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## Algebra I (2018 edition)

# Compound inequalities: OR

CCSS.Math:

Sal solves the compound inequality 5z+7<27 OR -3z≤18. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What is "P.E.M.D.A.S."?(2 votes)
- PEMDAS is the North American version of what we Europeans and Asians call BODMAS. All my peers here use PEMDAS, but I personally find that very confusing. It doesn't really matter what you use, you get the same answer.

Here are the full forms of the words

P- Parenthesis

E-exponents

M-multiplication

D-division

A-addition

S-subtraction

While BODMAS is

B-Bracket

O-of

D-division

M-multiplication

A-addition

S-subtraction

Hope it helped.

And uh... sorry for sending this 7 years late.(7 votes)

- I've taken the compound inequality practice a couple times now and haven't done well every time. I understand how to solve the problem, get each inequality's answer and plot it on a number line. What I don't understand is when to answer "No solutions", "All values of x are solutions", or the various answer choices based on the problem.(15 votes)
- Here's the scoop...

The word "OR" tells you to find the union of the 2 solution sets. The union is all the possible solutions from either inequality. So basically, a solution satisfies 1 or both of the individual inequalities.

There are 3 possible scenarios.

1) Solution is All real numbers. This is demonstrated in this video. You can see that the graph of the 2 inequalities ends up covering the entire number line.

2) The solution is 2 split intervals. For example: x<-2 OR x>0. The solution set is all numbers to the right of -2 combined with all the numbers larger than 0.

3) The solution is 1 interval. For example: x>-2 OR x>0. The union becomes x>-2 because this includes off the values from both inequalities.

If the inequality uses the word AND, then you need to find the intersection of the 2 solution sets. This is the values where the solution sets overlap (the values in common). Again, there are 3 scenarios.

1) Solution is No Solution. For example: x>5 AND x<0. These share no common values. When graphed, they have no overlap.

2) The 2 inequalities graph in opposite directions.

The solution is just where they overlap. For example: x>-2 AND x<0. The solution set is all numbers to the right of -2 up to the number 0. Basically, it is -2<x<0.

3) The 2 inequalities have graphs that go in the same direction. The solution becomes the shorter graph beause this is where they overlap. For example: x>-2 AND x>0. The intersection becomes x>0 because this includes the overlap (values in common) of both inequalities.

Hope this helps.

FYI - There is a video on union and intersection of sets. Use the search bar to find it. It may help you to understand the difference being OR (union) vs. AND (intersection).(4 votes)

- The test problems for these make no sense. On some of them the solution doesn't include numbers that both sets occupy on the number line, and for other ones they do like fpr x>5 or x <8 the solution will be all real numbers even though the numbers in between occupy the set, yet other times e.g. if, say, x>2 and x>4, then x>2 will be the solution even though someof the solutions will only be correct if the the number applies to both sets. What am I in missing?(6 votes)
- To understand this we need to look at the mathematical definitions of "and" and "or".

When we use "AND", then each statement must be true for all x.

Let's look at x>2 AND x>4.

If x< 2, say x=0 for example, then neither statement is true since 0>2 is not true and neither is 0>4, so x=0 CANNOT be in the solution set.

Lets say x=3 or x=4. Then 3>2 and 4>2 are TRUE but 3>4 and 4>4 are NOT, so neither 3 nor 4 are in the solution set. BUT, for any x>4, say 5, 5>2 is true and 5>4 is true, so 5 IS in the solution. So the answer in this case is all x such that x>4.

In the case of "OR", that means that AT LEAST ONE of the statements must be true.

Take x>5 or x<8. If x=0 then x<8 is satisfied, so even though x>5 is not satisfied, the statement x>5 or x<8 is true if x=0. Now, if x=7, then x>5 is true and x<8 is true, so the statement x>5 or x<8 when x=7 is also true and in this case, both conditions are satisfied (even though only one is required to be to make the OR true). Now, if x=10, then x>5 is true, x<8 is false, but since one of them is true, then x>5 or x<8 for x=10 is true.

Lets suppose that instead of x>5 or x<8, we had x<5 or x>8. In this case the statement would be false for all x when x=5,6,7,8. since for these values, neither x<5 or x>8 is true. But for any other x such that x is not a member of 5,6,7,8, the statement x<5 or x>8 is true.

Great Question!

Hope that helped.

Keep Studying.(18 votes)

- Why do we need compound inequalities? Do we use it in the real world?(4 votes)
- Compound inequalities are actually very important in life. Statisticians, business workers, engineers, and typical house-owners all use compound inequalities. They're kinda like systems of equations, which we use all the time in real life. But sometimes, our equations can't always equal something; they have to have constraints, like the value is greater than or less than, or something like that. Believe me, you will meet plenty of inequalities that can help you in life!(14 votes)

- Quick question.

