation y=3x+5
(c) Connecting the points with a straight line and adding labels.
Figure 3.5.16. Graphing y=3x+5
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Subsection 3.5.3 Writing a Slope-Intercept Equation Given a Graph

permalinkWe can write a linear equation in slope-intercept form based on its graph. We need to be able to calculate the line's slope and see its y-intercept.

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Checkpoint 3.5.17.
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Checkpoint 3.5.18.
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Subsection 3.5.4 Writing a Slope-Intercept Equation Given Two Points

permalinkThe idea that any two points uniquely determine a line has been understood for thousands of years in many cultures around the world. Once you have two specific points, there is a straightforward process to find the slope-intercept form of the equation of the line that connects them.

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Example 3.5.19.

Find the slope-intercept form of the equation of the line that passes through the points (0,5) and (8,āˆ’5).

Explanation

We are trying to write down \(y=mx+b\text{,}\) but with specific numbers for \(m\) and \(b\text{.}\) So the first step is to find the slope, \(m\text{.}\) To do this, recall the slope formula (3.4.3) from Section 3.4. It says that if a line passes through the points \((x_1,y_1)\) and \((x_2,y_2)\text{,}\) then the slope is found by the formula \(m=\frac{y_2-y_1}{x_2-x_1}\text{.}\)

Applying this to our two points \((\overset{x_1}{0},\overset{y_1}{5})\) and \((\overset{x_2}{8},\overset{y_2}{-5})\text{,}\) we see that the slope is:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{-5}-\substitute{5}}{\substitute{8}-\substitute{0}}\\ \amp=\frac{-10}{8}=-\frac{5}{4} \end{align*}

We are trying to write \(y=mx+b\text{.}\) Since we already found the slope, we know that we want to write \(y=-\frac{5}{4}x+b\) but we need a specific number for \(b\text{.}\) We happen to know that one point on this line is \((0,5)\text{,}\) which is on the \(y\)-axis because its \(x\)-value is \(0\text{.}\) So \((0,5)\) is this line's \(y\)-intercept, and therefore \(b=5\text{.}\) So, our equation is

\begin{equation*} y=-\frac{5}{4}x+5\text{.} \end{equation*}
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Example 3.5.20.

Find the slope-intercept form of the equation of the line that passes through the points (3,āˆ’8) and (āˆ’6,1).

Explanation

The first step is always to find the slope between our two points: \((\overset{x_1}{3},\overset{y_1}{-8})\) and \((\overset{x_2}{-6},\overset{y_2}{1})\text{.}\) Using the slope formula (3.4.3) again, we have:

\begin{align*} m\amp=\frac{y_2-y_1}{x_2-x_1}\\ \amp=\frac{\substitute{1}-\substitute{(-8)}}{\substitute{{-}6}-\substitute{3}}\\ \amp=\frac{9}{-9}\\ \amp=-1 \end{align*}

Now that we have the slope, we can write \(y=-1x+b\text{,}\) which simplifies to \(y=-x+b\text{.}\) Unlike in Example 3.5.19, we are not given the value of \(b\) because neither of our two given points have an \(x\)-value of \(0\text{.}\) The trick to finding \(b\) is to remember that we have two points that we know make the equation true! This means all we have to do is substitute either point into the equation for \(x\) and \(y\) and solve for \(b\text{.}\) Let's arbitrarily choose \((3,-8)\) to plug in.

\begin{align*} y\amp=-x+b\\ \substitute{-8}\amp=-(\substitute{3})+b\amp\text{(Now solve for }b\text{.)}\\ -8\amp=-3+b\\ -8\addright{3}\amp=-3+b\addright{3}\\ -5\amp=b \end{align*}

In conclusion, the equation for which we were searching is \(y=-x-5\text{.}\)

Don't be tempted to plug in values for \(x\) and \(y\) at this point. The general equation of a line in any form should have (at least one, and in this case two) variables in the final answer.

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Checkpoint 3.5.21.
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Checkpoint 3.5.22.
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Subsection 3.5.5 Modeling with Slope-Intercept Form

permalinkWe can model many relatively simple relationships using slope-intercept form, and then solve related questions using algebra. Here are a few examples.

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Example 3.5.23.

