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Question Video: Finding the Solution Set to a Linear Equation Mathematics

Find, in ℝ, the solution set of the equation (1/2 𝑥) − (1/4) = 2/5.

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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.

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