Question Video: Finding the Solution Set of a Linear Equation | Nagwa Question Video: Finding the Solution Set of a Linear Equation | Nagwa

# Question Video: Finding the Solution Set of a Linear Equation Mathematics • Second Year of Preparatory School

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Find, in ℝ, the solution set of the equation 5 − √(2)𝑥 = 3.

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### 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.

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