Multi-step inequalities 3 | Linear inequalities | Algebra I | Khan Academy
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0:00 - 0:02We're asked to solve for p.
-
0:02 - 0:05And we have the inequality
here negative 3p minus 7 -
0:05 - 0:07is less than p plus 9.
-
0:07 - 0:09So what we really want
to do is isolate the p -
0:09 - 0:11on one side of this inequality.
-
0:11 - 0:13And preferably the
left-- that just makes it -
0:13 - 0:14just a little easier to read.
-
0:14 - 0:16It doesn't have to be, but we
just want to isolate the p. -
0:16 - 0:18So a good step to
that is to get rid -
0:18 - 0:19of this p on the
right-hand side. -
0:19 - 0:21And the best way I can
think of doing that -
0:21 - 0:23is subtracting p from the right.
-
0:23 - 0:25But of course, if
we want to make sure -
0:25 - 0:27that this inequality is
always going to be true, -
0:27 - 0:29if we do anything to
the right, we also -
0:29 - 0:30have to do that to the left.
-
0:30 - 0:33So we also have to
subtract p from the left. -
0:33 - 0:36And so the left-hand side,
negative 3p minus p-- -
0:36 - 0:38that's negative 4p.
-
0:38 - 0:41And then we still have
a minus 7 up here-- is -
0:41 - 0:43going to be less than p minus p.
-
0:43 - 0:44Those cancel out.
-
0:44 - 0:46It is less than 9.
-
0:46 - 0:48Now the next thing
I'm in the mood to do -
0:48 - 0:51is get rid of this
negative 7, or this minus -
0:51 - 0:537 here, so that we
can better isolate -
0:53 - 0:55the p on the left-hand side.
-
0:55 - 0:57So the best way I can think
of to get rid of a negative 7 -
0:57 - 0:58is to add 7 to it.
-
0:58 - 1:00Then it will just
cancel out to 0. -
1:00 - 1:03So let's add 7 to both
sides of this inequality. -
1:03 - 1:05Negative 7 plus 7 cancel out.
-
1:05 - 1:08All we're left with
is negative 4p. -
1:08 - 1:12On the right-hand side, we
have 9 plus 7 equals 16. -
1:12 - 1:14And it's still less than.
-
1:14 - 1:16Now, the last step
to isolate the p -
1:16 - 1:18is to get rid of this
negative 4 coefficient. -
1:18 - 1:20And the easiest way I
can think of to get rid -
1:20 - 1:23of this negative
4 coefficient is -
1:23 - 1:26to divide both
sides by negative 4. -
1:26 - 1:28So if we divide this
side by negative 4, -
1:28 - 1:29these guys are
going to cancel out. -
1:29 - 1:31We're just going
to be left with p. -
1:31 - 1:33We also have to do it
to the right-hand side. -
1:33 - 1:35Now, there's one thing that
you really have to remember, -
1:35 - 1:38since this is an
inequality, not an equation. -
1:38 - 1:40If you're dealing
with an inequality -
1:40 - 1:44and you multiply or divide
both sides of an equation -
1:44 - 1:48by a negative number, you
have to swap the inequality. -
1:48 - 1:50So in this case, the less
than becomes greater than, -
1:50 - 1:52since we're dividing
by a negative number. -
1:52 - 1:56And so negative 4 divided by
negative 4-- those cancel out. -
1:56 - 2:00We have p is greater than
16 divided by negative 4, -
2:00 - 2:01which is negative 4.
-
2:01 - 2:04And we can plot this
solution set right over here. -
2:04 - 2:06And then we can
try out some values -
2:06 - 2:09to help us feel good about
the idea of it working. -
2:09 - 2:14So let's say this is negative
5, negative 4, negative 3, -
2:14 - 2:18negative 2, negative 1, 0.
-
2:18 - 2:19Let me write that a
little bit neater. -
2:22 - 2:24And then we can keep
going to the right. -
2:24 - 2:27And so our solution is p is
not greater than or equal, -
2:27 - 2:29so we have to
exclude negative 4. -
2:29 - 2:33p is greater than negative 4,
so all the values above that. -
2:33 - 2:38So negative 3.9999999 will work.
-
2:38 - 2:40Negative 4 will not work.
-
2:40 - 2:42And let's just try
some values out -
2:42 - 2:45to feel good that this is
really the solution set. -
2:45 - 2:49So first let's try out when
p is equal to negative 3. -
2:49 - 2:50This should work.
