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:**

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 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:**

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:**

*** y's have opposite
coefficients**

**y*'s dropped out

***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:**

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.**

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:**

*** 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.

No ordered triples to check.

**Final answer is no solution.**

Last revised on July 10, 2011 by Kim Seward.

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