This document discusses solving systems of linear equations with two or three variables. It explains that a system can have one solution, infinitely many solutions, or no solution. For two variables, the solutions are where lines intersect (one solution), coincide (infinitely many), or are parallel (no solution). For three variables, the solutions are where planes intersect (one point), lie on a line (infinitely many), or do not intersect (no solution). It demonstrates solving systems using substitution, elimination, and matrix methods, and discusses cases where a system has infinitely many or no solutions.

1. Consistency of linear
equations in two and
three variables
GROUP 1

2. What is a linear equation
A linear equation is an algebraic equation in which each term is either a constant or the product
of a constant and (the first power of) a single variable.
Linear equations can have one or more variables. Linear equations occur abundantly in most
subareas of mathematics and especially in applied mathematics. While they arise quite naturally
when modeling many phenomena, they are particularly useful since many non-linear equations
may be reduced to linear equations by assuming that quantities of interest vary to only a small
extent from some "background" state. Linear equations do not include exponents.

3. What is a solution to an equation
In mathematics, to solve an equation is to find what values fulfill a condition stated in the form
of an equation. These expressions contain one or more unknowns, which are free variables for
which values are sought that cause the condition to be fulfilled. To be precise, what is sought are
often not necessarily actual values, but, more in general, mathematical expressions. A solution
of the equation is an assignment of expressions to the unknowns that satisfies the equation; in
other words, expressions such that, when they are substituted for the unknowns, the equation
becomes an identity.

4. 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

5. 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
 
    
 

6. 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.
        









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.

7. 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  

8. 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.

9. 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)
                  















x
y
Any point on
this line is a
solution

10. 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.
        









x
y
There are
no points of
intersection

11. 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.

12. 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.

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

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

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

16. Steps to solving a system of equations in
3 Variables
Ensure that the equations are in standard form:
Ax + By + Cz = D
Remove any decimals or fractions from the equations.
Eliminate one of the variables using two of the three equations. Result will be a
new equation with two variables.
Eliminate the same variable using another set of two equations. Result will be a
second equation in two variables.
Solve the new system of two equations.
Using the solution for the two variables, substitute the values into one of the
original equations to solve for the third variable.
Check the solution set in the remaining two original equations.

17. Example system of three equations
Example: solve for x, y and z
First, we need to ensure that all equations are in standard
form, i.e. all variables are on the left side of the equation.
Note that equation #3 is not in standard form. We need to
get the y and z terms to the left side of the equation.

18. Put equations in Standard FormRewriting Equation 3 into standard form:
Add y to both sizes
Add 6z to both sides
Our 3 equations are now:
But, note that equations #1 contains a fraction, and #2 contains
decimals. If we eliminate the fractions and decimals, the
equations will be easier to work with.

19. Remove Fractions and Decimals
We can eliminate the fraction in #1 by multiplying both sides of the equation
by 4.
This is our new version of #1
• We can eliminate the decimals in #2 by multiplying both sides of the
equation by 100.
• So, Let’s divide both sides by 5, so we have smaller numbers to work with.
This becomes our new #2
Note that all terms have a common factor of 5

20. Eliminate One Variable
We have rewritten our three equations as follows:
Now, to solve, we need to first get two equations with two variables.
To do so, we eliminate one of the variables from two of the equations.
Then eliminate the same variable from another set of two equations.
In the first column, we
eliminated y from equations
1 & 2, resulting in equation A.
In the second column, we
eliminated y from equations 2
& 3, resulting in equation B.

21. Solve the system in two variables
To solve our new system of equations A & B, the first step is to eliminate one of the
variables
If we choose to eliminate x by addition, we must get the equations in a form such that
the coefficients of x in the two equations are inverses of one another (for example: +1
and -1 or +5 and -5). To do so we need to multiply each of the equations by some
factor.
Before we determine that factor, we need to determine what the resulting coefficient
of x will be. It will need to be a common multiple of the current coefficients, 8 and 10.
The least common multiple of 8 and 10 is 40, so we will want the coefficients of the x
terms to be 40 and -40.
If we multiply equation A by 5 and equation B by 4, the resulting coefficient of x will be
40. However, we need one of those coefficients to be negative, so we’ll multiply the
second equation by -4.
Equation A Equation B

22. Solve first for z, then for x
Add the new set of equations A and B to eliminate x, and then solve
for z:
Now substitute the value for of z into equation A or B to solve for x;
We are using equation B:
Add the two equations to eliminate x
Divide both sides by 102 and reduce
Substitute 1/6 for z
Simplify
Subtract 2 from both sides
Divide both sides by 10 and reduce

23. Use one of the original equations to
solve for y
Next we need to substitute the values of x and z into one of the
original equations to solve for y. We’ll use equation 2 (after
decimals were removed):
Substitute the values for x and z
Simplify
Subtract 3 from both sizes
Multiply by (-1)
The solution for the system of equations is the ordered triple:

24. Matrix methods…..

25. Equations with infinite solutions

26. In the last row of the above augmented matrix, we have ended up with all zeros on both sides of
the equations. This means that two of the planes formed by the equations in the system of
equations are parallel, and thus the system of equations is said to have an infinite set of
solutions. We solve for any of the set by assigning one variable in the remaining two equations
and then solving for the other two.

27. Equations with no solutions

28. In the last row, we ended up with the equation 0 = 6 which we know can't be true and so we
conclude that the system of equations has no solution.

29. Thanks