Algebra 25 – Linear Equations in the Real World

Hello. I’m Professor Von Schmohawk
and welcome to Why U. In the last several lectures, we introduced
several forms of linear equations whose graphs are lines in the xy-plane. In this chapter, we will use linear equations
to solve some real-world problems. Why Hello Hulk. I’m so happy to see that
you are gainfully employed this summer. I was looking for someone
to drain my pool tomorrow. Uh, sure Professor Ethylene but I’m supposed to clean
Mr. Squeegeeopolis’ pool tomorrow. Well I’m sorry Hulk, but I need it drained
by eight p.m. tomorrow. I’m having some very important guests over and the pool must be completely empty
by the time they arrive. Okay Professor, I’ll figure something out. Oh Hulk, you’re such a resourceful
young man. I knew I could count on you! Ahem. What’s the matter Hulk? Have you perhaps encountered
a small resource scheduling problem? Okay Geekman,
how much is this going to cost me? Well my normal short-notice subcontracting
fee is fifty dollars an hour but I’m running a special today …
only forty-nine buckeroonies an hour! Yeah? Well I’m running a special today
on knuckle sandwiches. Perhaps I didn’t mention
that you were eligible for the double-extra-special-very-best-buddy rate
of three dollars an hour? Well, just make sure you have that pool empty
by eight p.m. tomorrow. Yes sir, Mr. Moosemasher! Rats! It’s almost noon. I’d better get crackin! Start pool draining at twelve noon. Hmm, It’s one thirty
and the water level is ten feet. Hmm, three o’clock
and the water level is eight feet. Yikes! This pool better be empty by eight p.m. Well, it looks like Ace Algebra student
A.V. Geekman may be in hot water. Will the pool be empty by eight p.m.? If we graph the water depth versus time we might be able to figure out how long it
will take for the pool to completely drain. We will plot the amount of time
since the drain was first opened at noon on the horizontal axis and water depth on the vertical axis. We know that at one-thirty, one and one-half
hours after the drain was opened the water level was ten feet and at three o’clock
after three hours of draining the water level was eight feet. However, there is one problem. With only two data points
it is impossible to determine whether the water level is dropping
by the same amount every hour or if the water level drops more quickly
as time goes on or perhaps if the water level drops more slowly
with increasing time. Taking one additional measurement
a third time would allow us to determine which scenario
more accurately represents the water level as a function of time. Dang! Its six o’clock already and we still have four feet to go. We now can add a third data point to our graph
of water depth versus elapsed time since A.V. has determined that after six hours
the water depth is four feet. It looks like these three points
all fall on a line so maybe the function we are graphing is linear. We can verify this mathematically by calculating the slopes of the line segments
connecting the first and second points and the second and third points. If the slopes of both segments are the same
then all three points must lie on the same line. As we saw in the chapter on slope we can calculate the slope of a line
which passes through two points as the difference in the point’s y-coordinates,
delta-y divided by the difference in their x-coordinates,
delta-x. Dividing delta-y by delta-x we get negative four over three or a slope of negative four-thirds. Now let’s calculate the slope
of the other line segment. Once again, we take the difference
in the y-coordinates, delta-y and the difference in the x-coordinates, delta-x. Dividing delta-y by delta-x we get negative two over three-halves. We can simplify this complex fraction
by dividing negative two by three-halves. This is the same as multiplying negative two
times the reciprocal of three-halves, or two-thirds. Negative two times two-thirds
is negative four-thirds. So the slope of this line segment
is also negative four-thirds. Since the slopes of these two connected
line segments are the same all three points must lie on the same line. We can therefore assume that the water
in the pool is draining at a constant rate and that the water depth
as a function of time is linear. Since we know this line’s slope, and at least
one point which falls on the line we can write the equation for this line
using the point-slope form. If we choose the bottom point
which has coordinates (6,4) to use as the known point
in the point-slope formula then x-one is six y-one is four and the slope m, is negative four-thirds. Now that we have an equation
which represents water level, y as a function of elapsed time, x we can plug in different values for x to see what the water level will be
at those points in time. For example, an elapsed time, x, of zero corresponds to the time
when the drain was first opened. If we set x to zero then the resulting value of y
will tell us the initial depth of the pool. Completing the arithmetic, zero minus six is negative six. And negative four-thirds times negative six is positive twenty-four thirds or eight. Adding four to both sides of the equation allows us to cancel negative four on the left leaving us with “y equals eight plus four” or twelve. So the initial depth of the pool when the
elapsed time x, was zero, was twelve feet. If we would like to find out how long it will
take for the pool to be empty we can set the water level, y, to zero and solve the equation for x, to see what
the elapsed time is at that point. We start by setting the value of y to zero and we can then write “zero minus four” as negative four. Since we are trying to find the value of x we want to rearrange the equation
so that x is alone on one side. To do this, we must first eliminate
the negative four-thirds by multiplying both sides of the equation
by the reciprocal of negative four-thirds which is negative three-fourths. Since anything multiplied by its reciprocal
is one we can replace negative three-fourths
times negative four-thirds with one. Now, since everything in the parentheses
is multiplied by one we can eliminate the one and the parentheses. Completing the arithmetic on the left side negative three-fourths times negative four is twelve-fourths or three and adding six to both sides allows us to cancel the negative six
on the right leaving us with the equation
“three plus six equals x”. Adding three and six we have “nine equals x” or equivalently, “x equals nine”. So when y is zero, x is nine. Since x is the elapsed time and y is the water level this result tells us that
when the water level y, is zero the elapsed time x, will be nine hours. Since the drain was opened at noon
the pool will not be empty until nine o’clock. Excuse me Professor Von Schmohawk. Did I just hear you say that my pool
will not be empty until nine o’clock? Uh, yes Professor Ethylene. According to our calculations, that is correct. But I distinctly remember telling Mr. Moosemasher that my pool must absolutely be empty
by the time my guests arrive at eight. Uh, excuse me.
If I may interject here for a moment I believe that it is possible to calculate
the exact depth of water in your pool at the time your guests will arrive. If we start with the equation for water depth
versus elapsed time and set the elapsed time, x, to eight hours we can then solve this equation for y
which represents the water depth and thereby determine the depth
at exactly eight o’clock post meridiem. As we all know, eight minus six is two and negative four-thirds times two
is negative eight-thirds. Then adding four to both sides the fours on the left side cancel. We can then write four as twelve-thirds and negative eight-thirds plus twelve-thirds
is four-thirds. So according to my calculations the water depth at eight o’clock p.m.
will be exactly four-thirds feet … … in other words, one and one-third feet. Now, since there are twelve inches in a foot one third of a foot is
twelve inches divided by three or four inches. So, one and one-third feet is one foot, four inches. We can therefore deduce
that when your guests arrive your pool will contain exactly
one foot, four inches of water! Gulp! Uh, Professor Ethylene? How much longer do I have to entertain
your guests? Why you’re doing a splendid job A.V. I’ll let you know when it’s nine o’clock. Jeesh! Thank you A.V. It’s exactly nine o’clock and just as calculated the pool is empty! We have seen how linear equations
can be applied to real-world applications. In the next lecture we will see
how real-world quantities can be calculated using formulas, also referred to as
“literal equations” which relate two or more quantities.

