Video Transcript
Find all the values of 𝑥 that
satisfy seven 𝑥 plus four is greater than 11 and less than or equal to 32. Give your answer in interval
form.
This is what’s known as a
double-sided or compound inequality. In general, a compound inequality
contains at least two inequalities separated by either the word “or” or the word
“and”. Here, we have an expression, seven
𝑥 plus four, and the inequality tells us that the value of this expression is
greater than 11 and it’s less than or equal to 32. There’re two approaches we can take
to solving a double-sided inequality.
The first approach is to treat the
two parts of the inequality separately. So we have one inequality telling
us that seven 𝑥 plus four is greater than 11 and another telling us that seven 𝑥
plus four is less than or equal to 32. We then solve each inequality. For the inequality on the left, the
first step is just subtract four from each side, giving seven is less than seven
𝑥. We can then divide each side of
this inequality by seven to give one is less than 𝑥 or 𝑥 is greater than one. So, we’ve solved our first
inequality.
To solve the second, we also
subtract four from each side, giving seven 𝑥 is less than or equal to 28 and then
divide both sides by seven to give 𝑥 is less than or equal to four. We’ve found then that the value of
𝑥 is greater than one and less than or equal to four. So we must make sure to put these
two parts of the solution back together again at the end. We can write our solution as the
double-sided inequality 𝑥 is greater than one and less than or equal to four. As an interval then, this would
have endpoints of one and four.
And now we need to consider the
type of brackets or parentheses for each end. At the lower end, the sign is a
strict inequality; 𝑥 is strictly greater than one. So the value one is not included in
the solution set. And so, our interval is open at the
lower end. However, at the upper end, we have
a weak inequality, 𝑥 is less than or equal to four. So the value of four is included in
the solution set, and our interval is closed at its upper end. And so, we have our answer to the
problem. The solution set for this
inequality is the interval from one to four, open at its lower end and closed at its
upper end.
Now, notice that the steps involved
in solving these two separate inequalities were exactly the thing. In each case, we subtracted four
first and then divided by seven. So, in fact, there was no need for
us to treat the two parts of this inequality separately. The second and probably more
efficient approach, then, is to keep all three parts of the inequality together. In this case, we solve in exactly
the same way but we must make sure that we perform the same operation to all three
parts of the inequality. We begin by subtracting four from
each part. 11 minus four is seven. Seven 𝑥 plus four minus four is
seven 𝑥. And 32 minus four is 28. So we now have the statement seven
𝑥 is greater than seven and less than or equal to 28.
We then divide each part of our
inequality by seven, giving the statement 𝑥 is greater than one and less than or
equal to four, which we notice is identical to the solution we had using our first
method. We would then express this in
interval notation in exactly the same way. This method is certainly quicker,
but we must make sure that we treat all three parts of the inequality
identically. So if we’re subtracting four, we
must make sure we do it from every part. And if we’re dividing by some
number, in this case seven, again we need to do it to every part.
