Download Section 4-1: System of linear equations in two variables Solving a

Section 4-1: System of linear equations in
two variables
Solving a system of linear equations
graphically
In chapter 1 we learned how to solve a linear
equation in one variable, now we have two
equations in two variables, such as:
2x  y  1
 x  2y  7
To solve this is to find values for x, y such
that they satisfy both equations. We know
that each equation represents a line, so the
solution is exactly the intersection, which is
(3, 5) from the graph below, i.e., x = 3, y =
5.
We see three cases when solving a linear
system graphically:
Solving a system of linear equations by
substitution
Substitution means that you solve for one
variable from one equation and plug into the
other equation. For example: Given
2x  y  1
 x  2y  7
Solving a system of linear equations using
elimination by addition
Addition means that you multiply each
equation (or only one equation) with a
number then add both equations to get rid of
a variable.
3x  2 y  8
Example 1. Solve by addition:
2 x  5 y  1
Solution:
x  2y  8
Example 2. Solve
2 x  4 y  4
Solution:
Note: if you graph these two equations, you
will see two __________lines!
Can you tell directly from the system?
2x  6 y  4
Example 3. Solve
3x  9 y  6
Solution:
Note: If you graph you see two lines
coincide. In next section we’ll write infinite
number of solutions in the form of
parameters.
How do you tell immediately?
Application Problems:
Example 4 (supply and demand)
Suppose that the supply and demand for
printed baseball caps for a particular
p  0.4q  3.2
week are
, where p is the
p  1.9q  17
price in dollars and q is the quantity in
hundreds.
a. Find the supply and demand (to the
nearest unit) if baseball caps are $4
each. Discuss the stability of the
baseball cap market at this price level.
b. Find the supply and demand (to the
nearest unit) if baseball caps are $9
each. Discuss the stability of the
baseball cap market at this price level.
c. Find the equilibrium price and
quantity.
d. Graph the two equations in the same
coordinate system and identify the
equilibrium point, supply curve, and
demand curve.
Solution: a. Plug 4 into p in both
equations: 44  0.14.q9q317.2 we get supply q is 2
and demand q is 6.84, since they are in
hundreds, so we have 200 and 684
correspondingly. Since supply quantity is
much less than demand quantity, the
price is going up.
b. Similar to part a we get 1450 for
supply and 421 for demand. Since supply
is much more than demand, the price is
going down.
c. Solve the linear system to get
equilibrium: q = 6 (i.e. 600) and p =
$6.50
d.
Section 4-2: Using augmented matrices to
solve a linear system
2  4 0 
A matrix is the form A  
,

6 1  5
which is called matrix of size 2 3 (2 rows
and 3 columns), where the entries
a21  6, a13  0 .
A square matrix is a matrix with same
number of rows and columns, such as
1  3
6 0  ;


a column matrix is a matrix with only one
0
column like  3  ;
 
 2
a row matrix is a matrix with only one row,
as 9 4  3.
In section 4-1 we used addition to solve a
linear system. It works well when we have
two or three variables, but when we have
more than 3 variables, it’s not a very
efficient way. In this section we will use
matrices to do it, and this method works
well for any size of linear system.
x  2y  3
Consider the linear system
.
2 x  3 y  1
We’ll explain how each step in solving by
addition corresponds to each step in solving
by augmented matrix method.
3 row operations: (‘ ’ means ‘replace’)
1. Swap two rows: Ri  R j
2. Multiply a row by a nonzero constant:
kRi  Ri ( k  0 )
3. Multiply a row with a constant and add
it to another row: cRi  R j  R j
Ex1. (a linear system with exactly one
solution)
2 x1  3 x2  6
Solve
1 by augmented matrix.
3 x1  4 x2 
2