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# Compound inequalities: OR

CCSS Math: HSA.REI.B.3

## 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.