CONCEPT ABOUT OFSystems of Linear Equations

1. 11 12 1
21 22 2
a x a y b
a x a y b
+ =
+ =
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
Two Equations Containing Two Variables
The first two cases are called consistent since there are
solutions. The last case is called inconsistent.

2. 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.
11 12 1
21 22 2
a x a y b
a x a y b
+ =
+ =
Case 1: The lines
intersect at a point
(x, y),
the solution.
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.
dependent
consistent consistent
independent
inconsistent

3. 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.
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).
Substitute this for y in the first equation.4 3y x= −
2 3 1
4 3
x y
x y
+ = −
− =
( )2 3 4 3 1x x+ − = − Now we only have the x variable and we
solve for it.
14 8x =
4
7
x = Substitute this for x in one of the equations to find y.
Easiest here since we already solved for y.
4 5
4 3
7 7
y
 
= − = − ÷
 

4. 2 3 1
4 3
x y
x y
+ = −
− =
So our solution is
4 5
,
7 7
 
− ÷
 
This means that the two lines intersect at
this point. Let's look at the graph.
−4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
x
y
4 5
,
7 7
 
− ÷
 
We can check this by
subbing in these values
for x and y. They should
make each equation true.
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.

5. 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.
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.12 3 9x y− =
The y's are eliminated.
Substitute this for x in one of the equations to find y.
3 3
2 3 1x y+ = −
14 8x =
4
4 3
7
y
 
− = ÷
 
2 3 1
4 3
x y
x y
+ = −
− =
4
7
x =
5
7
y = −

6. 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 6 15
2 5
x y
x y
− = −
− + =
Let's multiply the second equation by 3 and
add to the first equation to eliminate the x's.3 3
3 6 15
3 6 15
x y
x y
− = −
− + =
0 0=
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.

7. Now to get a solution, you chose any real
number for y and x depends on that choice.
2 5x y= −
If y is 0, x is -5 so the
point (-5, 0) is a solution
to both equations.
3 6 15
2 5
x y
x y
− = −
− + =
If y is 2, x is -1 so the
point (-1, 2) is a solution
to both equations.
So we list the solution as:
2 5 where is any real numberx y x= −
Let's solve the second equation for x. (Solving for x
in either equation will give the same result)
−9 −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 9 10
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
7
8
9
x
y
Any point on
this line is a
solution

8. Let's try another one: 2 5 1
4 10 5
x y
x y
− + =
− =
Let's multiply the
first equation by 2
and add to the
second to eliminate
the x's.
2 2
4 10 2x y− + =
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.
−4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
x
y
There are
no points of
intersection

9. 3333231
2232221
1131211
bzayaxa
bzayaxa
bzayaxa
=++
=++
=++
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
Three Equations Containing Three Variables
As before, the first two cases are called consistent since
there are solutions. The last case is called inconsistent.

10. Planes intersect at a point: consistent with
one 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.

11. Planes intersect in a line: consistent
system called dependent with an infinite
number of solutions

12. Three parallel planes: no intersection so
system called inconsistent with no solution

13. No common intersection of all three
planes: inconsistent with no solution

14. 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.
732
10223
42
=+−
−=−+−
=++
zyx
zyx
zyx
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.
coefficient is a 1 here so it will be easy to work with

15. 732
10223
42
=+−
−=−+−
=++
zyx
zyx
zyx 732 =+− zyx
keep over here for later use
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.
( )7322 =+−− zyx
1055 −=− zy
1055 −=− zy
put this equation up with
the other one we kept
14642 −=−+− zyx
42 =++ zyx

16. 732
10223
42
=+−
−=−+−
=++
zyx
zyx
zyx 732 =+− zyx
( )7323 =+− zyx 21963 =+− zyx
10223 −=−+− zyx
1174 =+− zy
1055 −=− zy
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.
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.

17. 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
732
10223
42
=+−
−=−+−
=++
zyx
zyx
zyx 732 =+− zyx
keep over here for later use
1174 =+− zy
1055 −=− zy
We can add this to the one’s we’ve
kept up in the corner
1055 −=− zy 402020 −=− zy
553520 =+− zy
1515 =z
1=z
1=z

18. 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.
732
10223
42
=+−
−=−+−
=++
zyx
zyx
zyx 732 =+− zyx
keep over here for later use1055 −=− zy
Now we know both y and z we can sub
them in the first equation and find x
( ) 10155 −=−y
1=z
55 −=y 1−=y
( ) ( ) 71312 =+−−x
75 =+x 2=x
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.

19. I’m going to “keep” this one since it
will be easy to use to eliminate x’s
from others.
Let’s do another one:
143
52
032
=−−
=++−
=−−
zyx
zyx
zyx
52 =++− zyx
032
10242
=−−
=++−
zyx
zyx
10=+ zy
10=+ zy
we’ll keep
this one
143
15363
=−−
=++−
zyx
zyx
1622 =+ zy
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
If we multiply the equation we kept
by 2 and add it to the first equation
we can eliminate x’s.
If we multiply the equation we kept
by 3 and add it to the last equation
we can eliminate x’s.

20. 143
52
032
=−−
=++−
=−−
zyx
zyx
zyx 52 =++− zyx
10=+ zy
10=+ zy
we’ll multiply the first
equation by -2 and add these
together
1622 =+ zy
Oops---we eliminated the y’s alright but the z’s ended
up being eliminated too and we got a false equation.
2022 −=−− zy
1622 =+ zy
40 −=
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.

21. Let’s do another one:
454
22
12
=++
=+−
=++
zyx
zyx
zyx
let's “keep” this one since it will be
easy to use to eliminate x’s from
others.
12 =++ zyx
22
2422
=+−
−=−−−
zyx
zyx
033 =−− zywe’ll keep
this one
454
4844
=++
−=−−−
zyx
zyx
If we multiply the equation we kept
by -2 and add it to the second
equation we can eliminate x’s.
033 =−− zy
Now we’ll use the 2 equations we have with y and z to
eliminate the y’s.
033 =−− zy
If we multiply the equation we kept
by -4 and add it to the last equation
we can eliminate x’s.

22. 454
22
12
=++
=+−
=++
zyx
zyx
zyx 12 =++ zyx
033 =−− zy
multiply the first equation by -1
and add the equations to
eliminate y.
033 =−− zy
033 =−− zy
033 =+ zy
033 =−− zy
00 =
Oops---we eliminated the y’s alright but the z’s ended up
being eliminated too but this time we got a true equation.
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.

23. 454
22
12
=++
=+−
=++
zyx
zyx
zyx 12 =++ zyx
033 =−− zy
zz =
First we just put z = z since it can be any real number.
Now solve for y in terms of z.
033 =−− zy
zy −=
Now sub it -z for y in first equation and solve for x in
terms of z.
12 =+− zzx
zx −=1
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
or any other real
number so there are
infinitely many
solutions.