Content

In this course you will learn at which point two straight lines intersect. The concepts of the intersection point as well as the linear system of equations (with two unknown variables) are introduced. At the end you will be able to solve such a system using elimination by equating coefficents.

Prerequisites

You should know what a linear equation is and how to transform it. Furthermore, it is important that you can read points from a coordinate system and use them again. Since you will be dealing with straight lines and their equations in this course, you should know how to set up such a straight line equation and draw it.

Duration

This course takes about 1.5 hours.

Click on "Start race!" to view the race.

On the x-axis you see the elapsed time since the start,

on the y-axis the distance already covered. On the left side you can also see the current position of the hare and the tortoise.

As you could observe from the animation, the hare was able to overtake the tortoise. So on their race track there must be a meeting point of the two. This point is thus the answer to the question:

"When and where do the animals meet?"

You will find the answer to this question in several ways in this course.

Quick exercise:

Write down the two linear function equations for the graphs of the tortoise and hare.

Since points in the coordinate system are always expressed by $\mathsf{x}\backslash sf\; x$ and $\mathsf{y}\backslash sf\; y$, the axis "time" is set as $\mathsf{x}\backslash sf\; x$-axis and "distance travelled" as $\mathsf{y}\backslash sf\; y$-axis.

The straight line equations from the story now look like this:

• Tortoise: $\mathsf{y}=\frac{\mathsf{1}}{\mathsf{2}}\mathsf{x}+\mathsf{3}\backslash sf\; y\; =\; \backslash frac\{1\}\{2\}x\; +\; 3$

• Hare: $\mathsf{y}=\frac{\mathsf{3}}{\mathsf{4}}\mathsf{x}\backslash sf\; y\; =\; \backslash frac\{3\}\{4\}x$

So the point where two straight lines cross is the meeting point mentioned above. This is called an intersection point in mathematics. By reading off the intersection point from the graph, you can find the solution to the problem.

Check the intersection point

Once you have read an intersection point from the graph, you can substitute it into your two straight line equations to check it. The coordinates of the intersection are $\mathsf{P}(\mathsf{12}\mathsf{\mid }\mathsf{9})\backslash sf\; P\backslash ;(\backslash color\{\#CC0000\}\{12\}|\backslash color\{\#009999\}\{9\})$.

Tortoise: $\mathsf{y}=\frac{\mathsf{1}}{\mathsf{2}}\mathsf{x}+\mathsf{3}\backslash sf\; y\; =\; \backslash frac\{1\}\{2\}x\; +\; 3$

Hare: ${\displaystyle \mathsf{y}=\frac{\mathsf{3}}{\mathsf{4}}\mathsf{x}}\backslash displaystyle\; \backslash sf\; y\; =\; \backslash frac\{3\}\{4\}x$

Insert the coordinates of the intersection.

Tortoise: ${\displaystyle \mathsf{9}=\frac{\mathsf{1}}{\mathsf{2}}\cdot \mathsf{12}+\mathsf{39}=\mathsf{21}\cdot \mathsf{12}+\mathsf{3}}\backslash displaystyle\; \backslash sf\; \backslash color\{\#009999\}\{9\}=\backslash frac\{1\}\{2\}\backslash cdot\; \backslash color\{\#cc0000\}\{12\}\; +39=21\cdot 12+3$

Hare: ${\displaystyle \mathsf{9}=\frac{\mathsf{3}}{\mathsf{4}}\cdot \mathsf{12}}\backslash displaystyle\; \backslash sf\; \backslash color\{\#009999\}\{9\}=\backslash frac\{3\}\{4\}\backslash cdot\; \backslash color\{\#cc0000\}\{12\}$

If you get two true statements, you have read off the correct intersection.

Tortoise: $\mathsf{9}=\mathsf{9}\backslash sf\; \backslash color\{\#009999\}\{9\}=9$

Hare: ${\displaystyle \mathsf{9}=\mathsf{9}}\backslash displaystyle\; \backslash sf\; \backslash color\{\#009999\}\{9\}=9$

Since both statements are true, you know that you have determined the intersection correctly.

