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".