-
2:50 - 2:52The way I've drawn it,
this is in our solution -
2:52 - 2:55set. p equals negative 3
is greater than negative 4. -
2:55 - 2:56So let's try that out.
-
2:56 - 3:00We have negative 3
times negative 3. -
3:00 - 3:02The first negative
3 is this one, -
3:02 - 3:04and then we're saying
p is negative 3. -
3:04 - 3:06Minus 7 should be less
than-- instead of a p, -
3:06 - 3:08we're going to
putting a negative 3. -
3:08 - 3:10Should be less than
negative 3 plus 9. -
3:10 - 3:12Negative 3 times
negative 3 is 9, -
3:12 - 3:17minus 7 should be less than
negative 3 plus 9 is 6. -
3:17 - 3:189 minus 7 is 2.
-
3:18 - 3:212 should be less than 6,
which, of course, it is. -
3:21 - 3:24Now let's try a value that
definitely should not work. -
3:24 - 3:27So let's try negative 5.
-
3:27 - 3:29Negative 5 is not
in our solution set, -
3:29 - 3:30so it should not work.
-
3:30 - 3:34So we have negative 3
times negative 5 minus 7. -
3:34 - 3:39Let's see whether it is
less than negative 5 plus 9. -
3:39 - 3:43Negative 3 times negative
5 is 15, minus 7. -
3:43 - 3:47It really should not be
less than negative 5 plus 9. -
3:47 - 3:51So we're just seeing if p
equals negative 5 works. -
3:51 - 3:5515 minus 7 is 8.
-
3:55 - 3:58And so we get 8 is
less than 4, which -
3:58 - 4:01is definitely not the case.
-
4:01 - 4:03So p equals negative
5 doesn't work. -
4:03 - 4:05And it shouldn't work, because
that's not in our solution set. -
4:05 - 4:07And now if we really want
to feel good about it, -
4:07 - 4:09we can actually try
this boundary point. -
4:09 - 4:12Negative 4 should
not work, but it -
4:12 - 4:14should satisfy the
related equation. -
4:14 - 4:15When I talk about
the related equation, -
4:15 - 4:20negative 4 should satisfy
negative 3 minus 7 -
4:20 - 4:22is equal to p plus 9.
-
4:22 - 4:25It'll satisfy this, but
it won't satisfy this. -
4:25 - 4:28Because when we get the
same value on both sides, -
4:28 - 4:30the same value is not
less than the same value. -
4:30 - 4:31So let's try it out.
-
4:31 - 4:34Let's see whether
negative 4 at least -
4:34 - 4:37satisfies the related equation.
-
4:37 - 4:41So if we get negative 3
times negative 4 minus 7, -
4:41 - 4:44this should be equal
to negative 4 plus 9. -
4:44 - 4:48So this is 12 minus 7 should
be equal to negative 4 plus 9. -
4:48 - 4:49It should be equal to 5.
-
4:49 - 4:51And this, of course, is true.
-
4:51 - 4:535 is equal to 5.
-
4:53 - 4:54So it satisfies the
related equation, -
4:54 - 4:56but it should not satisfy this.
-
4:56 - 4:58If you put negative
4 for p here-- -
4:58 - 4:59and I encourage you to do so.
-
4:59 - 5:01Actually, we could
do it over here. -
5:01 - 5:03Instead of an equals
sign, if you put it -
5:03 - 5:07into the original inequality--
let me delete all of that-- -
5:07 - 5:09it really just becomes this.
-
5:09 - 5:12The original inequality
is this right over here. -
5:12 - 5:15If you put negative
4, you have less than. -
5:15 - 5:18And then you get 5 is less
than 5, which is not the case. -
5:18 - 5:20And that's good,
because we did not -
5:20 - 5:21include that in
the solution set. -
5:21 - 5:23We put an open circle.
-
5:23 - 5:25If negative 4 was included,
we would fill that in. -
5:25 - 5:27But the only reason why
we'd include negative 4 -
5:27 - 5:29is if this was
greater than or equal. -
5:29 - 5:33So it's good that this does
not work, because negative 4 -
5:33 - 5:35is not part of our solution set.
-
5:35 - 5:38You can kind of view
it as a boundary point.
- Title:
- Multi-step inequalities 3 | Linear inequalities | Algebra I | Khan Academy
- Description:
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Multi-Step Inequalities
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Fran Ontanaya edited English subtitles for Multi-step inequalities 3 | Linear inequalities | Algebra I | Khan Academy |