### Video Transcript

Find the solution set of the
inequality five π₯ minus one over 10 is less than negative two π₯ plus five which is
less than π₯ plus three over two in the real numbers. Give your answer in interval
notation.

Now, this is a relatively complex
question. We have a double-sided inequality,
we have some fractions, and we also have our variable π₯ appearing in each part of
this inequality. Letβs first consider how we could
solve this problem by treating the two parts of the inequality separately. So we have the single-sided
inequality five π₯ minus one over 10 is less than negative two π₯ plus five and
another single-sided inequality negative two π₯ plus five is less than π₯ plus three
over two.

Through our first inequality, weβd
begin by multiplying each side by 10 to eliminate the fractions. And we now have five π₯ minus one
is less than negative 20π₯ plus 50. We can then add one to each side of
this inequality. Now, we still have π₯ terms
appearing on each side. So next, we want to group all of
the π₯βs on the same side. And if we add 20π₯ to each side,
weβll have a positive number of π₯βs on the left-hand side of our inequality. Adding 20π₯ to each side of the
inequality, then, gives 25π₯ is less than 51. Finally, dividing both sides of our
inequality by 25 gives π₯ is less than 51 over 25. And weβll keep this answer in
fractional form.

So weβve solved the first part, and
now letβs consider how we can solve the second part. First, we can multiply both sides
by two to eliminate the fractions, giving negative four π₯ plus 10 is less than π₯
plus three. We can then subtract three from
each side giving negative four π₯ plus seven is less than π₯ and then add four π₯ to
each side of the inequality, giving seven is less than five π₯. Finally, we divide by five, giving
seven over five is less than π₯ or π₯ is greater than seven over five.

So, weβve solved the two parts of
our inequality separately. But actually, we can combine these
answers. We can express this as a
double-sided inequality, π₯ is greater than seven over five and less than 51 over
25. The question, though, asked us to
give our answer an interval notation. We can write this then as the
interval with endpoints of seven over five and 51 over 25. And as each of the inequalities are
strict inequalities, meaning neither of the endpoints are included in the interval,
this means that our interval will be open at both ends. So we have our answer to the
problem. The solution set of this inequality
is the open interval from seven over five to 51 over 25.

Now, although the steps involved in
solving our two separate inequalities were different here, it is actually possible
to solve this inequality by keeping it all together, although this wonβt always be
the case. As the denominators involved in the
fractions are 10 and two and as two is a factor of 10, we can eliminate each of the
denominators by multiplying each part of our inequality by 10. Which takes us to five π₯ minus one
is less than negative 20π₯ plus 50, which is less than five π₯ plus 15. In that final part, the factor of
two in the numerator has canceled with the factor of two in the denominator. But we still have a factor of five
to multiply π₯ plus three by.

Now, we can subtract 50 from each
part of the inequality to give five π₯ minus 51 is less than negative 20π₯, which is
less than five π₯ minus 35. But at this stage, we notice that
we still have π₯ terms involved in each part of our inequality. We need to group them all together
in the center. And in this instance, because the
number of π₯βs on the far left of the inequality, thatβs five π₯, is the same as the
number of π₯βs on the far right of our inequality, we can do this by subtracting
five π₯ from each part. However, had the number of π₯βs not
been the same on the two sides of the inequality, we wouldnβt have been able to do
this. And weβd need to have used our
first method.

We now have negative 51 is less the
negative 25π₯, which is less than negative 35. Finally, we can divide each part of
our inequality by negative 25. But we must be really careful here
because we should remember that when we multiply or divide an inequality by a
negative number, we have to reverse the direction of the inequality sign. So each of our less than signs must
become greater than signs. We therefore have negative 51 over
negative 25 is greater than π₯, which is greater than negative 35 over negative
25. In each of our fractions, the
negatives in the numerators and denominators will cancel out. And the fraction 35 over 25 can be
simplified to seven over five.

Writing our inequality the other
way round so that the smallest number is on the left, we have seven over five is
less than π₯ is less than 51 over 25, which is the same as the solution we found
using our previous method. So we can convert our answer to
interval notation in exactly the same way.