Hello. I'm Professor Von Schmohawk and welcome to Why U.
In the previous lecture, we saw a race between a tortoise and a hare.
Although the hare decided to spend the first four hours of the race
reading a chapter in his Algebra book
he soon caught up with the tortoise.
Graphing their distances from the starting point as functions of time
we saw that the tortoise and hare each travelled at constant speeds
the tortoise starting first
followed four hours later by the hare.
Since the hare's speed was greater than the tortoise's
once he started, he quickly caught up
eventually passing the tortoise sometime between five and six hours after the race had begun.
At the end of the lecture we derived equations for these graphs.
In this lecture, we will see how to use these equations
to mathematically determine the coordinates of their intersection point
and thus the exact time
and distance
when the tortoise and hare meet.
There are several different mathematical methods
which can be used to find the points of intersection of two graphs.
In this lecture, we will demonstrate a method called "substitution".
We know that the coordinates of any point on an equation's graph
represent one possible solution to the equation.
So the set of all points which make up the graph
represents every possible solution to the equation.
Likewise, for a different equation
the set of all the points which make up that graph
represents every possible solution to that equation.
Any point which lies on both graphs, therefore represents a solution to both equations.
We can find this intersection point if we find values for x and y
which simultaneously satisfy both equations.
Each equation tells us something about the relationship between x and y.
For example, the second equation tells us that the value of y
must be equal to one-half the value of x.
Since we know that the value of y must be "one-half x"
we can substitute "one-half x" for y in the first equation.
This gives us an equation with a single variable, x.
We can then solve this equation to find the value of x.
Multiplying both sides by 2
on the left we get "2 times one-half x"
or x
and on the right, distributing the 2 to the terms in parentheses
we get "2 times 2x"
or 4x
minus 2 times 8
or 16.
Adding 16 to both sides
and subtracting x from both sides
we get "16 equals 4x minus x"
or "16 equals 3x".
Finally, dividing both sides by 3
gives us "sixteen-thirds equals x"
or "x equals sixteen-thirds".
Now that we know the value of x
we can set the value of x in either equation to sixteen-thirds
and find the corresponding value of y.
Either equation will give us the same answer
so since the second equation is the simplest, we will use it to find the value of y.
Setting x to sixteen-thirds
we see that y is one-half times sixteen-thirds
or eight-thirds.
We can check our calculated values for x and y
by substituting them for x and y in the original equations
and verifying that they create true statements.
So the values of x and y which satisfy both equations
are sixteen-thirds and eight-thirds.
These are the values of the x and y-coordinates of the intersection point of the two graphs.
Converting these fractions to mixed numbers
we see that this point represents an elapsed time of five and one-third hours
and a distance of two and two-thirds miles from the starting point.
So five hours and twenty minutes after the race has begun
the tortoise and hare meet two and two-thirds miles from the starting point.
This method of solving a system of equations is called "substitution".
This name is appropriate, since as we saw in this example
the value of y represented in terms of x in one equation
is substituted for the variable y in a second equation.
Since only the variable x remains in that equation
it can then be solved to find a unique value for x
which can then be substituted back into either equation to find a unique value for y.
In the next lecture, we will introduce a second method of solving systems of equations
called "elimination".