Compound inequalities: AND
Sal solves the compound inequality 3y+7<2y AND 4y+8>-48. Created by Sal Khan and Monterey Institute for Technology and Education.
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- If we got that y < -7; and y > -14, is it correct to state it like this: -14 < y < -7 ??(24 votes)
- Yes, that is the most common way of expressing that type of solution. You could also just leave it as an "and" statement as you did in the first half of your comment, or even put it in set and interval notation. They all mean the same thing, and it just comes down to the visual representation that you click with the best (or that your teacher or test requests).(39 votes)
- what is the difference between compound inequalities one and two(10 votes)
- The first equation he did in "Compound Inequalities 1 video" had a "or" in between the two inequalities while this video has a "and" in it.
"Or" means it can satisfy either one of the inequalities or both while "and" means it has to satisfy both inequalities for x to qualify. You can do this by inserting a number that falls in the number line that you made for x, or after finding "x" inserting one of the numbers that x could be.(27 votes)
- I am very confused. How do you answer the questions? I already know how to break down an inequality to x<7 from 3x+4>25 or something like that, but if it gives me a question like:
2x+3≥7 OR 2x+9>11
Where inequality 1 breaks down to x≥2 and in equality 2 breaks down to x>1, how would the answer be
x>1? I don't get it, because couldn't the answer very well be x≥2 because it says OR?? I am very confused. Please help me.(9 votes)
- This is a very good question! In a problem where it says OR, either of the equalities or both equalities can satisfy the equation. In this instance, x>1, when graphed on a number line, and since the equality is greater than 1, x>1 definitely satisfies both equalities because its line encompasses the other equality's line. I would encourage you to make a number line and graph the two equalities to visualize them. This may help to alleviate your confusion.(9 votes)
- what happens when you divide a negative by a negative and isolate it?(0 votes)
- Here's a chart to help you.
By the way, N means negative and P is positive.
ta da!(27 votes)
- Wait, how are AND inequalities different from OR inequalities?(2 votes)
- Let me explain with an example. Let's take two inequalities
If the two inequalities are joined by AND, both of the inequalities must be satisfied by the values of x. In other words, both the inequalities must be true at the same time.
x<3 AND x>1 means x must be smaller than 3 and x must be larger than 1. Clearly x must lie between 1 and 3 so x∈(1,3).
If the two inequalities are joined by OR, the inequality will be true even if the value of x is true for one inequality and false for the other inequality.
x<3 OR x>1 means that x is less than 3 or x is greater than 1. Since any one of these possibilities is true for every real number, x∈R.
In essence, when using AND to join 2 inequalities we take the intersection of the solution sets of the 2 inequalities and when using OR to join 2 inequalities we take the union of the solution sets of the 2 inequalities.(7 votes)
- Is this concept still doable for a 5th grader? Because I am and my teacher assigned me to do this based on my test scores. Even after I watched this video I find it pretty difficult.(3 votes)
- Algebra I is usually taught in highschool but you can still learn it in fifth.(3 votes)
- i dont get it lol(2 votes)
- Here's the basic approach...
1) Solve each individual inequality.
2) The word "AND" is telling you to find the intersection of the two solution sets. This is all the values in common between the 2 sets. If you graph the 2 inequalities, the intersection is where the 2 graphs overlap. If there is no intersection (no values in common, or no overlap in the graphs), then there is no solution.
Hope this helps.(5 votes)
- How can you tell if all values of x can be solutions in the exercise after this video?(2 votes)
- You can put the values of x into the inequality as a solution and work it out. If the remaining(or simplified) part of the equation is correct, then that solution out of the number of solutions you got for x in the inequality is correct.(4 votes)
- What is the answer? I am lost. :-((2 votes)
- The answer set is -14 < y < -7 as shown in the graph.(2 votes)
- why do you change the ineqlity when you divide by a negative(2 votes)
- If you have - x > 5, add x to both sides and subtract 5 to get -5 > x, flip the equation around to get x < -5. So dividing by -1 flips the inequality.(1 vote)
Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this "and" is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side, and we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7. Those cancel out. We just have y is less than 0 minus 7, which is negative 7. So that's one of the constraints. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be a 56, so this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4, and we don't have to swap the inequality since we're dividing by a positive number. So it's divide both sides by 4 over here. So we get y is greater than-- what is 56/4, or negative 56/4? Let's see. 40 is 10 times 4, and then we have another 16 to worry about. So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this "and" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So let's look at this, less than negative 7. So not including negative 7, so we'll do an open circle around negative 7, and less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint-- and y has to be greater than negative 14. So you make a circle around negative 14, and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. And we can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here, 8, 9, this is negative 10. That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction, so we're getting less negative. And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works. Now, let's just verify some things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We've got 3 times is 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this condition right over here if we put a 0 over here. If you put a negative 15 over here, it should violate this condition right over here because it wasn't in this guy's solution set. Anyway, hopefully you found that useful.