Now the hare and the tortoise race again, but this time the tortoise has a lead of $\mathsf{4}\mathsf{m}\backslash sf\; 4\; \backslash ;\; \{m\}$.

The straight line equation for the tortoise and the straight line equation for the hare are now given by:

Tortoise: ${\displaystyle \mathsf{y}=\frac{\mathsf{1}}{\mathsf{2}}\mathsf{x}+\mathsf{4}}\backslash displaystyle\; \backslash sf\; y=\; \backslash frac\{1\}\{2\}x\; +4$

Hare: ${\displaystyle \mathsf{y}=\frac{\mathsf{3}}{\mathsf{4}}\mathsf{x}}\backslash displaystyle\; \backslash sf\; y\; =\; \backslash frac\{3\}\{4\}x$

Again, we want to find out at what time $\mathsf{x}\backslash sf\; x$ the hare overtakes the tortoise.

a) Draw the equations of the straight lines in the coordinate system.

b) Read off the intersection point from the drawing.

c) Interpret what this intersection point means for the meeting point of both animals.

Under Construction

If the hare leaves his friend too big a head start at the beginning, the hare may never overtake the tortoise or the meeting point may be so late that your page is not big enough to capture the meeting point.

The graphical solution is therefore not always useful.

How do you proceed in such a case?

In the above example, you worked with two straight lines whose equations you know. One possible way to write a system of equations is to write the equations one below the other and number them with Roman numerals. This looks as follows:

What is a linear system?

• A linear equation is an equation depending linearly on one or more unknowns (variables), such as $xx$ and $yy$.

• Several linear equations with the same variables, for example $xx$ and $yy$, form a system of linear equations. The goal is to find values for the variables such that all equations are satisfied simultaneously.

To understand how the different solution methods work, you should first think about what is the determining feature of an intersection.

Think a bit

How do you detect an intersection?

Graphical peculiarity

Computational peculiarity

At the intersection point both equations of the equation system are fulfilled!

So if you plug the intersection point into both equations, you get the same, true statement!

Intersection point:$P(1\mathrm{\mid }3)\backslash \; P\; \backslash ;\; (\backslash color\{\#cc0000\}\{1\}|\backslash color\{\#009999\}\{3\})$

Line g:

$y=-\frac{2}{2}x+\frac{9}{2}y=-\backslash frac\{2\}\{2\}x\; +\; \backslash frac\{9\}\{2\}$

$3=-\frac{3}{2}\cdot 1+\frac{9}{2}\backslash color\{\#CC0000\}\{3\}=-\backslash frac\{3\}\{2\}\backslash cdot\; \backslash color\{\#009999\}\{1\}\; +\; \backslash frac\{9\}\{2\}$

$3=3\backslash color\{\#CC0000\}\{3\}=3$

Line h:

$y=2x+1y=2x\; +1$

$3=2\cdot 1+1\backslash color\{\#CC0000\}3=2\backslash cdot\; \backslash color\{\#009999\}\{1\}\; +1$

$3=3\backslash color\{\#CC0000\}\{3\}=3$

The intersection point is a point that lies on both lines. You can use this to calculate the intersection of the two functions.

The method used for this is called elimination by equating coefficients.

How does elimination by equating work?

As an example, consider two straight lines from the previous slides in the course. Write them down as a system of linear equations:

Since you are looking for a point where the two $yy$-values are equal, the right part of the equation must also have the same value:

Now you can equate the two right sides:

Exercise

Solve the first equation for $xx$!

Solve the first equation for $xx$ .

Plug that $xx$ into one of both equations

To demonstrate that it really doesn't matter which of the two equations you use, here are both equations.

$y=2\cdot 1+1\backslash color\{\#FF6600\}y\; =\; 2\; \backslash cdot\; \backslash color\{\#660099\}\{1\}\; +1$

$y=3\backslash color\{\#FF6600\}y\; =\; 3$

$y=-\frac{3}{2}\cdot 1+\frac{9}{2}\backslash color\{\#FF6600\}y\; =\; -\backslash frac\{3\}\{2\}\; \backslash cdot\; \backslash color\{\#660099\}\{1\}\; +\; \backslash frac\{9\}\{2\}$

$y=3\backslash color\{\#FF6600\}y\; =\; 3$

So the intersection point of both lines is $P(1\mathrm{\mid }3)P(\backslash color\{\#660099\}\{1\}|\backslash color\{\#FF6600\}3)$!

