Great catch! yes x=-5 is an extraneous solution and should have been labeled as so. Looks like I was quick to finish the video and did not slow down to check. Thanks again.
@@jesuswalk1012 No problem. My advice is that, whenever you are solving rational equations, you should put every term on one side, combine fractions, and factorize the numerators and denominators. Once you cancel the common factors, the zeros of the numerator become exactly the only solutions to the equation, with no extraneous solutions, and those 0s are given for free if you complete all three steps, because the factorization gives you the solutions, and you do not need to do any algebra, because once the common factors disappear, the denominator becomes actually irrelevant.
You substitute the solutions into the equation, and those that lead to contradictions ar extraneous. However, I do want to point out that it is entirely possible to solve rational equations without ever introducing extraneous solutions, so there is no need to actually check for them if you use the appropriate solving method. What you do is 1. put all the fractions on side of the equation, so the other side is 0. 2. You rewrite the side with the sum of fractions as a single fraction. 3. You look for the common factor between the numerator and denominator, and you divide them out. 4. Let the numerator be equal to 0. 5. Solve the resulting equation, which is just a polynomial equation. This will give you all the solutions AND no extraneous solutions, and this works for ANY rational equation.
For example, to solve 2·x/(x + 5) - (x^2 - x - 10)/(x^2 + 8·x + 15) = 3/(x + 3), you add -3/(x + 3) to both parts of the equation. This results in the equivalent equation 2·x/(x + 5) - 3/(x + 3) - (x^2 - x - 10)/(x^2 + 8·x + 15) = 0. On the left side, there is a sum of fractions, and on the right side, there is 0, so the first step is complete. The next step is to rewrite the left side as a single fraction. How do you do this? Notice that x^2 + 8·x + 15 = (x + 5)·(x + 3). So 2·x/(x + 5) - 3/(x + 3) - (x^2 - x - 10)/(x^2 + 8·x + 15) = 2·x/(x + 5) - 3/(x + 3) - (x^2 - x - 10)/[(x + 5)·(x + 3)] = 2·x·(x + 3)/[(x + 5)·(x + 3)] - 3·(x + 5)/[(x + 5)·(x + 3)] - (x^2 - x - 10)/[(x + 5)·(x + 3)] = [2·x·(x + 3) - 3·(x + 5) - (x^2 - x - 10)]/[(x + 5)·(x + 3)] = (x^2 + 4·x - 5)/[(x + 5)·(x + 3)]. Therefore, the equation to solve is (x^2 + 4·x - 5)/[(x + 5)·(x + 3)] = 0. The second step is complete, and the next step is to find the common factors of the numerator and denominator. Notice that x^2 + 4·x - 5 = (x + 5)·(x - 1), so (x + 5)·(x - 1)/[(x + 5)·(x + 3)] = 0. Since they have a common factor of x + 5, you can simply divide it out, because (x + 5)/(x + 5) is the polynomial 1 for not x = 5. Therefore, (x - 1)/(x + 3) = 0. The fourth step is to multiply both parts of the equation by x + 3, which creates an equivalent equation, because 1/(x + 3) is never equal to 0. This creates the equation, equivalent to the original, x - 1 = 0. The final step is to solve this equation, which is trivial: x = 1. Done. Notice how this method has a short of amount of simple steps that are applicable to every equation, and built into the method is the discarding of extraneous solutions, so that you never actually obtain extraneous solutions you have to check for. How nice!
Once again you, in your zeal, made a mistake. The moan from your students was an obvious indicator that you are losing credibility. It was painful to hear and watch. Slow down. I am an ex Math teacher/College lecturer.
