12 अपऑन सेवेन ब्रैकेट में एक्स माइनस 5 इज इक्वल टू 4 प्लस एक्स
Answers
12 on seven brackets x minus 5 is equal to 4 plus x .
Answer:
Step-by-step explanation:
Let's do some compound inequality problems, and these are just inequality problems that have more than one set of constraints. You're going to see what I'm talking about in a second. So the first problem I have is negative 5 is less than or equal to x minus 4, which is also less than or equal to 13. So we have two sets of constraints on the set of x's that satisfy these equations. x minus 4 has to be greater than or equal to negative 5 and x minus 4 has to be less than or equal to 13. So we could rewrite this compound inequality as negative 5 has to be less than or equal to x minus 4, and x minus 4 needs to be less than or equal to 13. And then we could solve each of these separately, and then we have to remember this "and" there to think about the solution set because it has to be things that satisfy this equation and this equation. So let's solve each of them individually. So this one over here, we can add 4 to both sides of the equation. The left-hand side, negative 5 plus 4, is negative 1. Negative 1 is less than or equal to x, right? These 4's just cancel out here and you're just left with an x on this right-hand side. So the left, this part right here, simplifies to x needs to be greater than or equal to negative 1 or negative 1 is less than or equal to x. So we can also write it like this. X needs to be greater than or equal to negative 1. These are equivalent. I just swapped the sides. Now let's do this other condition here in green. Let's add 4 to both sides of this equation. The left-hand side, we just get an x. And then the right-hand side, we get 13 plus 14, which is 17. So we get x is less than or equal to 17. So our two conditions, x has to be greater than or equal to negative 1 and less than or equal to 17. So we could write this again as a compound inequality if we want. We can say that the solution set, that x has to be less than or equal to 17 and greater than or equal to negative 1. It has to satisfy both of these conditions. So what would that look like on a number line? So let's put our number line right there. Let's say that this is 17. Maybe that's 18. You keep going down. Maybe this is 0. I'm obviously skipping a bunch of stuff in between. Then we would have a negative 1 right there, maybe a negative 2. So x is greater than or equal to negative 1, so we would start at negative 1. We're going to circle it in because we have a greater than or equal to. And then x is greater than that, but it has to be less than or equal to 17. So it could be equal to 17 or less than 17. So this right here is a solution set, everything that I've shaded in orange. And if we wanted to write it in interval notation, it would be x is between negative 1 and 17, and it can also equal negative 1, so we put a bracket, and it can also equal 17. So this is the interval notation for this compound inequality right there. Let's do another one. Let me get a good problem here. Let's say that we have negative 12. I'm going to change the problem a little bit from the one that I've found here. Negative 12 is less than 2 minus 5x, which is less than or equal to 7. I want to do a problem that has just the less than and a less than or equal to. The problem in the book that I'm looking at has an equal sign here, but I want to remove that intentionally because I want to show you when you have a hybrid situation, when you have a little bit of both. So first we can separate this into two normal inequalities. You have this inequality right there. We know that negative 12 needs to be less than 2 minus 5x. That has to be satisfied, and-- let me do it in another color-- this inequality also needs to be satisfied. 2 minus 5x has to be less than 7 and greater than 12, less than or equal to 7 and greater than negative 12, so and 2 minus 5x has to be less than or equal to 7. So let's just solve this the way we solve everything. Let's get this 2 onto the left-hand side here. So let's subtract 2 from both sides of this equation. So if you subtract 2 from both sides of this equation, the left-hand side becomes negative 14, is less than-- these cancel out-- less than negative 5x. Now let's divide both sides by negative 5. And remember, when you multiply or divide by a negative number, the inequality swaps around.