### Video Transcript

Find, in the set of real numbers,
the solution set of the equation one-half 𝑥 minus one-quarter is equal to
two-fifths.

In this question, we are given an
equation in a variable 𝑥 and asked to find the solution set of this equation over
the set of real numbers.

To do this, we can begin by
recalling that the solution set to an equation over the set of real numbers is the
set of all real values that satisfy the given equation. In this case, we want to find the
set of all real values of 𝑥 that satisfy this equation.

In an equation, both sides of the
equation are equal. So we know that one-half 𝑥 minus
one-quarter is equal to two-fifths. If we apply the same operations to
both sides of the equation, then they will remain equal. We can use this idea to isolate 𝑥
on one side of the equation. To isolate 𝑥 on the left-hand side
of the equation, we can start by adding one-quarter to both sides of the
equation. On the left-hand side of the
equation, we have negative one-quarter plus one-quarter is zero. So we just have one-half 𝑥. And on the right-hand side of the
equation, we have two-fifths plus one-quarter.

We can simplify the right-hand side
of the equation by adding the fractions together. We note that the lowest common
multiple of the two numbers is 20. So we want to rewrite both the
fractions to have a denominator of 20. We can do this by multiplying the
numerator and denominator of the first fraction by four and the numerator and
denominator of the second fraction by five. We get eight over 20 plus five over
20, which we can rewrite by adding their numerators as 13 over 20.

To isolate 𝑥 on the right-hand
side of the equation, we can either divide both sides of the equation by one-half or
multiply both sides of the equation through by two. Of course, these are equivalent,
since we divide by a fraction by multiplying by its reciprocal and the reciprocal of
one-half is two. This gives us that 𝑥 is equal to
13 over 20 multiplied by two.

We can then cancel the shared
factor of two in the numerator and denominator to obtain 𝑥 equals 13 over 10. We can verify that this is a
solution to the equation by substituting 𝑥 equals 13 over 10 back into the given
equation and evaluating.

We are not done yet, however, since
we are asked to find the solution set of the equation over the set of real
numbers. This means that we need to check
that our solutions are real numbers and give our answer as a set. Of course, we know that 13 over 10
is a real number, since it is a rational number. And we showed that this was the
only solution to the equation. So the solution set of the given
equation over the set of real numbers is just the set containing 13 over 10.