# Algebra 41 – Using Systems of Equations Versus One Equation 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.

### 4 thoughts on “Algebra 41 – Using Systems of Equations Versus One Equation”

• January 30, 2015 at 8:16 pm

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

• February 12, 2015 at 9:52 am

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

• February 22, 2015 at 10:13 am
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