Determine whether the system has no solutions
or infinite solutions.
So they're telling us right from the get go that we're not
going to have one nice, clean solution here.
So let's just try to solve this system.
And so to solve the system we want
to eliminate one variable first.
So let's say we want to focus on eliminating,
let's eliminate the x's.
So if we want to eliminate the x's,
if I multiply this first equation over here
by negative 4, then this will become a negative 8x
which should be able to cancel out with this 8x over here.
So let me multiply it by negative 4.
So times negative 4.
So 2 times negative 4 is negative 8x.
Negative 4y times negative 4 is plus 16y.
Negative 4 times z is negative 4z.
And then that is going to be equal to negative 4 times 3
is negative 12.
And then let me just rewrite the second equation
right over here.
We have 8x minus 2y plus 4z is equal to 7.
And now we can add these two equations
and hopefully eliminate the x terms.
Negative 8x plus 8x, they eliminate.
16y minus 2y is 14y.
And then negative 4z plus 4z, those also eliminate.
We were able to eliminate and x and a z in one fell swoop.
And so we get 14y is equal to-- we
have negative 12 plus 7 is negative 5.
We have negative 5, just like that.
And actually, we could solve this.
This is one equation and one unknown.
But we haven't used the third constraint yet.
We haven't used this constraint over here.
So we have-- we could actually solve it.
Just divide both sides by 14 and we get y is equal to negative 5
over 14.
So that's interesting, but it doesn't necessarily
tell us much about the solution just yet.
You want to be very careful about back substituting
and all of that, because we haven't
used all of the constraints yet.
We still haven't used this third constraint.
So our goal was to eliminate x's from the get go.
So let's still try to eliminate x's using these second two
constraints and see if we get anything interesting.
And the reason why I'm being especially wary
is because they're telling us that this thing either
has no solutions or an infinite number of solutions.
So let's use these second two constraints.
And we can eliminate the x here if we
keep the first equation the same.
And we multiply the second equation by 2,
because this will become a negative 8x which
will eliminate.
So our first equation is just 8x minus 2y plus 4z is equal to 7,
just rewrote it.
And then the second equation over here.
I'm going to multiply it by 2.
Negative 4x times 2 is negative 8x.
y times 2 is plus 2y, positive 2y.
Negative 2z times 2 is negative 4z.
And then negative 14 times 2 is negative 28.
And something very interesting is about to happen.
When you use these two constraints, when
you try to find the intersection, essentially,
between these two planes, you get something very bizarre.
8x minus 8x is 0.
Negative 2y plus 2y is another 0.
4z minus 4z, another 0.
And 7 minus 28 is negative 21.
So using these two constraints, you
get this kind of nonsensical statement or inconsistent
statement that 0 is equal to negative 21.
This obviously can never happen.
So this is, we end up with no solution.
If we got something like 0 is equal to 0, or 21
is equal to 21, or negative 21 is equal to 21,
then it would mean that we have the same exact planes which
would mean we have an infinite number solutions.
In this circumstance right here, we have the same coefficients.
But when you add them up in this weighted way
you get a different result.
So these are actually going to be parallel planes.
So if you were to graph this plane and this plane,
one would look like that and then
the other would look like this.
And so that the only way that they could intersect
is if 0 is equal to negative 21.
And we know that's impossible.
They never, ever intersect these two planes.
So regardless of this first constraint and all of the stuff
we did here, these two planes will never intersect,
these two right over here.
And so we have no solutions.