Uber is a ride-sharing company. Its pricing in Portland factors in how much time and how many miles a trip takes. But if you assume that rides average out at a speed of 30 mph, then their pricing scheme boils down to a base of $7.35 for the trip, plus $3.85 per mile. Use a slope-intercept equation and algebra to answer these questions.

  1. How much is the fare if a trip is 5.3 miles long?

  2. With $100 available to you, how long of a trip can you afford?

Explanation

The rate of change (slope) is \(\$3.85\) per mile, and the starting value is \(\$7.35\text{.}\) So the slope-intercept equation is

\begin{equation*} y=3.85x+7.35\text{.} \end{equation*}

In this equation, \(x\) stands for the number of miles in a trip, and \(y\) stands for the amount of money to be charged.

If a trip is \(5.3\) miles long, we substitute \(x=5.3\) into the equation and we have:

\begin{align*} y\amp=3.85x+7.35\\ \amp=3.85(\substitute{5.3})+7.35\\ \amp=20.405+7.35\\ \amp=27.755 \end{align*}

And the \(5.3\)-mile ride will cost you about \(\$27.76\text{.}\) (We say ā€œabout,ā€ because this was all assuming you average 30 mph.)

Next, to find how long of a trip would cost \(\$100\text{,}\) we substitute \(y=100\) into the equation and solve for \(x\text{:}\)

\begin{align*} y\amp=3.85x+7.35\\ \substitute{100}\amp=3.85x+7.35\\ 100\subtractright{7.35}\amp=3.85x\\ 92.65\amp=3.85x\\ \divideunder{92.65}{3.85}\amp=x\\ 24.06\amp\approx x \end{align*}

So with \(\$100\) you could afford a little more than a \(24\)-mile trip.

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Checkpoint 3.5.24.
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Reading Questions 3.5.6 Reading Questions

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1.

How does ā€œslope-intercept formā€ get its name?

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2.

What are two phrases you can use for ā€œbā€ in a slope-intercept form line equation?

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3.

Explain the two basic steps to graphing a line when you have the equation in slope-intercept form. (Not counting the step where you draw and label the axes and ticks.)

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Exercises 3.5.7 Exercises

Review and Warmup
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1.

Evaluate 2A+10a for A=1 and a=āˆ’8.

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2.

Evaluate āˆ’5B+3C for B=7 and C=āˆ’9.

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3.

Evaluate

y2āˆ’y1x2āˆ’x1

for x1=14, x2=āˆ’4, y1=āˆ’5, and y2=2:

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4.

Evaluate

y2āˆ’y1x2āˆ’x1

for x1=18, x2=āˆ’19, y1=12, and y2=āˆ’9:

Identifying Slope and y-Intercept
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5.

Find the line’s slope and y-intercept.

A line has equation y=2x+7.

This line’s slope is .

This line’s y-intercept is .

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6.

Find the line’s slope and y-intercept.

A line has equation y=3x+4.

This line’s slope is .

This line’s y-intercept is .

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7.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’8xāˆ’10.

This line’s slope is .

This line’s y-intercept is .

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8.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’7xāˆ’4.

This line’s slope is .

This line’s y-intercept is .

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9.

Find the line’s slope and y-intercept.

A line has equation y=xāˆ’4.

This line’s slope is .

This line’s y-intercept is .

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10.

Find the line’s slope and y-intercept.

A line has equation y=x+2.

This line’s slope is .

This line’s y-intercept is .

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11.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’x+5.

This line’s slope is .

This line’s y-intercept is .

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12.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’x+7.

This line’s slope is .

This line’s y-intercept is .

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13.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’89x+1.

This line’s slope is .

This line’s y-intercept is .

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14.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’27x+10.

This line’s slope is .

This line’s y-intercept is .

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15.

Find the line’s slope and y-intercept.

A line has equation y=12x+8.

This line’s slope is .

This line’s y-intercept is .

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16.

Find the line’s slope and y-intercept.

A line has equation y=14xāˆ’7.

This line’s slope is .

This line’s y-intercept is .

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17.

Find the line’s slope and y-intercept.

A line has equation y=1+5x.

This line’s slope is .

This line’s y-intercept is .

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18.

Find the line’s slope and y-intercept.

A line has equation y=āˆ’7+6x.

