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# Multi-step inequalities

Sal solves several multi-step linear inequalities. Created by Sal Khan.

Video transcript

Let's do a few more problems
that bring together the concepts that we learned
in the last two videos. So let's say we have the
inequality 4x plus 3 is less than negative 1. So let's find all of the
x's that satisfy this. So the first thing I'd like to
do is get rid of this 3. So let's subtract 3 from both
sides of this equation. So the left-hand side is just
going to end up being 4x. These 3's cancel out. That just ends up with a zero. No reason to change the
inequality just yet. We're just adding and
subtracting from both sides, in this case, subtracting. That doesn't change the
inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3
is negative 4. And then we'll want to-- let's
see, we can divide both sides of this equation by 4. Once again, when you multiply
or divide both sides of an inequality by a positive number,
it doesn't change the inequality. So the left-hand
side is just x. x is less than negative 4
divided by 4 is negative 1. x is less than negative 1. Or we could write this
in interval notation. All of the x's from negative
infinity to negative 1, but not including negative
1, so we put a parenthesis right there. Let's do a slightly
harder one. Let's say we have 5x is greater
than 8x plus 27. So let's get all our x's on the
left-hand side, and the best way to do that is subtract
8x from both sides. So you subtract 8x
from both sides. The left-hand side becomes
5x minus 8x. That's negative 3x. We still have a greater
than sign. We're just adding or subtracting
the same quantities on both sides. These 8x's cancel out and you're
just left with a 27. So you have negative 3x
is greater than 27. Now, to just turn this into an
x, we want to divide both sides by negative 3. But remember, when you multiply
or divide both sides of an inequality by a negative
number, you swap the inequality. So if we divide both sides of
this by negative 3, we have to swap this inequality. It will go from being
a greater than sign to a less than sign. And just as a bit of a way that
I remember greater than is that the left-hand side
just looks bigger. This is greater than. If you just imagine this height,
that height is greater than that height right there,
which is just a point. I don't know if that confuses
you or not. This is less than. This little point
is less than the distance of that big opening. That's how I remember it. But anyway, 3x over
negative 3. So now that we divided both
sides by a negative number, by negative 3, we swapped the
inequality from greater than to less than. And the left-hand side, the
negative 3's cancel out. You get x is less than
27 over negative 3, which is negative 9. Or in interval notation, it
would be everything from negative infinity to negative
9, not including negative 9. If you wanted to do it
as a number line, it would look like this. This would be negative 9, maybe
this would be negative 8, maybe this would
be negative 10. You would start at negative 9,
not included, because we don't have an equal sign here, and
you go everything less than that, all the way down, as we
see, to negative infinity. Let's do a nice,
hairy problem. So let's say we have 8x minus
5 times 4x plus 1 is greater than or equal to negative 1
plus 2 times 4x minus 3. Now, this might seem very
daunting, but if we just simplify it step by step, you'll
see it's no harder than any of the other problems
we've tackled. So let's just simplify this. You get 8x minus-- let's
distribute this negative 5. So let me say 8x, and then
distribute the negative 5. Negative 5 times 4x
is negative 20x. Negative 5-- when I say negative
5, I'm talking about this whole thing. Negative 5 times 1 is negative
5, and then that's going to be greater than or equal
to negative 1 plus 2 times 4x is 8x. 2 times negative 3
is negative 6. And now we can merge these two
terms. 8x minus 20x is negative 12x minus 5 is greater
than or equal to-- we can merge these constant terms.
Negative 1 minus 6, that's negative 7, and then we
have this plus 8x left over. Now, I like to get all my x
terms on the left-hand side, so let's subtract 8x from both
sides of this equation. I'm subtracting 8x. This left-hand side, negative
12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason
to change the inequality just yet. All we're doing is simplifying
the sides, or adding and subtracting from them. The right-hand side becomes--
this thing cancels out, 8x minus 8x, that's 0. So you're just left
with a negative 7. And now I want to get rid
of this negative 5. So let's add 5 to both sides
of this equation. The left-hand side, you're just
left with a negative 20x. These 5's cancel out. No reason to change the
inequality just yet. Negative 7 plus 5, that's
negative 2. Now, we're at an interesting
point. We have negative 20x is greater
than or equal to negative 2. If this was an equation, or
really any type of an inequality, we want to divide
both sides by negative 20. But we have to remember, when
you multiply or divide both sides of an inequality by a
negative number, you have to swap the inequality. So let's remember that. So if we divide this side by
negative 20 and we divide this side by negative 20, all I did
is took both of these sides divided by negative 20, we have
to swap the inequality. The greater than or equal to has
to become a less than or equal sign. And, of course, these cancel
out, and you get x is less than or equal to-- the
negatives cancel out-- 2/20 is 1/10. If we were writing it in
interval notation, the upper bound would be 1/10. Notice, we're including it,
because we have an equal sign, less than or equal, so we're
including 1/10, and we're going to go all the way down
to negative infinity, everything less than
or equal to 1/10. This is just another way
of writing that. And just for fun, let's
draw the number line. Let's draw the number
line right here. This is maybe 0, that is 1. 1/10 might be over here. Everything less than
or equal to 1/10. So we're going to include the
1/10 and everything less than that is included in
the solution set. And you could try out any
value less than 1/10 and verify that it will satisfy
this inequality.