00:01
In this problem, we are asked to solve the rational inequality and write the solution set in interval notation.
00:08
Our first step will be figuring out what value of x will make the numerator equal to zero, and what value of x will make the denominator equal to zero.
00:16
We need to first move positive 2 from the right -hand side to the left -hand side of our inequality.
00:28
So now that we've moved our 2 from the right -hand side to the left -hand side, we can multiply our negative 2 by a factor of x -minus 6 over.
00:38
X minus 6 to achieve a common denominator.
00:52
And i'm just going to move up here to keep working on our algebra.
00:57
So now what we want to do is combine both of our fractions and simplify our numerators.
01:03
We can see that we have a negative sign here and a 2 outside of this quantity.
01:09
So that means that negative 2 is going to be distributed across to the x and the negative 6.
01:16
So when we do that, we get x minus, 2x plus 12 in the numerator, all over x minus 6.
01:28
And then we can simplify a little bit further, so we end up with positive 15 minus 2x, all divided by x minus 6, is less than or equal to 0.
01:40
So now that we have our simplified inequality here, we can find out what values of x will make the numerator and denominator equal to 0.
01:49
We can recognize in the numerator, here that when x is equal to positive 15 halves, the numerator will be zero.
01:59
And when x is equal to positive six, the denominator will be equal to zero.
02:04
So these two x values are called end points, and they're used to break up our solution set into intervals.
02:11
We can now construct a table using these intervals and employ test points to determine which intervals fall into our solution set.
02:18
Here we have our table with the intervals that are dependent on the two end points that we determined above.
02:24
We can test a value that's within each interval, but not equal to either endpoint of the interval in order to see if the test point makes our inequality true or false.
02:34
And if the inequality is true within a certain interval, that means that that interval is included in our solution set...