This line’s slope is .

This line’s y-intercept is .

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19.

Find the line’s slope and y-intercept.

A line has equation y=7āˆ’x.

This line’s slope is .

This line’s y-intercept is .

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20.

Find the line’s slope and y-intercept.

A line has equation y=8āˆ’x.

This line’s slope is .

This line’s y-intercept is .

Graphs and Slope-Intercept Form
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29.

Graph the equation y=4x.

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30.

Graph the equation y=5x.

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31.

Graph the equation y=āˆ’3x.

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32.

Graph the equation y=āˆ’2x.

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33.

Graph the equation y=52x.

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34.

Graph the equation y=14x.

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35.

Graph the equation y=āˆ’13x.

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36.

Graph the equation y=āˆ’54x.

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37.

Graph the equation y=5x+2.

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38.

Graph the equation y=3x+6.

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39.

Graph the equation y=āˆ’4x+3.

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40.

Graph the equation y=āˆ’2x+5.

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41.

Graph the equation y=xāˆ’4.

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42.

Graph the equation y=x+2.

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43.

Graph the equation y=āˆ’x+3.

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44.

Graph the equation y=āˆ’xāˆ’5.

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45.

Graph the equation y=23x+4.

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46.

Graph the equation y=32xāˆ’5.

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47.

Graph the equation y=āˆ’35xāˆ’1.

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48.

Graph the equation y=āˆ’15x+1.

Writing a Slope-Intercept Equation Given Two Points
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49.

Find the following line’s equation in slope-intercept form.

The line passes through the points (2,12) and (4,20).

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50.

Find the following line’s equation in slope-intercept form.

The line passes through the points (4,21) and (1,6).

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51.

Find the following line’s equation in slope-intercept form.

The line passes through the points (āˆ’5,13) and (3,āˆ’3).

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52.

Find the following line’s equation in slope-intercept form.

The line passes through the points (2,āˆ’14) and (āˆ’1,1).

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53.

Find the following line’s equation in slope-intercept form.

The line passes through the points (5,āˆ’12) and (4,āˆ’11).

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54.

Find the following line’s equation in slope-intercept form.

The line passes through the points (1,āˆ’6) and (āˆ’2,āˆ’3).

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55.

Find the following line’s equation in slope-intercept form.

The line passes through the points (7,11) and (21,19).

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56.

Find the following line’s equation in slope-intercept form.

The line passes through the points (āˆ’9,āˆ’3) and (9,7).

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57.

Find the following line’s equation in slope-intercept form.

The line passes through the points (āˆ’5,13) and (āˆ’10,19).

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58.

Find the following line’s equation in slope-intercept form.

The line passes through the points (6,āˆ’4) and (3,3).

Applications
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59.

A gym charges members $40 for a registration fee, and then $38 per month. You became a member some time ago, and now you have paid a total of $534 to the gym. How many months have passed since you joined the gym?

months have passed since you joined the gym.

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60.

Your cell phone company charges a $29 monthly fee, plus $0.15 per minute of talk time. One month your cell phone bill was $98. How many minutes did you spend talking on the phone that month?

You spent talking on the phone that month.

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61.

A school purchased a batch of T-shirts from a company. The company charged $3 per T-shirt, and gave the school a $60 rebate. If the school had a net expense of $960 from the purchase, how many T-shirts did the school buy?

The school purchased T-shirts.

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62.

Laney hired a face-painter for a birthday party. The painter charged a flat fee of $55, and then charged $5.50 per person. In the end, Laney paid a total of $181.50. How many people used the face-painter’s service?

people used the face-painter’s service.

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63.

A certain country has 277.29 million acres of forest. Every year, the country loses 3.51 million acres of forest mainly due to deforestation for farming purposes. If this situation continues at this pace, how many years later will the country have only 136.89 million acres of forest left? (Use an equation to solve this problem.)

After years, this country would have 136.89 million acres of forest left.

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64.

Irene has $77 in her piggy bank. She plans to purchase some Pokemon cards, which costs $1.45 each. She plans to save $65.40 to purchase another toy. At most how many Pokemon cards can he purchase?

Write an equation to solve this problem.

Irene can purchase at most Pokemon cards.

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65.

