Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 20: Solving Systems of Linear Equations
in Three Variables
Answer/Discussion
to 1a

Note that the numbers in ( ) are equation
numbers. They will be used throughout the problems for reference
purposes. 
No simplification needed here. Let's go on to the next step. 
Note that each equation is missing a variable. For example, equation
(1) is missing z. We can use this to our advantage. We can
use the equation in it's original form as an equation with a variable eliminated.
I choose to eliminate z.
Since z is already eliminated from the
first equation we will use that first equation in its original form for
this step: 

*z is already
eliminated from eq. (1) 
Now we can't just stop here for of two reasons. First, we would
be stuck because we have one equation and two unknowns. Second, when
we solve a system it has to be a solution of ALL equations involved and
we have not incorporated the third equation yet. Let's do that now. 
We are still going after eliminating z,
this time I want to use the second and the third equations.
Since the z terms of the second and third equations are already opposites,
we can go right into adding them. 

*z's have opposite
coefficients
*z's dropped out 
Putting the two equations that we have found together we now have a
system of two equations and two unknowns, which we can solve just like
the ones shown in tutorial 19 (Solving a System of Linear Equations in
Two variables) . You can use either elimination or substitution.
I'm going to go ahead and stick with the elimination method to complete
this.
Let's first put those equations together: 

*Put equations found in steps 2 and 3
together into one system 
I'm going to choose y to eliminate.
I can either multiply the first equation by 1 or the second, either way
will create opposites in front of the y terms.
Multiplying equation (4) by 1 and then adding the equations together
we get: 

*Mult. eq. (4) by 1
*y's have opposite
coefficients
*y's dropped out 

*Inverse of mult. by 2 is div. by 2 
If we go back one step to the system that had two equations and
two variables and plug in ½ for x in equation (1) we would
get: 

*Eq. (1)
*Plug in 1/2 for x
*Inverse of add 1 is sub. 1

Now we need to go back to the original system and pick any equation
to plug in the two known variables and solve for our last variable .
I choose equation (2) to plug in 1 for y that we found: 

*Eq. (2)
*Plug in 1 for y
*Inverse of add 1 is sub. 1 
You will find that if you plug the ordered triple (1/2, 1, 2) into
ALL THREE equations of the original system, that this is a solution to
ALL THREE of them.
(1/2, 1, 2) is a solution to our system.
(return to
problem 1a) 
Answer/Discussion
to 1b

Note that the numbers in ( ) are equation
numbers. They will be used throughout the problems for reference
purposes. 
No simplification needed here. Let's go on to the next step. 
Let's start by picking our first variable to eliminate. I'm going
to pick z to eliminate. We need to do
this with ANY pair of equations.
Let's first eliminate z using the first
and second equations. This process is identical to how we approached
it with the systems found in Tutorial 19: Solving a System of Linear Equations
in Two variables . If I multiply 2 times the first equation, then
the z terms will be opposites of each other
and ultimately drop out.
Multiplying 2 times the first equation and then adding that to the
second equation we get: 

*Mult. eq (1) by 2
*z's have opposite
coefficients
*All three variables drop out AND
we have a FALSE statement 
Wow, that was quick. All of our variables dropped out and we
ended up with a FALSE statement. You know what that means?
No solution. 
Since we have no solution, there is no value to be found here. 
Since we have no solution, there is no value to be found here. 
Since we have no solution, there is no value to be found here. 
Last revised on July 10, 2011 by Kim Seward.
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