Add the following and simplify the sum if possible.x2−4x2−x−6+x+2x2−x−6\frac{x^2-4}{x^2-x-6}+\frac{x+2}{x^2-x-6}

x − 1 x − 3 \frac{x-1}{x-3}

Add the following and simplify the sum if possible. x 2 − 4 x 2 − x − 6 + x + 2 x 2 − x − 6 \frac{x^2-4}{x^2-x-6}+\frac{x+2}{x^2-x-6}

This problem asks us to add the following and simplify the sum if possible. So we have x squared minus four over x squared minus x minus six, and then we have all of this plus x plus two over x squared minus x minus six. Now what we need to do for rational expressions when we're adding or subtracting is look for common denominators. Now I can see we have the same quadratic, the same denominator, so that means we can just combine the numerators. So we have x squared minus four for the first rational expression, then we have x plus two for the second, then all of this is gonna be over the common denominator, x squared minus x minus six. Now notice in the numerator, we don't have any values, we don't have any numbers that are being multiplied on the outside of these parentheses, nor do we have any negative signs or subtractions. So this whole thing is just gonna be x squared minus four plus x plus two, and then the denominator is going to remain x squared minus x minus six. Now what can I do from here? Well what I can do is combine things and simplify in the numerator. Now I notice we have an x squared, that's one term, and then we have an x, that's another term. Then I notice we have a negative four and positive two. Negative four plus two will give us negative two. So we're gonna end up with x squared plus x, and then the negative two will just be a minus two that we can write like that. And we're gonna have x squared minus x minus six. Now it may be tempting to think we can stop here, but we need to simplify this as far as it will go. But how can we simplify farther from this point? Well, notice I end up with two quadratics in the numerator and the denominator. Now I'm gonna write these separately. So we have x squared plus x minus two that's happening in the numerator, and then we have x squared minus x minus six in the denominator. And what I need to do is factor both of these. So I need to think about a pair of numbers for the numerator, the the top quadratic, that is going to multiply to give me negative two, and it's going to add to give me positive one. There's an implicit one that we have right here. So what's going to multiply to give me negative two and add to give me positive one? Well, if we think about through all these pairs of numbers, that's going to be two and negative one. Right? Because two times negative one gives us negative two, and then two plus negative one gives us positive one. So we've gone ahead and found this pair of numbers. Now what I'm going to do is also find the pair of numbers for the other quadratic, x squared minus x minus six. I need to find a pair of numbers that will multiply to give me negative six, and they'll add to give me negative one. What pair of numbers is that going to be? Well if I think about it, that's going to be two and negative three. And we can just check two times negative three, that's negative six. Two plus negative three, that's negative one. So we have found this pair of numbers. So now that we found each of these pairs, we can continue on by factoring what we have. So we're going to keep the, we're going to have this fraction bar here and then we're going to have in the numerator, we're going to have the factored version of this. We found the pair of numbers was two and negative one, so it's going to be x plus two, but positive two, and x minus one, the negative one. Now in the denominator, we got positive two and negative three. So we're going to have x plus two and then x minus three, that's what happens when we factor. And notice here that the x plus two cancels entirely. So all we end up with is x minus one over x minus three. And this cannot simplify down any farther, so this right here would be our final answer. So that's how you can solve these problems with rational expressions. As you can see, it's oftentimes the simplifying step that takes a long time, and you will often have to do multiple steps to get to your fully simplified answer. So make sure whatever answer you end up with, it's something that cannot be factored or broken down any farther and something that will not simplify any farther. So So I hope you found this video helpful, and let's move on.

Add the following and simplify the sum if possible. x 2 − 4 x 2 − x − 6 + x + 2 x 2 − x − 6 \frac{x^2-4}{x^2-x-6}+\frac{x+2}{x^2-x-6}

x − 1 x − 3 \frac{x-1}{x-3}

8. Rational Expressions and Equations

Adding and Subtracting Rational Expressions with Common Denominators

Beginning Algebra

8. Rational Expressions and Equations

Adding and Subtracting Rational Expressions with Common Denominators

Struggling with Beginning Algebra?

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Multiple Choice

Add the following and simplify the sum if possible.
$\frac{x^{2} - 4}{x^{2} - x - 6} + \frac{x + 2}{x^{2} - x - 6}$

$\frac{x + 2}{x - 3}$

$\frac{x + 2}{x + 3}$

$\frac{x - 1}{x - 3}$

$\frac{x - 1}{x + 2}$

Verified step by step guidance

Identify the common denominator in the given expression: both fractions have the denominator $x^2 - x - 6$.

Factor the quadratic denominator $x^2 - x - 6$ into two binomials: $x^2 - x - 6 = (x - 3)(x + 2)$.

Rewrite each fraction with the factored denominator: $\frac{x^2 - 4}{(x - 3)(x + 2)} + \frac{x + 2}{(x - 3)(x + 2)}$.

Factor the numerator $x^2 - 4$ as a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$, so the first fraction becomes $\frac{(x - 2)(x + 2)}{(x - 3)(x + 2)}$.

Since both fractions have the same denominator, combine the numerators over the common denominator: $\frac{(x - 2)(x + 2) + (x + 2)}{(x - 3)(x + 2)}$. Then factor and simplify the numerator if possible.

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Struggling with Beginning Algebra?

Add the following and simplify the sum if possible.x2−4x2−x−6+x+2x2−x−6\(\frac{x^2-4}{x^2-x-6}\)+\(\frac{x+2}{x^2-x-6}\)

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Add the following and simplify the sum if possible.
$\frac{x^{2} - 4}{x^{2} - x - 6} + \frac{x + 2}{x^{2} - x - 6}$