Hello. I’m Professor Von Schmohawk

and welcome to Why U. So far we have discussed methods of solving

systems of two linear equations in two variables including the substitution method

and the elimination method. We also showed how a system of linear equations

could be used to find the time and place where two runners starting at

different times in a race will meet. But what other types of problems can be solved

using systems of linear equations? Let’s start by looking at a few problems

from earlier lectures which we solved without creating

systems of equations. One problem, from the lecture

“Solving Problems with Linear Equations” involved finding the dimensions

of three identical dog pens. In this problem we have exactly

180 feet of fence all of which must be used to

enclose and separate the pens. And we are told that each pen’s length

must be twice its width, plus 3 feet. We solved this problem

by creating a single variable, w which we used to represent

the width of a pen. We then created an expression

for the total length of fence by adding the 4 fence segments of length “2w + 3” plus the 6 shorter segments of length “w” which we knew must total 180 feet. Since this equation

has only a single variable, w it can be solved to find the value of w. So although this problem has two unknowns,

the length and width of a pen we were able to express both unknowns

in terms of a single variable, w. We can then combine these expressions

to create a single equation. But instead of creating

a single equation with one variable this problem can also be solved by creating

a system of two equations in two variables. To do this, we assign two variables,

“L” and “w” to represent the length and width of a pen. We can then create two equations each of which states one relationship

between the variables. The first equation states that

the 4 length segments plus the 6 width segments equals 180 feet. The second equation states that the length

of a pen is twice its width plus 3 feet. This gives us a system of

two equations in two variables. Solving this system

using the elimination method we first rewrite the second equation

so that both equations are in standard form. We can now eliminate either variable,

L or w by choosing an appropriate multiplier

for the second equation before adding the equations. If we choose to eliminate the variable w,

we multiply the second equation by 3. Now, when we add the equations the “w” terms disappear giving us the equation “7L=189” or dividing both sides by 7 “L=27”. We can then substitute this value

into either of the original equations to find the value for w. Since the bottom equation

is the simpler of the two we will use it to calculate the value of w. Subtracting 27 from both sides we eliminate that term on the left. And completing the arithmetic we get “negative 2w=negative 24” or “w=12”. So this problem containing two unknowns can be solved by creating

a single equation with a single variable expressing both unknowns

in terms of that one variable. Or we can solve the problem by creating

a system of two equations in two variables where each unknown is represented

by a different variable. Now lets look at another example from the lecture “Solving Motion Problems

with Linear Equations”. In this problem, a car and a bus

are each travelling at constant speeds in opposite directions on a highway. Three hours after they pass each other,

they are 360 miles apart. If the car is travelling

ten miles per hour faster than the bus how fast is each vehicle travelling? We solved this problem by creating

a single equation with a single variable, s which represented the speed of the bus. We then represented the speed of the car

as “s + 10” since the car’s speed is

ten miles per hour faster than the bus. We know that both vehicles

travelled for 3 hours. So knowing that

speed times time equals distance we wrote expressions for

the distance travelled by each vehicle. In 3 hours the bus travelled “3s” miles and the car travelled

3 times “s + 10” miles. Since the sum of these distances must equal

the distance between the two vehicles or 360 miles we could then create a single equation

which can be solved to find the value of s. Solving this equation

we get “s=55” which tells us that the bus’s speed is 55 miles per hour and the car’s speed is “55 + 10” or 65 miles per hour. So just as in the previous example we were able to express both unknowns

in terms of a single variable and write expressions for the distances travelled

by each vehicle using only that one variable. These expressions were then combined

to create a single equation. But once again, this problem can also be solved by creating a system of

two equations in two variables. To do this, we assign the variables

“s” and “r” to represent the speeds

of the bus and the car. Since distance is equal to speed times time the distances travelled for the bus and car

are “3s” miles and “3r” miles. We can then create two equations each of which states one relationship

between the variables. The first equation states

that the distance travelled by the car plus the distance travelled by the bus is equal to 360 miles. The second equation states

that the speed of the car is 10 miles per hour faster

than the speed of the bus. This gives us a system of

two equations in two variables. Solving this system

using the elimination method we first rewrite the second equation

so that both equations are in standard form. We can now eliminate either variable,

