Download 7.1+System+of+Linear+Equations

Two Equations Containing Two Variables
a11 x  a12 y  b1
a21 x  a22 y  b2
The first two cases are called consistent since there are
solutions. The last case is called inconsistent.
The solution will be one of three cases:
1. Exactly one solution, an ordered pair (x, y)
2. A dependent system with infinitely many solutions
3. No solution
a11 x  a12 y  b1
a21 x  a22 y  b2
independent
With two equations and two variables,
the graphs are lines and the solution (if
there is one) is where the lines
intersect. Let’s look at different
possibilities.
dependent
consistent
Case 1: The lines
intersect at a point
(x, y),
the solution.
inconsistent
consistent
Case 2: The lines
coincide and there
are infinitely many
solutions (all points
on the line).
Case 3: The lines
are parallel so there
is no solution.
If we have two equations and variables and we want to
solve them, graphing would not be very accurate so we will
solve algebraically for exact solutions. We'll look at two
methods. The first is solving by substitution.
The idea is to solve for one of the variables in one of the equations
and substitute it in for that variable in the other equation.
2 x  3 y  1
Let's solve for y in the second equation. You
can pick either variable and either equation, but
go for the easiest one (getting the y alone in the
second equation will not involve fractions).
4x  y  3
4 x  3 5 Substitute this for y in the first equation.
4
y

y4
3  
 
7
7 here since we already solved for y.
Easiest
2 x  3  4 x  3  1
14 x  8
4
x
7
Now we only have the x variable and we
solve for it.
Substitute this for x in one of the equations to find y.
2 x  3 y  1
4 5

So our solution is  ,  
7 7
4x  y  3
This means that the two lines intersect at
this point. Let's look at the graph.
y
We can check this by
subbing in these values
for x and y. They should
make each equation true.








4 5
 , 
7 7









x

4  5
2    3     1
7  7
4  5
4       3
7  7
8 15
  1
7 7
16 5
 3
7 7
Yes! Both equations are
satisfied.
Now let's look at the second method, called the method of
elimination.
The idea is to multiply one or both equations by a constant (or
constants) so that when you add the two equations together, one of
the variables is eliminated.
2 x  3 y  1
3
4x  y  3 3
12 x  3 y  9
2 x  3 y  1
14 x  8
4
x
7
Let's multiply the bottom equation by 3. This
way we can eliminate y's. (we could instead
have multiplied the top equation by -2 and
eliminated the x's)
Add first equation to this.
The y's are eliminated.
4
4   y  3
7
5
y
7
Substitute this for x in one of the equations to find y.
So we arrived at the same answer with either method, but
which method should you use?
It depends on the problem. If substitution would involve
messy fractions, it is generally easier to use the
elimination method. However, if one variable is already or
easily solved for, substitution is generally quicker.
With either method, we may end up with a surprise. Let's see
what this means.
3 x  6 y  15
3 x  2 y  5 3
3 x  6 y  15
3 x  6 y  15
00
Let's multiply the second equation by 3 and
add to the first equation to eliminate the x's.
Everything ended up eliminated. This tells us
the equations are dependent. This is Case 2
where the lines coincide and all points on the
line are solutions.
3 x  6 y  15
x  2 y  5
x  2y 5
Let's solve the second equation for x. (Solving for x
in either equation will give the same result)
Now to get a solution, you chose any real
number for y and x depends on that choice.
y
If y is 0, x is -5 so the
point (-5, 0) is a solution
to both equations.








If y is 2, x is -1 so the
point (-1, 2) is a solution
to both equations.

Any point on
this line is a
solution
        










x






So we list the solution as:
x  2 y  5 where x is any real number
Let's try another one:2 2 x  5 y
1 2
4 x  10 y  5
4 x  10 y  2
Let's multiply the
first equation by 2
and add to the
second to eliminate
the x's.
07
This time the y's were eliminated
too but the constants were not.
We get a false statement. This
tells us the system of equations
is inconsistent and there is not
an x and y that make both
equations true. This is Case 3,
no solution.
y

















