Example
1 Solving
the following inequalities. Give both inequality and interval notation forms of the solution.
(a) – 2(m – 3) < 5(m + 1) – 12
(b)
2(1 – x) + 5 ≤ 3(2x – 1)
Solution
Solving single linear inequalities follow
pretty much the same process for solving linear equations. We will simplify both sides, get all
the terms with the variable on one side and the numbers on the other side, and then
multiply/divide both sides by the coefficient of the variable to get the solution. The one thing that you’ve got
to remember is that if you multiply/divide by a negative number then switch the direction of
the inequality.
(a) – 2(m – 3) < 5(m + 1) – 12
There really isn’t much to do here other than follow
the process outlined above.
– 2(m – 3) < 5(m + 1) – 12
– 2m + 6 < 5m + 5 – 12
–7m < –13
You did catch the fact that the direction of the
inequality changed here didn’t you? We divided by a “–7” and so we had to change the direction. The inequality form of the
solution is
.
The interval notation for this solution is,
.
.
(b) 2(1 – x) + 5 ≤ 3(2x – 1)
Again, not much to do here.
2(1
– x) + 5 ≤ 3(2x – 1)
2 – 2x
+ 5 ≤ 6x – 3
10 ≤ 8x
Now, with this inequality we ended up with the variable
on the right side when it more traditionally on the left side. So, let’s switch things around to get the
variable onto the left side. Note however, that we’re going to need also switch the direction of the
inequality to make sure that we don’t change the answer. So, here is the inequality notation for the
inequality.
The interval notation for the solution is
.
.
Example 2 Solve each of the following
inequalities. Give both inequality and interval notation forms for the solution.
(a) – 6 ≤ 2(x –
5) < 7
(c) – 14 < – 7(3x +
2) < 1
Solution
(a) – 6 ≤ 2(x –
5) < 7
The process here is fairly similar
to the process for single inequalities, but we will first need to be careful in a couple of
places. Our first step in this case will be to clear any parenthesis in the
middle term.
– 6 ≤ 2x – 10) < 7
Now, we want the x all by itself in the middle term and
only numbers in the two outer terms. To do this we will
add/subtract/multiply/divide as needed. The only thing that we need to remember
here is that if we do something to middle term
we need to do the same thing to BOTH of the out terms. One of the more common mistakes at this
point is to add something, for example, to the middle and only add it to one of the two sides.
Okay, we’ll add 10 to all three parts and then
divide all three parts by two.
4 ≤ 2x < 17
That is the inequality form of the
answer. The interval notation form of the answer is
.
.
In this case the first thing that we
need to do is clear fractions out by multiplying all three parts by 2. We will then proceed as we
did in the first part.
– 6 < 3(2 – x) ≤ 10
– 6 < 6 – 3x ≤ 10
– 12 < –3x ≤ 4
Now, we’re not quite done here, but
we need to be very careful with the next step. In this step we need to divide all three
parts by -3. However, recall that whenever we divide both sides of an
inequality by a negative number we need to
switch the direction of the inequality. For us, this means that both of the inequalities will need
to switch direction here.
So, there is the inequality form of the solution. We will need to be careful with the interval notation for the solution. First, the interval notation is NOT
. Remember that in interval notation the smaller number must always go on the left side!
Therefore, the correct interval notation for the solution is.
Note as well that this does match up with the
inequality form of the solution as well. The inequality is telling us that x
is any number between 4 and
oritself
and this is exactly what the interval notation is telling us.
oritself
and this is exactly what the interval notation is telling us.
Also, the inequality could be flipped around to get the
smaller number on the left if we’d like to. Here is that form,
When doing this make sure to correctly deal with the
inequalities as well.
(c) – 14 < – 7(3x +
2) < 1
Not much to this one. We’ll proceed as we’ve done the
previous two.
– 14 < – 21x – 14 < 1
0 < – 21x < 15
Don’t get excited about the fact that one of the sides
is now zero. This isn’t a problem. Again, as with the last part, we’ll be
dividing by a negative number and so don’t forget to switch the direction of
the inequalities.
Either of the inequalities in the second row will work
for the solution. The interval notation of the solution is
.
.
Example 3 If – 1 < x < 4 then determine a and b in
a < 2x + 3 < b.
Solution
This is easier than it may appear at first. All we are
really going to do is start with the given inequality and then manipulate the middle term to look like the second
inequality. Again, we’ll need to remember that whatever we do to the middle term we’ll also need to do
to the two outer terms.
So, first we’ll multiply everything by 2.
– 2 < 2x < 8
Now add 3 to everything.
1 < 2x + 3 < 11
We’ve now got the middle term identical to the second
inequality in the problems statement and so all we need to do is pick off a and
b. From this inequality we can see that a = 1 and b = 11.
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