You have just seen how the equation procedure was applied in a specific example. Now we will look again in general how the procedure works.

Basic requirement

The two equations in your system of equations must both be solved for the same term! In this term there should be an unknown (often $xx$ or $yy$), which is then no longer on the other side.

Example 1

Example 2

Step 1

Put together the two sides of the equations that are not equal.

Example 1

Example 2

Step 2

Solve for the other variable.

Example 1

Example 2

Step 3

Plug the first solution into one of both equations.

Example 1

In $\mathrm{I}\backslash mathrm\{I\}$:

$y=2\cdot \frac{4}{5}+1y=\; 2\backslash cdot\; \backslash frac\{4\}\{5\}\; +1$

$y=\frac{13}{5}y=\backslash frac\{13\}\{5\}$

$P(\frac{4}{5}\mathrm{\mid }\frac{13}{5})P\; \backslash ;\; (\backslash frac\{4\}\{5\}|\backslash frac\{13\}\{5\})$

Example 2

In $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$:

$P(2\mathrm{\mid }\frac{5}{3})P\; \backslash ;\; (2|\backslash frac\{5\}\{3\})$

Consider the linear system

$\backslash ;$

Calculate the coordinates of the intersection point of the two lines using elimination by equationg coefficients! Berechne die Koordinaten des Schnittpunktes der zwei Geraden, indem du dazu das Gleichsetzungsverfahren benutzt!

Solution

To apply the elimination by equationg coefficients, you must solve the two equations either for $xx$ or for $yy$. Since a line equation has the form $y=mx+ty\; =\; mx\; +\; t$, it is easier to solve the equations for $yy$.

Keep in mind how to transform equations.

Solve for one variable

Apply the elimination

The elimination by equating coefficients ensures that only one of the two variables remains.

To do this, you have just solved both equations for $yy$. You can then set equal both equations, since in a linear system of equations the variables $xx$ and $yy$ satisfy both equations $\mathrm{I}\backslash mathrm\{I\}$ and $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$.

Now plug $xx$ into $\mathrm{I}\backslash mathrm\{I\}$ or $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$. To make the calculation easier, we may take the first equation.

$\mathrm{I}y=5\cdot 0-1\backslash mathrm\{I\}\; \backslash qquad\; y\; =\; 5\; \backslash cdot\; \backslash color\{\#CC0000\}\{0\}\; -1$

$\mathrm{I}y=-1\backslash mathrm\{I\}\backslash qquad\; y=-1$

So the intersection point of both lines is $P(0\mathrm{\mid }-1)P\; \backslash ;\; (0|-1)$.

Intersection point

You learned in this course what an intersection point is and how to read it off.

Linear system of equations

You now know that a linear system of equations consists of two (or more) linear equations. These equations depend on the same variables (usually denoted $xx$ and $yy$).

• Number the equations with Roman numerals so that you don't lose track of them.

• You are looking for the solution of the linear system of equations, which is the point $(x\mathrm{\mid }y)(x|y)$ at which all equations yield true statements. That means: If you plugthe point into the equations, something like $7=77=7$ comes out. (Hint: Plug $x=2x\; =\; 2$ and $y=0y=0$ into the above system).

• If you transform all equations of the sysetm into straight line equations of the form $y=m\cdot x+ty=m\backslash cdot\; x+t$, the solution of the equation system is exactly the intersection of these straight lines.

Elimination by equating coefficients

If you cannot read off the intersection or cannot read it off precisely enough, you can calculate it. To do this, you use the elimination by equating coefficients, which gives you the solution of the system in three steps:

1. Equating: Set equal equations $\mathrm{I}\backslash mathrm\{I\}$ and $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ .

2. Solving: Solve the resulting equation for one variable.

3. Plugging in: By substituting into the equation $\mathrm{I}\backslash mathrm\{I\}$ or $\mathrm{I}\mathrm{I}\backslash mathrm\{II\}$ you get the value of the other variable.