x

There are
no points of
intersection
Three Equations Containing Three Variables
a11x  a12 y  a13 z  b1
a21x  a22 y  a23 z  b2
a31x  a32 y  a33 z  b3
As before, the first two cases are called consistent since
there are solutions. The last case is called inconsistent.
The solution will be one of three cases:
1. Exactly one solution, an ordered triple (x, y, z)
2. A dependent system with infinitely many solutions
3. No solution
With two equations and two variables, the graphs were
lines and the solution (if there was one) was where the
lines intersected. Graphs of three variable equations are
planes. Let’s look at different possibilities. Remember the
solution would be where all three planes all intersect.
Planes intersect at a point: consistent with
one solution
Planes intersect in a line: consistent
system called dependent with an infinite
number of solutions
Three parallel planes: no intersection so
system called inconsistent with no solution
No common intersection of all three
planes: inconsistent with no solution
We will be doing elimination to solve these systems. It’s
like elimination that you learned with two equations and
variables but it’s now the problem is bigger.
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
coefficient is a 1 here so it will be easy to work with
Your first strategy would be to choose one equation to
keep that has all 3 variables, but then use that equation to
“eliminate” a variable out of the other two. I’m going to
choose the last equation to “keep” because it has just x.
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
x  2 y  3z  7
Now use the third equation multiplied through by whatever
it takes to eliminate the x term from the first equation and
add these two equations together.
In this case when added to eliminate the x’s you’d need a -2.
 2x  2 y  3z  7
 2 x  4 y  6 z  14
2x  y  z  4
put this equation up with
the other one we kept
5 y  5 z  10
2x  y  z  4
x  2 y  3z  7
 3 x  2 y  2 z  10
5 y  5 z  10
x  2 y  3z  7
Now use the third equation multiplied through by whatever
it takes to eliminate the x term from the second equation
and add these two equations together.
In this case when added to eliminate the x’s you’d need a 3.
3x  2 y  3z  7
3x  6 y  9 z  21
 3x  2 y  2 z  10
We won’t “keep” this equation, but
we’ll use it together with the one we
“kept” with y and z in it to eliminate
the y’s.
 4 y  7 z  11
2x  y  z  4
 3 x  2 y  2 z  10
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
z 1
x  2 y  3z  7
So we’ll now eliminate y’s from the 2 equations in y and z
that we’ve obtained by multiplying the first by 4 and the
second by 5
5 y  5 z  10
 4 y  7 z  11
20 y  20 z  40
 20 y  35 z  55
We can add this to the one’s we’ve
kept up in the corner
15z  15
z 1
2x  y  z  4
x  2 y  3z  7
 3 x  2 y  2 z  10
x  2 y  3z  7
5 here
y  5for
z
10
keep over
later
use
z 1
Now we are ready to take the equations in the corner and
“back substitute” using the equation at the bottom and
substituting it into the equation above to find y.
5 y  51  10
5 y  5
y  1
Now we know both y and z we can sub
them in the first equation and find x
x  21  31  7
x5  7
x2
These planes intersect at a
point, namely the point (2, -1, 1).
The equations then have this
unique solution. This is the
ONLY x, y and z that make all 3
equations true.
 x  2y  z  5
y  z  10
Let’s do another one:
2x  3y  z  0
 x  2y  z  5
3x  4 y  z  1
I’m
going
to “keep”
this onewe
since
If we
multiply
the equation
keptit
will be easy to use to eliminate x’s
by 3
2 and add it to the last
first equation
equation
from
others.
we can
eliminate x’s.
 2 x  4 y  2 z  10
2x  3 y  z  0
we’ll keep y  z  10
this one
 3 x  6 y  3 z  15
3x  4 y  z  1
2 y  2 z  16
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
 x  2y  z  5
y  z  10
2x  3y  z  0
 x  2y  z  5
3x  4 y  z  1
y  z  10
This means the equations are
inconsistent and have no solution.
The planes don’t have a common
intersection and there is not any
(x, y, z) that make all 3 equations true.
2 y  2 z  16
we’ll multiply the first
equation by -2 and add these
together
 2 y  2 z  20
2 y  2 z  16
0  4
Oops---we eliminated the y’s alright but the z’s ended
up being eliminated too and we got a false equation.
Let’s do another one:
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
x  y  2z  1
 3 y  3z  0
let'sIf “keep”
this one
since it will
be
we
multiply
the
equation
we
kept
If
we
multiply
the
equation
we
kept
easy
to
use
to
eliminate
x’s
from
by
-2
and
add
it
to
the
second
by -4 and add it to the last equation
others.
equation
we can x’s.
eliminate x’s.
we
can eliminate
 2 x  2 y  4 z  2
 4 x  4 y  8 z  4
2x  y  z  2
we’ll keep  3 y  3 z  0
this one
4 x  y  5z  4
 3 y  3z  0
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
x  y  2z  1
 3 y  3z  0
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
 3 y  3z  0
 3 y  3z  0
This means the equations are
consistent and have infinitely many
solutions. The planes intersect in a
line. To find points on the line we can
solve the 2 equations we saved for x
and y in terms of z.
multiply the first equation by -1
and add the equations to
eliminate y.
3 y  3z  0
 3 y  3z  0
00
Oops---we eliminated the y’s alright but the z’s ended up
being eliminated too but this time we got a true equation.
x  y  2z  1
2x  y  z  2
4 x  y  5z  4
x  y  2z  1
 3 y  3z  0
zz
First we just put z = z since it can be any real number.
Now solve for y in terms of z.
Now sub it -z for y in first equation and solve for x in
terms of z.
The solution is (1 - z, - z, z) where z is any real number.
For example: Let z be 1. Then (0, -1, 1) would be a solution.
Notice is works in all 3 equations.
But so would the point
you get when z = 2 or 3
any other real
 3 y  3z  0
x  z  2z  1 or
number so there are
infinitely many
x 1  z
y  z
solutions.