Transcript Week 7

Systems of two
equations (and more)
Solving systems of several
equations
Supply and demand
Systems of two equations

Today, we use algebra to solve several
equations with as many unknown variables


Although in theory this can be used to solve an
arbitrarily large system, we’ll limit ourselves to
2-3 equations/unknowns
This is basically just an extension on what
we saw last week:

The aim is to modify the system of equations
into a series of single – variable equations that
we know we can solve
Systems of two equations
Notation: Equations with several
unknowns
Solving a simple system of equations
A practical example: supply and
demand
Equations with several unknowns

Last week we saw the notation used for
unknowns inside and equation:
2 x  6

We also introduced the idea that several
components of the equation could be unknown,
including parameters
2 x  a

With such an equation you can’t find a solution
for “x”: you need more information
Equations with several unknowns

This extra information is provided by a second
equation, which helps to specify “a”
a  23

Replacing in the first equation, one can now
solve for “x”
2 x  a
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As a result, you have the value of both “x” and
“a”
Equations with several unknowns

There are a few elements of notation to
consider:
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
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There is no distinction between unknown
variables, parameters, etc: all are “unknowns”
Unknowns all have the same notation, typically
“x,y,z” in mathematics (not necessarily so in
economics)
The system of equations is indicated by an
“accolade”
2  x  y

 y  23
Systems of two equations
Notation: Equations with several
unknowns
Solving a simple system of equations
A practical example: supply and
demand
Solving a simple system of equations

The system considered in the previous section
is rather simple:
2  x  y

 y  23


In particular the 2nd equation is trivial!!
What about a more complicated system?
 x  10  4 y  5

2 x  3  11y
Solving a simple system of equations
 x  10  4 y  5

2 x  3  11y

This system can be solved by isolating an
unknown in one equation, then substituting it
in the other equation



You then have a single equation with a single
unknown
This method (the substitution method) is the
simplest, and it works best for small systems (2-3
equations)
For larger system, other (faster) methods are used
Solving a simple system of equations
 x  10  4 y  5

2 x  3  11y

Step 1: isolate one of the variables.

Lets isolate “x” in the 1st equation
 x  4 y  15

2 x  3  11y

Step 2: replace in the other equation
 x  4 y  15

2  4 y  15   3  11 y
Solving a simple system of equations
 x  4 y  15

2  4 y  15   3  11 y

We now have a single equation (the 2nd) with a
single unknown (y)

Lets rearrange and solve the 2nd equation for y:
 x  4 y  15

8 y  30  3  11y
 x  4 y  15

33  3 y
Solving a simple system of equations
 x  4 y  15

 y  11

Step 3 : replace in the 1st equation

This gives us again a single equation with unknown x
 x  4 11  15

 y  11
 x  59

 y  11
Systems of two equations
Notation: Equations with several
unknowns
Solving a simple system of equations
A practical example: supply and
demand
Supply and Demand
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
Supply and demand on a market provide a good
example of how systems of equations can be
used in economics
On a market (say the market for computers)
economists want to know 2 variables:


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The quantity of computers available (Q)
The price of a computer (P)
Supply and demand provide the 2 equations
required to solve the system
Supply and Demand
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Supply : There is a positive relation between the
quantity supplied and the price:

The higher the price, the more computer
manufacturers will want to sell
Q s  400  P

Demand: There is a negative relation between
the quantity demanded and the price

The higher the price, the fewer computers people
will be willing to buy:
Qd  1600  2 P
Supply and Demand
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The system is completed by a 3rd trivial
equation: the market equilibrium equation
Qs  Qd

The full system is:
Q s  400  P
 d
Q  1600  2 P
Q s  Q d
