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Hello, this is Wendy Lightheart and for this lesson we are going to learn
how to solve linear inequalities.
Let's start by looking at some linear inequalities.
The first one is read
"x is less than 5".
Its English interpretation is "the set of all real numbers that are
strictly less than 5".
This means that any real number in that set
makes this inequality a true statement.
The next one reads "x is greater than
or equal to -2".
Its English interpretation is "the set of all real numbers that are either
greater than -2 or equal to -2".
Notice that a line under the less than or greater than sign
indicates that that number itself is included in that set.
Now let's look at
how these inequalities can be graphed on a real number line.
We will also look at how to write the interval notation for each inequality.
Recall that the
first inequality says
that x is any real number strictly less than 5.
We show this with an arrow pointing
from 5
toward all numbers less than five,
which are the numbers to the left of 5.
Since 5 is not included in the set, we put a parenthesis at 5.
You can think of it as getting really close to 5 but never actually
reaching 5 as the parenthesis curves away from 5.
The interval notation for this set follows naturally from its graph.
You can think of it as naming the range of numbers the set includes.
Since it points to the left, we say it goes from negative infinity to 5,
but not including 5.
This is how to notate that with interval notation.
We list the left endpoint first, which is in this case is negative infinity
and list the right endpoint next,
separating the endpoints with a comma.
Note that infinity and negative infinity
are not actual numbers so we will always use parentheses next to the
infinity symbols.
However, the grouping symbol we use next to an endpoint that is an actual number, like 5,
will depend on whether or not the number is included in the set.
If it is not included in the set,
we use a parenthesis next to that number.
In the next example, we will see what to use when the number is to be included
in the set.
Recall that the next inequality is the set of all real numbers greater than or
equal to -2.
Thus, the graph will have an arrow that indicates just that.
Notice that the arrow starts at -2
and then points to all numbers greater than -2.
A square bracket is placed at -2 to
indicate that this number is included in the set.
This is the same
symbol we will use in the interval notation.
Note that this notation says that -2 is the left endpoint of this
set and it is included in the set.
Then the right endpoint is infinity
and remember we always use a parenthesis next to infinity.
The nice thing about solving inequalities
is that it includes the same steps that we used to solve linear equations,
with only one exception.
And that is
whenever you multiply or divide
by a negative number
you must
flip the direction of the inequality sign.
So let's solve some inequalities.
Remember that the first step to solving a linear equation is to simplify both
sides
and that's going to be the same for solving linear inequalities.
That's the first thing we need to do here is to distribute the -8
to get rid of the parentheses.
Then, we need to combine like terms.
Once both sides are simplified
we collect the variable terms to one side and the constant terms to the other.
So we can move the constant terms to the left by subtracting 28
from both sides.
Now we have just one more step to isolate the x
and that is to divide both sides by 8.
Notice that we haven't multiplied or divided both sides by a
negative number yet.
So we do not need to flip the direction of the sign.
Simplifying both sides we obtain our solution
that reads -1 is less than or equal to x.
Or in other words,
x is greater than or equal to -1.
Next we'll graph and write the interval notation for this solution set.
As you can see the arrow starts at -1
and includes -1 with the bracket.
Then it points to all numbers greater than -1.
Thus, our intervall is from and including -1
to positive infinity.
Let's look at a fun one with fractions.
Remember that we can clear the fractions by multiplying every term
on both sides by the least common denominator,
which is 6 in this case.
So we multiply each term
by 6.
Then we simplify each term.
Notice that the factions have been eliminated.
Also notice that both sides are simplified completely so we're ready to
collect variable terms to one side
and constant terms to the other.
Let's say we move the variable terms to the left.
We can do that by subtracting 3x from both sides.
Now we must move the constants to the right
and we can do that by subtracting 15 from both sides.
Notice that the we almost have x isolated,
but we're still going have to get rid of that negative sign.
Let's continue this on the next slide.
To get rid of the negative sign on a variable remember
we can divide those sides by -1.
However, whenever we multiply or divide both sides by a negative number
we must
flip the sign.
So the greater than or equal to sign will be
flipped to become a less than or equal to sign.
The division
gives us our solution
of x is less than or equal to -3.
Let's graph and write the interval notation for the solution set.
Think of what kind of symbol will be placed at -3
and which direction the arrow will be pointing.
If you pictured a bracket at -3
and an arrow pointing to the left of -3, then you're right!
The left endpoint is negative infinity
and the right endpoint is -3.
We use a bracket because we are including -3 in that set.
This is the interval notation for this solution set.
Okay. Let's try one more example.
For this one we need to use the distributive property on one side to
simplify
and combine like terms on the other side to simplify.
So we will distribute the 3 to get 3x + 3 on the left
and combine like terms
to get 3x + 7 on the right.
Now that both sides are simplified,
we need to collect variable terms to one side and constants to the other.
Let's say we move the variable terms to the left so we can do that
by subtracting
3x from both sides.
Notice that the variables will cancel out on both sides.
Whenever this happens, just like the equations,
if the remaining statement is false
then there is no solution.
Thus, this inequality has no solution because 3 is not greater than 7.
On the other hand, if the variable terms cancel out and the remaining statement
is true
then the solution set is all real numbers.
Well that's it for today.
Hopefully, you've become proficient in enough at solving linear equations
that solving linear inequalities
will be a piece of cake.