Video Transcript
Find, in the set of real numbers,
the solution set of the equation five minus root two 𝑥 is equal to three.
In this question, we are given a
linear equation in one variable and asked to find its solution set over the set of
real numbers. We can begin by recalling that this
is the set of all real values of 𝑥 that satisfy the equation. This means that we want to isolate
𝑥 on one side of the equation to determine the possible values of 𝑥 that satisfy
the equation.
We can begin by subtracting five
from both sides of the equation. On the left-hand side of the
equation, we have five minus five equals zero. And on the right-hand side, we have
three minus five is negative two. So, we are left with negative root
two 𝑥 is equal to negative two. We can then isolate 𝑥 by dividing
both sides of the equation by negative root two. This gives us that 𝑥 is equal to
negative two over negative root two. We can simplify the expression for
𝑥 by canceling the shared factor of negative one to get two over root two.
We then want to rationalize the
denominator. We do this by multiplying both the
numerator and denominator of the fraction by root two. This does not change the value
since it is equivalent to multiplying by one. We then obtain two root two over
root two squared, which is equal to two root two over two. We can then cancel the shared
factor of two in the numerator and denominator to get root two.
We are not done yet, since we are
asked to find the solution set of the equation over the set of real numbers. We have shown that root two is the
only real solution of the equation. So, our answer is the set
containing only the square root of two.
