Just as we saw happen with linear equations, sometimes linear inequalities will also have fractions, variables on both sides of the symbol, or maybe even both at the same time, but we're actually going to address them the exact same way we did with linear equations by getting rid of those fractions using the least common denominator and then getting all of our variables to one side. So let's go ahead and take a look at an example.

So here I have 14 times x plus 2 is greater than or equal to 112 minus 13x. Now the very first thing we want to do is get rid of those fractions, which we can do by multiplying the entire inequality by the least common denominator. Now I have 3 denominators here: 4, 12, and 3. So my least common denominator is going to be 12. So let's go ahead and multiply our entire inequality by 12. Remember that whenever we're multiplying by our least common denominator, we need to make sure to carry that into every single term in our equation or in this case, our inequality. So let's go ahead and expand this out.

So this becomes 124x plus 2 is greater than or equal to a negative 1212 and then carrying that 12 to our very last term −123x. Let's go ahead and simplify. So 124 is just 3, so this becomes 3 times x + 2 is greater than or equal to a negative 1. And then negative 123 is going to give me negative 4 times x. Okay.

Now we need to look at a couple of different things here. The first thing that we need to do from this point is to go ahead and distribute this 3. So I need to distribute this 3 into my x and my 2. So this is going to become 3x plus 6 is greater than or equal to negative one minus 4 x. Okay. Now since we have variables on both sides, this is where we need to move all of our variables to one side, all of our constants to the other, the same way we did with linear equations. So I want to move this 4 x to my left side and move my negative 6 to my right side. And I can do that by adding 4 x to both sides, you'll, of course, cancel on that right side, and then subtracting 6 from both sides, canceling it on the left and moving it to the right. Okay. So 3x + 4x is going to give me 7x. I still have my inequality symbol greater than or equal to. And then on that right side, I have negative 1 minus 6 which will give me negative 7. Okay. Last step here, we're going to go ahead and isolate x by dividing both sides by 7. Now I am simply left with x ≥ −1 and this is my solution.

But remember, we want to express our solution set in interval notation and graph it. So let's go ahead and graph first so that we can better visualize this solution. So my solution is x ≥ −1. I know that since it's greater than or equal to, I need to choose a solid circle, a closed circle, for my endpoint of negative one. And then since it's anything that's greater than or equal to negative 1, I'm just going to use an arrow to indicate that that goes all the way to infinity. Okay. Now that I can visualize it better, what is this in interval notation? Well, interval notation for this, since I know that that negative one endpoint is included, I'm going to use a square bracket to enclose that endpoint of negative one, and then that goes all the way to anything greater than negative one, which is just infinity which I always enclose with a parenthesis. And this is my final answer and my final graph. Let me know if you have any questions.