Last hour’s graphing of points and lines from equations with an x and a y variable showed you one way to find many ordered pairs that are true for a given equation. Plotting the intercepts allowed you to draw the line that contains each equation’s solutions.

This hour you will learn how to use a line’s slope and key points to write that line’s equation—in some ways, the reverse of the process you learned last hour—along with how to solve real-life problems about direct and inverse variation.

The graph of the equation y - 2x = 6 follows. When x = 0, y = 6, and when y = 0, x = -3. These two key points are the ordered pairs (0,6) and (-3,0). (0,6) is the y-intercept, and (-3,0) is the x-intercept.

The slope of this line is the ratio of rise to run → 6 to 3 → 2.

The preceding graph came from the equation y - 2x = 6.

If you move some of the terms and isolate the y variable, the equation would be this:

When the equation is written as y = 2x + 6, the 2 is the slope and the 6 is the y-intercept of the graph.

Looking at the intercepts and slope of the next two graphs might help you see this is not a coincidence—there’s definitely a close connection between the numbers in the equation and the slope and y-intercept:

Did you notice that the slope and the y-intercept from each graph also appear in the equation? (The x-intercept is helpful for graphing, but it isn’t included in the equation.)

In each equation, the y is isolated on the left side of the equals sign. The x with its coefficient is on the right side, followed by a number, called a constant. The slope is the coefficient of the x term, and the y-intercept is the constant. This leads to the slope/intercept formula for the graph of a line:

The following graphs give additional examples of possible m and b values, as shown by the slope and y-intercepts of these lines:

From the equation for line C, y = 3x - 1, you can see that m = 3 and b = -1. The graph of line C does have a y-intercept of -1 and a slope of 3.

From the equation for line D, , you can see that and b = 2. The graph of line D does have a y-intercept of 2 and a slope of .

The important part of this connection between the line and the equation is that if you know the slope and y-intercept for any line, you can write the equation for that line.

You might know the line’s slope and y-intercept from examining the graph of the line, or you might simply be told what the slope and y-intercept are. Either way, the equation for the line is based on y = mx + b.

For example, the equation for the line with slope of 4 and y-intercept of -3 is:

And the equation for the line with slope of and y-intercept of is: (This line would be rather difficult to graph, but it could be done.)

Now try to write the equation based on y = mx + b for each of the following lines, using their slope and y-intercept for m and b. Check your answers on pages 216-217.

The equation of a line can easily be written when you can find the slope and y-intercept from the graph or when you are told the values of the slope and y-intercept. Your algebra skills can also help you write the equation when the information you have is not quite so complete.

If you know a line’s slope (m) and y-intercept (b), you can write its equation with the help of y = mx + b. However, if all you know is the slope of the line and a random point on that line, how do you know the b value and the y-intercept?

You have at least two choices:

In the first choice, graphing is certainly possible, but it depends on your abilities with graph paper, straightedges, and so on. This could be challenging if the slope is a fraction or if the given point is something like (-8,28).

The second choice—using algebra to help you find the b value—makes perfect sense because you have been spending many hours learning algebra, and now it can be put to a good use. Let’s try it.

Consider the following information about a line:

Of these variables, you know from the point (4,6) that x is 4 and y is 6. You are also told that the slope is 3, so m is 3. Put these values in for y, m, and x:

Solve this equation for b:

You now know that b = -6. Put b = -6 along with m = 3 into y = mx + b, and you have the equation you are looking for:

You could verify this by graphing this equation, and you would see that a line with y-intercept of -6 and slope of 3 does contain the point (4,6), but this verification isn’t necessary. Your algebra skills will serve you well in this and many other equation-writing problems.

Follow the solution steps on the next example. Find the equation for the line that goes through the point (1,4) and has a slope of -2.

In this problem, the ordered pair (1,4) tells you that x = 1 and y = 4. The slope is -2, so m = -2. You need to find the b value. Put the values you know into y = mx + b:

Because m = -2 and b = 6, the equation for the line is y = -2x + 6.

Now try to write the equation for each of the following lines, checking your answers on page 217.

When you know the point on a line and the line’s slope, substituting the values for x, y, and m into y = mx + b gives you a way to know what b equals, and that leads you directly to knowing the equation of the line.

You can now write the equation of a line from two different methods: analyzing its graph to find its slope and its y-intercept or knowing a point on the line and the line’s slope. Depending on what you know about any given line, one of these methods can work. However, what if you only know two points that are on a line? Again, your algebra skills will help you find the equation.

Consider the line that goes through the points (1,2) and (3,6). Can you find the equation that describes this line?

You know the slope-intercept formula is y = mx + b, where (x,y) is a point on the line, m is the slope of the line, and b is the y-intercept of the line. But knowing two of the points on the line doesn’t really seem that helpful.

However, you can find the slope by using the rise-to-run ratio from Hour 15. With the two given points—(1,2) and (3,6)—the rise is 2 - 6 = -4, and the run is 1 - 3 = -2:

Because you are given two of the points on this line, choose one point and put its x and y values into y = mx + b. Let’s choose the point (1,2). The equation that can be solved to find b follows:

With m = 2 and b = 0, the equation for the line containing the points (1,2) and (3,6) is y = 2x + 0 or simply y = 2x.

What if you had chosen the point (3,6)? The slope is still 2, and the equation would be this:

With either point on the line, the equation is the same: y = 2x + 0 or y = 2x.

Follow the solution steps in the next example.

Find the equation of the line that contains the points (-1,4) and (3,-8).