“s” or “r” by choosing an appropriate multiplier for

the second equation before adding the equations. If we choose to eliminate s,

we multiply the second equation by 3. Now when we add the equations the “s” terms disappear giving us the equation “6r=390” or dividing both sides by 6,

“r=65”. We can then substitute this value

into either of the original equations to find the value for s. Since the bottom equation

is the simpler of the two we will use it to calculate the value of s. Subtracting 65 from both sides eliminates this term on the left. And completing the arithmetic we get “negative s=negative 55” or “s=55”. So this problem containing two unknowns can be solved by creating

a single equation with a single variable expressing both unknowns

in terms of that one variable. Or we can solve this problem by creating

a system of two equations in two variables where each unknown

is represented by a different variable. The method which is easiest

depends upon the problem. The two examples which we have seen so far could be solved using either method

with similar difficulty. However, in some cases

creating a system of equations is clearly the easier method. For example, lets solve a problem

involving different combinations of items which cost different prices. John, Denny, Michelle, and Cass go to a movie. They spend 69 dollars

on 4 movie tickets and 3 drinks. But Roger, David, Chris, Gene, and Michael

are also going to see the same movie and they spend 88 dollars

on 5 movie tickets and 4 drinks. So given the information that

4 tickets plus 3 drinks cost 69 dollars and 5 tickets plus 4 drinks cost 88 dollars is it possible to calculate the pricing

at the theater for tickets and drinks? Solving this problem by creating

a system of equations is straightforward. If we let “t” represent the price of a ticket and “d” represent the price of a drink we can easily create a system of two equations

in two variables, “t” and “d”. We can then use the elimination method

to eliminate a variable. For instance,

if we wish to eliminate the “d” terms we multiply the top equation by 4

and the bottom equation by negative 3. Completing the arithmetic we then add the equations which eliminates the variable, d leaving us with the equation “t=12”. We can then substitute this value

into either of the original equations to find the value of d. Since the top equation

has the smallest coefficients we will use it to calculate the value of d. Subtracting 48 from both sides eliminates this term on the left. And completing the arithmetic we get “3d=21” or “d=7”. Since d represents the price of a drink and t represents the price of a ticket we know that drinks cost 7 dollars each and tickets cost 12 dollars each. So how could we have solved this same problem

using a single equation with only one variable? In order to do this,

we would have to write one unknown either the price of tickets or drinks in terms of the other, so that one of

the variables can be eliminated. So let’s say we rewrite the first equation so that the price of a ticket is written

in terms of the price of a drink. This expression for the price of a ticket could then be substituted

into the second equation to eliminate the price of a ticket

from that equation. This gives us a single equation with one unknown,

the price of a drink which can be represented

with a single variable, d. But all we have done is to eliminate one of

the variables through substitution. This was basically no different than solving

the system of two equations in two variables using the substitution method. But what if instead of two groups of people

who buy tickets and drinks there are three groups of people

who buy tickets, drinks, and popcorn? This problem now has three unknowns the price of a ticket the price of a drink and the price of popcorn which we could represent by three variables “t” “d” and “p”. This problem would be very difficult to set up

using a single equation with one variable. However, the solution can be calculated

in a straightforward manner by using a system of three equations

in three variables. In the next several lectures, we will introduce

methods of solving systems such as these and see how systems of linear equations

in three variables can be visualized in 3-dimensional space.

Do you recognize the two groups of movie-goers at 11:01?

One of the most eloquent explanations of this topic that I have ever heard. Thank you, again.

Can you make Science videos?

What makes define width , length and height ?

which one is which ? i am so confused. because all these concept is depends on the point where are look at it. can you help me ?