41 thoughts on “Algebra 25 – Linear Equations in the Real World

  1. Great lecture! Keep them coming… PS, it would be great if organisations like Khan Academy could help fund your projects… I really think yours are well explained. You should request funding perhaps?

  2. We have seen how mixed numbers spoil algebra… 4/3 = 1⅓ and I read 4/3 = 1*(1/3) = 1/3 Arggg!!! [Couldn't you write "1 + 1/3"? we should only have a single not written operator and world wide* that operator is multiplication]

    And then you say "1 foot 4 inches" that would be 48 inches^2 just because you neglect to write "1 foot + 4 inches" (I would have accepted "1 foot and 4 inches" or "1 foot with 4 inches").

    Yes, I'm being picky at notation.

    *: Americans of the Unites States you had to be.

  3. I personally like the rather lengthy set up to the problem.  It makes it more "real-world."  In reality problems aren't discovered and solved quickly. The tutu and skateboard are excellent touches. Please start working on the next video while I work my 360s.

  4. Are you going to make some science or history videos? I would like to see some history of language videos. The subject of linguistic history is very intriguing to me.

  5. My gosh, thank you! Linear equations are a bit confusing, but you all explained it perfectly.

    Also..Geekman's glass defies continuity xD
    2:27 – 2:36

  6. I like the video very much, its good way to explain things. But I am sorry to tell you, that when draining a tank it is not a linear equation, it is exponential. In the beginning the water pressure is very high at the plug. When only one feet left the pressure is lower. So it will take even more time to drain the pool – poor guy.

    Here is how to calculate it:
    Im not sure when I learned it but it might have been when I studying engineering.

  7. I though that professor Ethylene wanted to clean the pool but apparently wanted to actually use it as a skate park!

  8. I've never seen a below ground swimming pool with a drain plug. Therefore you would need to calculate volume of water over time to pump it out.

  9. 20yrs back math teacher taught this ..but iy made sense today…suscribed.
    .i would have been in AI long back if it was so intersting…modern education system in india country sucks

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