Mastering Linear Equations: Graphing Time-Distance Relationships
Unlock the power of linear equations through graphing. Discover how to analyze time-distance relationships, interpret slopes, and apply these skills to real-world scenarios. Enhance your problem-solving abilities today!

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Now Playing:Graph linear relations – Example 0a
Intros
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  1. What is interpolation and extrapolation?
Examples
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  1. A train travels at a speed of 120km/h. The equation that represents the relationship of distance (d) and time (t) is d=120t.
    1. Plot the equation on a graph.

    2. Calculate the distance that the train travels in 4.5 hours.

Identifying proportional relationships
Notes
We will learn how to plot linear equations, and then use the graphs as an aid to answer questions about the linear relations. Alternatively, we can find out the linear equations from their graphs. In addition, we will learn how to plot graphs and determine their equations when we are given the table of values of linear relations. We will also explore the concepts of interpolating and extrapolating data from graphs of linear equations.

How to solve an equation by graphing

A linear equation gives us a straight line when we graph out the points on it. They have a linear relationship. So how do we go about solving an equation by graphing? In this lesson, we'll specifically be focusing on linear equations.

How to graph linear equations

The first thing you'll have to do in order to graph a linear equation is to put the equation into the slope intercept form. When writing linear equations, you can write it a variety of ways. The slope intercept form looks like this:

y = mx + b

You may need to rearrange some of the terms around to help you isolate y to the left side of the equation before you can get it to look like the slope intercept form equation. Once you are done this, you can move on to step 2.

For this second step, plot out the y-intercept of the equation. Find out what y's coordinate is when x is equalled to 0. This will give you the y-intercept. At times, you may find that your equation doesn't give you a y-intercept (which is times when there is no "b" in your slope intercept form equation). In that case, your first point will be at 0,0 instead of at the y-intercept.

Thirdly, we'll find the second point that we'll be noting down. This will be found through the slope. When you have an equation in the form of a slope-intercept form, the "m" is your slope. The slope m is equalled to rise/run: the rise is how far it goes on the y-axis and the run is how far it goes on the x-axis. Using this knowledge, let it help you find where the second point is based off your first point. So for example, in step 2, if you found that a point was at (0, -2) and your "m" was 1, then your riserun\frac{rise}{run} is equalled to 11\frac{1}{1}. This means you'd move 1 point up on the y-axis, and 1 point to the right on the x-axis. You'll end up with the second point being at (1,-1).

For the fourth and last step, all you'll have to do is draw a line between the two points you've found from the above steps. Make sure that your graph is point the right way depending on if the slope is positive, negative, 0 or undefined.

Example problems

It's time to put the steps we just learned into use and see how to graph equations. Here are some example problems for you to walk through.

Question 1:

A train travels at a speed of 120km/h. The equation that represents the relationship of distance (d) and time (t) is d=120t.

Plot the equation on a graph

Solution:

The equation in this question is d=120t, which resembles y=mx+b. This means our equation is in slope intercept form.

Equation matches with the slope formula
Equation in slope form

So, draw the x and y axis

Draw out the x and y axis with labels
Draw x and y axis

120 is the slope. Slope riserun=1201\frac{rise}{run} = \frac{120}{1}

Since it is 120km/hour, it means the train moves 120km every hour. So the graph will look like this

Train moving at 120 km/h
Graph of the speed of a train

Question 2:

A train travels at a speed of 120km/h. The equation that represents the relationship of distance (d) and time (t) is d=120t.

Calculate the distance that the train travels in 4.5 hours.

Solution:

Plug the number into the equation

d = 120t

d = 120(4.5)

= 540 km

Question 3:

The graph shows the relationship of the temperature of water, c, and time, t, as the water boils.

Solving linear equations by graphing
Solving linear equations by graphing

What is the equation for the relationship?

Solution:

Try to write the equation in slope intercept form y = mx + b

So, look for the slope from the graph:

Look for the slope of the graph with a straight line
Find the slop of the graph

m = slope = riserun=202=10\frac{rise}{run} = \frac{20}{2} = 10

So the equation is: c = 10t

Question 4:

The graph shows the relationship of the temperature of water, c, and time, t, as the water boils.

Solving linear equations by graphing
Solving linear equations by graphing

Find the temperature of the water after 7 minutes of boiling.

Solution:

From the graph, we know that at 7 minutes the temperature is 70°

Look for the temperature after seven minutes
Find the temperature after 7 minutes

We can also get the answer algebraically

c = 10t

c = 10(t)

c = 70°C

Calculate the temperture after 7 minutes algebrically.
Calculation of temperature algebraically

Question 5

Plot the graph for each set of values. Then, find the equation that represents each relationship.

x

Y

-2

-3

-1

0

0

3

1

6

2

9

Solution:

Draw the appropriate x and y axis from the above table of values. This is a method to graphing linear equations.

Draw the x and y axis according to the table of values provided above
Draw x and y axis according to the data table

Plot the point based on the table, and draw the line

Plot the points according to the data table and draw the line
Plot the points and draw the line

Now, let's determine the equation.

It will be in this form: y = mx + b. We need to know the "m" and the "b", where m = slope. To find this we can calculate using the slope equation y2y1x2x1\frac{y_{2} - y_{1}}{x_{2} - x_{1}}.

slope = y2y1x2x1=6310=31=3\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - 3}{1 - 0} = \frac{3}{1} = 3

Calculate the slope using the slope formula algebraically
Calculation of slope algebraically

For "b", it is actually the y-intercept. From the graph, we know the y intercept = 3. Also, we can find the "b" by looking at the table of values because the x value of the coordinate of the y-intercept is always = 0. We can look at the table of values for x = 0, which gives us y = 3.

So the equation: y = 3x + 3

If you wanted to look more into graphing lines check out this online interactive simulation that touches on slope intercept form equations.

To further solidify your understanding of this lesson, check out these lessons on general linear function slope equations, the slope point form linear equations, and how to graph linear functions using a table of values. Then try out some word problems dealing with graphing linear functions.