00:01
Okay, here we have x minus 5 over x is less than 1.
00:07
Now, they want us to solve for this.
00:09
And the first thing to do is figure out what x cannot be.
00:13
Remember that the denominator can never be zero because that would make it undefined.
00:17
So in this case, x cannot equal zero.
00:24
Because if we plug in zero for x, then the denominator would be zero, making the inequality undefined.
00:32
Next, let's solve for x.
00:38
To solve, we need to first remove the denominator, and we're going to do that by multiplying by the denominator.
00:43
So multiply both sides by x.
00:52
When that happens on the left side, x is pretty much canceled.
00:59
We have x minus 5 on the left, drop down the less than, and then one times x is x.
01:09
Now here, something peculiar happens.
01:13
We have to get the variable to that left side, and to do that, we'd have to subtract x.
01:22
But when that happens, we have negative 5 is less than 0.
01:31
So really, we only have one number to work with.
01:34
We have the 0 that we got at the very beginning of the problem.
01:38
So we're going to put that on the interval or on the number line.
01:42
I'm sorry.
01:45
So now we have to test out a number in each interval.
01:49
So on the left interval, we can test out by plug in negative 1.
01:54
On the right side, we could test it out by plugging in a positive...