By your cell phone contract, you pay a monthly fee plus $0.05 for each minute you spend on the phone. In one month, you spent 290 minutes over the phone, and had a bill totaling $32.50.

Let x be the number of minutes you spend on the phone in a month, and let y be your total cell phone bill for that month, in dollars. Use a linear equation to model your monthly bill based on the number of minutes you spend on the phone.

  1. This line’s slope-intercept equation is .

  2. If you spend 150 minutes on the phone in a month, you would be billed .

  3. If your bill was $42.50 one month, you must have spent minutes on the phone in that month.

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66.

A company set aside a certain amount of money in the year 2000. The company spent exactly $38,000 from that fund each year on perks for its employees. In 2002, there was still $702,000 left in the fund.

Let x be the number of years since 2000, and let y be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

  1. The linear model’s slope-intercept equation is .

  2. In the year 2010, there was left in the fund.

  3. In the year , the fund will be empty.

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67.

A biologist has been observing a tree’s height. This type of tree typically grows by 0.24 feet each month. Fourteen months into the observation, the tree was 15.06 feet tall.

Let x be the number of months passed since the observations started, and let y be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

  1. This line’s slope-intercept equation is .

  2. 27 months after the observations started, the tree would be feet in height.

  3. months after the observation started, the tree would be 25.86 feet tall.

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68.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way. Each minute, they lose 7.7 grams. Five minutes since the experiment started, the remaining gas had a mass of 331.1 grams.

Let x be the number of minutes that have passed since the experiment started, and let y be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

  1. This line’s slope-intercept equation is .

  2. 34 minutes after the experiment started, there would be grams of gas left.

  3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

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69.

A company set aside a certain amount of money in the year 2000. The company spent exactly the same amount from that fund each year on perks for its employees. In 2003, there was still $617,000 left in the fund. In 2007, there was $425,000 left.

Let x be the number of years since 2000, and let y be the amount of money, in dollars, left in the fund that year. Use a linear equation to model the amount of money left in the fund after so many years.

  1. The linear model’s slope-intercept equation is .

  2. In the year 2010, there was left in the fund.

  3. In the year , the fund will be empty.

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70.

By your cell phone contract, you pay a monthly fee plus some money for each minute you use the phone during the month. In one month, you spent 300 minutes on the phone, and paid $15.50. In another month, you spent 360 minutes on the phone, and paid $16.40.

Let x be the number of minutes you talk over the phone in a month, and let y be your cell phone bill, in dollars, for that month. Use a linear equation to model your monthly bill based on the number of minutes you talk over the phone.

  1. This linear model’s slope-intercept equation is .

  2. If you spent 150 minutes over the phone in a month, you would pay .

  3. If in a month, you paid $17.60 of cell phone bill, you must have spent minutes on the phone in that month.

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71.

Scientists are conducting an experiment with a gas in a sealed container. The mass of the gas is measured, and the scientists realize that the gas is leaking over time in a linear way.

Seven minutes since the experiment started, the gas had a mass of 248.2 grams.

Thirteen minutes since the experiment started, the gas had a mass of 204.4 grams.

Let x be the number of minutes that have passed since the experiment started, and let y be the mass of the gas in grams at that moment. Use a linear equation to model the weight of the gas over time.

  1. This line’s slope-intercept equation is .

  2. 34 minutes after the experiment started, there would be grams of gas left.

  3. If a linear model continues to be accurate, minutes since the experiment started, all gas in the container will be gone.

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72.

A biologist has been observing a tree’s height. 14 months into the observation, the tree was 16 feet tall. 20 months into the observation, the tree was 16.9 feet tall.

Let x be the number of months passed since the observations started, and let y be the tree’s height at that time, in feet. Use a linear equation to model the tree’s height as the number of months pass.

  1. This line’s slope-intercept equation is .

  2. 27 months after the observations started, the tree would be feet in height.

  3. months after the observation started, the tree would be 21.85 feet tall.

Challenge
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73.

Line S has the equation y=ax+b and Line T has the equation y=cx+d. Suppose a>b>c>d>0.

  1. What can you say about Line S and Line T, given that a>c? Give as much information about Line S and Line T as possible.

  2. What can you say about Line S and Line T, given that b>d? Give as much information about Line S and Line T as possible.