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Algebra 1
Course: Algebra 1 > Unit 2
Lesson 6: Compound inequalitiesCompound 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).(40 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.(29 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.(10 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.(10 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.
N*P=N
P*N=N
N*N=P
P*P=P
ta da!(29 votes)
where did -10 come from?(2 votes)
It's just a random number he chose that was between -14 and -7.(12 votes)
- I'm having a lot of trouble with the exercise after this. I don't have a problem with solving the inequalities themselves, but I don't understand the part when it asks if there's no answer, or if all values of x are answers, or if x is more than/less than y, etc. Can someone help me?(4 votes)
I don't understand where you got those numbers for the number line I am very confused do I need that or was it just to help in something?(3 votes)
To solve a compound inequality, you start by solving each individual inequality. Then, the word "AND" or "OR" tells you the next step to take.
AND tells you to find the intersection of the two solution sets. An intersection is the values in common or the overlap of the two sets. This is why it is common to graph the 2 original inequalities. From the graph, you can quickly identify what values are in common because it is where the graphs overlap.
OR tells you to find the union of the two solution sets. A union combines all solutions from the original inequalities into one solution set. If a value works for either inequality or both inequalities, it goes in the union.
For more info on Intersections vs Unions, try this video: https://www.khanacademy.org/math/statistics-probability/probability-library/basic-set-ops/v/intersection-and-union-of-sets
Hope this helps.(7 votes)
- Wait, how are AND inequalities different from OR inequalities?(2 votes)
- Let me explain with an example. Let's take two inequalities
x<3, x>1
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 have almost 1000000 energy points.(4 votes)
nice, keep going(1 vote)
Video transcript
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.