Is a 'OR' and 'AND' work the same way as it's in programming. Where OR mean if either one is true, and AND means both has to be true ?(7 votes)- Yes, OR is when either have to be true and AND is when both have to be true.(7 votes)

- Can someone summarize what Sal is trying to say here? I don't completely understand why both z<4 and z≥-6 will satisfy both inequalities. Help would be greatly appreciated! Thanks in advance! :)(4 votes)
- So for z<4 and z≥-6, Sal is basically saying that z is less than four, but greater than or equal to -6. Some possible solutions could be -5, -4, -3, -2, -1, 0, 1, 2, or 3.(4 votes)

- I don't get the answer "all values of x"

The help section implies that it is everything on the graph, but if I go up to, let's say 10000, it can disprove one of the equations.

On a side note, when it is all values of x, does the value of x have to be the same for both of the inequalities?(5 votes) - isnt 5/5 or five divided by five , one and not zero?(3 votes)
- Yes that is correct. You can even check it with a calculator.(4 votes)

- I am so confused on this. :( For this:

−15x+60≤105 AND 14x+11≤−31

why is the answer x = -3 ?

wouldn't it be "all values of x are solutions" ?

Every time I think I get this lesson I get one wrong. Why are inequalities so elusive and confusing? It seems like the rules are always changing! I am sad. Can someone help plz :((2 votes)- I think I get it now.

It is -3 because that is the only one that would make the inequality true. Any other value wouldn't be AND, it would make it OR.

Maybe inequalities aren't as a elusive as I thought they were :)(5 votes)

- When graphing a union, does the graph have to include arrows in both directions? For example, if you were graphing x≤1 or x>7, would the graph have to have arrows in both directions of the number line? I graphed the number line but found that 2 through 7 doesn't qualify. Does this mean that any numbers
**besides**2 through 7 will qualify if not both, but one of the inequalities? Also, how is this related to unions if unions are supposed to include the whole number line? Any help would be extremely appreciated! Thank you! :)(2 votes)- yes you would have to have arrows on the end, but your interpretation is slightly off. It would be all numbers except those between 1 (inclusive) and 7 (exclusive). If you say 2 through 7, you have all the numbers between 1 and 2 (1.5,1.75, etc.) included when they should not be. Graphing would be a closed circle at 1 going to negative infinity and an open circle on 7 going to positive infinity. I do not know what you mean by unions supposed to include the whole number line, this does not have to be true as in your example.(4 votes)

## Video transcript

Solve for z. 5z plus 7 is less
than 27 or negative 3z is less than or equal to 18. So this is a
compound inequality. We have two conditions here. So z can satisfy this or z
can satisfy this over here. So let's just solve each
of these inequalities. And just know that z can
satisfy either of them. So let's just look at this. So if we look at just
this one over here, we have 5z plus 7
is less than 27. Let's isolate the z's
on the left-hand side. So let's subtract 7 from both
sides to get rid of this 7 on the left-hand side. And so our left-hand side
is just going to be 5z. Plus 7, minus 7--
those cancel out. 5z is less than
27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides
of this inequality by 5. And we don't have to swap
the inequality because we're dividing by a positive number. And so we get z
is less than 20/5. z is less than 4. Now, this was only
one of the conditions. Let's [? look at ?] the
other one over here. We have negative 3z is
less than or equal to 18. Now, to isolate the
z, we could just divide both sides of this
inequality by negative 3. But remember, when
you divide or multiply both sides of an inequality
by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide
it by negative 3. And then you have 18. We're going to divide
it by negative 3. But we're going to
swap the inequality. So the less than or equal
will become greater than or equal to. And so these guys cancel out. Negative 3 divided
by negative 3 is 1. So we have z is greater than
or equal to 18 over negative 3 is negative 6. And remember, it's this
constraint or this constraint. And this constraint right
over here boils down to this. And this one boils down to this. So our solution set--
z is less than 4 or z is greater than
or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than
4 or z is greater than or equal to negative 6. It can satisfy
either one of these. And this is kind of
interesting here. Let's plot these. So there's a number
line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4
is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about
z being less than 4. We would put a circle around
4, since we're not including 4. And it'd be everything
less than 4. Now let's think
about what z being greater than or equal to
negative 6 would mean. That means you can
include negative 6. And it's everything-- let me
do that in different color. It means you can
include negative 6. I want to do that--
oh, here we go. It means you include negative 6. Let me do it in a
more different color. Do it in orange. So z is greater than
equal to negative 6. Means you can
include negative 6. And it's everything greater
than that, including 4. So it's everything
greater than that. So what we see is
we've essentially shaded in the
entire number line. Every number will meet either
one of these constraints or both of them. If we're over here,
we're going to meet both of the constraints. If we're a number
out here, we're going to meet this constraint. If we're a number
down here, we're going to meet this constraint. And you could just try it
out with a bunch of numbers. 0 will work. 0 plus 7 is 7, which
is less than 27. And 3 times 0 is less than 18,
so it meets both constraints. If we put 4 here, it should only
meet one of the constraints. Negative 3 times 4 is negative
12, which is less than 18. So it meets this constraint, but
it won't meet this constraint. Because you do 5
times 4 plus 7 is 27, which is not less than 27. It's equal to 27. Remember, this is an or. So you just have to meet
one of the constraints. So 4 meets this constraint. So even 4 works. So it's really the
entire number line will satisfy either one or
both of these constraints.