Solution: Find the slope of the line and then choose one of the given points for the x and y values in y = mx + b:

Choose the point (-1,4). The equation becomes:

With m = -3 and b = 1, the equation for the line containing the points (-1,4) and (3,-8) is y = -3x + 1.

A good way to verify this answer is to put the point you didn’t choose, (3,-8), into the equation:

The point (3,-8) does make the equation true and validates your answer.

Now try to write the equation for each of the following lines, checking your answers on page 217.

Next you’ll finish this hour by seeing some real-life applications of equations that include both x and y variables.

In one of the earlier examples, the equation for the line that contains the points (1,2) and (3,6) is y = 2x + 0 or y = 2x. The ordered pair (1,2) shows that when x = 1, y = 2; the other ordered pair (3,6) shows that when x = 3, y = 6.

Can you predict the y value for the ordered pair if x = 5? The equation y = 2x certainly helps you see that if x = 5, then y = 10, because putting 5 in for x gives the following equation:

Think about the real-life situation of paying $2 per jar for strawberry jam at the farmer’s market. You know without doing any heavy algebra that three jars will cost $6, four jars will be $8, and so on.

Buying jars of this jam is a perfect example of direct variation. The cost ( y) rises constantly as you buy more jars (x), and it rises at a constant rate—each additional jar you buy, you pay $2 more. Another way to say this is that y varies directly as x.

The cost of the jam is based on the algebra equation y = 2x, where x is the number of jars and y is the total cost, and there is no price break for volume purchases here.

Buying jars of jam can be described as an example of “y varies directly as x” because as x increases, so does y; as x decreases, so does y.

Think of the algebra side of this story: the equation y = 2x has a slope (m) of 2 and a y-intercept (b) of 0. The graph of this equation would go through (0,0). With a slope of 2, the line would rise two units for every one unit of run because slope is the ratio of rise to run.

The line would rise to the right because the slope is positive, and the line would never end. There is an end to the number of jars the farmer can sell—when they’re sold out for the day—but the graphed line goes on infinitely, in both directions.

The graphed line also includes all the ordered pairs between the whole number values, such as (2.5,5) and (3.4,6.8). In reality, you can’t buy 2.5 jars of jam, but if you could, you would have to pay $5. A more realistic graph of the cost of the jam would be a series of dots beginning with the ordered pair (0,0)—if you buy no jars, you pay no money—and then (1,2), (2,4), (3,6), and so on.

When you can say that y varies directly as x, the general form of the equation is y = kx. When k is 5, the graphed line will be steeper than when k is 2, and y will increase more rapidly when k is 5 than when k is 2.

In many areas of research, business, and finance, direct variation plays an important role. For example, the weekly payroll costs for a company vary directly as the total number of full-time employees who are working that week. Last week, the payroll was $31,500 for 45 employees who are each paid on the same pay scale. To meet the expected increase in orders for their product next week, five people who had been laid off during a slowdown are called in to work. What will the payroll costs be?

To set up this situation as a direct variation equation y = kx, you need to know what the k value is. You know that when x is 45, y is 31,500. Put these values into the equation and solve for k:

The equation for this situation is y = 700x.

To find the payroll costs for five additional people, the x value will be 50:

The payroll costs for the 50 employees next week will be $35,000. This direct variation example depends on each employee receiving equal pay, which is not typical.

A more common direct variation problem is the cookie recipe example. The amount of chocolate chips in a batch of cookies varies directly with the number of cookies. For a large party coming up, you are going to bake 150 cookies, but your recipe will make only 60 cookies and calls for 2 cups of chocolate chips. What amount of chips will you need for the party recipe?

This is a direct variation problem, but be careful of the order. The chips vary directly with the number of cookies, so use y = kx, where y = 2 and x = 60. Solve for k:

The equation you will now use is . Put 150 in for x:

You’ll need 5 cups of chips for the 150 cookies.

Now try the following direct variation problems, checking your answers on page 218.

In many real-life situations, as one thing increases, another decreases. The following examples give you algebraic equations that apply in this type of variation problem.

When you order a large pizza to share with friends, the more people who want a piece, the smaller each piece will be. This is a great example of inverse variation. As the number of people increases, the size of the slice decreases, which is not at all like direct variation.

Consider the following plans for setting up the chairs for an outdoor wedding. In each row, the space between the chairs gets smaller as the number of chairs in the row increases. If 30 chairs fill the row when they are 12 inches apart, how many chairs can be in the row when they are 8 inches apart?

The inverse equation is xy = k. In this example, x is the number of chairs (30) and y is the space (12) between the chairs. Put 30 in the equation for x, 12 for y, and solve for k:

With k = 360, the equation for the chairs/spaces problem is xy = 360. Put 8 in for y to find the number of chairs:

There can be 45 chairs in each row if they are 8 inches apart.

Another example of inverse variation follows:

The inverse variation equation is xy = k. In this example, x is the speed of your car (50) and y is the time (2). Put 50 in the equation for x and 2 for y, and solve for k:

With k = 100, the equation for the problem is xy = 100. Put 60 in for x to find the time for the trip if you go 60 miles per hour:

If you go 60 miles per hour, it will take you about 1.7 hours, and 1.7 · 60 is 102 minutes, so the time will be about 18 minutes less than the two hours it takes when you go 50 miles per hour.

Although the graph of the direct variation equation is always a straight line, the inverse variation graph is a curve. So the two types of variation not only apply to different situations, but their equations and graphs are also different.

Now try to solve this inverse variation problem, checking your answer on page 218.

If you must use 2 Newtons of force on a crowbar that is 6 feet long, how much force will it take if the crowbar is 4 feet long?

Based on what you’ve learned in Hour 16, do the following problems. Check your answers with the solution key in Appendix B.