Add the following expressions and simplify if possible:2x2−1+3x+1\frac{2}{x^2-1}+\frac{3}{x+1}

3 x − 1 ( x + 1 ) ( x − 1 ) \frac{3x-1}{\left(x+1\right)\left(x-1\right)} ( x + 1 ) ( x − 1 ) 3 x − 1

Add the following expressions and simplify if possible: 2 x 2 − 1 + 3 x + 1 \frac{2}{x^2-1}+\frac{3}{x+1}

This problem asks us to add the following expressions and simplify if possible. So notice I have a two over x squared minus one, and I'm adding that to a three over x plus one. When dealing with rational expressions where we need to add them, it's going to be important that we get common denominators. And one of the most straightforward ways, or I should say straightforward first steps to do, when dealing with rational expressions is to factor everything that can be factored. Now looking at this first fraction, or this first rational expression I see, I can factor this denominator. We have x squared minus one, which is gonna be the same thing as x squared minus one squared. Now the reason I'm writing one as one squared is cause notice we then get a difference in squares, which we have a shortcut to factor. So x squared minus one squared is the same thing as x plus one times x minus one. Now I'm going to go ahead and add this to the second fraction, which we have this three over x plus one. Now what I need to do is figure out what my least common denominator is, and the least common denominator is going to be the product of all unique prime factors that we see in the denominators. Looking at this rational expression here, I see that we have an x plus one. We also have an x plus one over here, so I only need to write the x plus one once. That's a unique prime factor. I can also see we have x minus one. That only shows up in one place, so it is a unique prime factor, x minus one. So x plus one times x minus one is the LCD. One of our rational expressions has this denominator already, but the other one does not. So what that means is we need to get this LCD in the second rational expression. So to do that, I'm gonna multiply the numerator and denominator by the missing factor. I can see clearly we have an x plus one. What we don't have is an x minus one. So you need to multiply the denominator by x minus one, but also multiply the numerator by x minus one. Now in the numerator, we're going to have everything added together, so it's gonna be two plus three times x minus one. And in the denominator, we're just gonna have the common denominator we figured out, x plus one times x minus one, the LCD. Now at this point, all I need to do is simplify what I see in the numerator. So we'll have two plus, and then distributing this three into the parentheses, we'll get three x minus three all over x plus one times x minus one. And then in this case, I can actually combine this two and this negative three to give me negative one, so we'll get three x plus negative one or minus one all over x plus one times x minus one. There's not a way to simplify things farther from here, so this would be our final answer to this problem. Hope you found this helpful.

Add the following expressions and simplify if possible: 2 x 2 − 1 + 3 x + 1 \frac{2}{x^2-1}+\frac{3}{x+1}

3 x − 1 ( x + 1 ) ( x − 1 ) \frac{3x-1}{\left(x+1\right)\left(x-1\right)} ( x + 1 ) ( x − 1 ) 3 x − 1

3 x + 1 ( x + 1 ) ( x − 1 ) \frac{3x+1}{\left(x+1\right)\left(x-1\right)} ( x + 1 ) ( x − 1 ) 3 x + 1

3 x − 1 x 2 − 1 \frac{3x-1}{x^2-1} x 2 − 1 3 x − 1

3 x − 1 x 2 + 1 \frac{3x-1}{x^2+1} x 2 + 1 3 x − 1

8. Rational Expressions and Functions

Adding and Subtracting Rational Expressions with Different Denominators

Intermediate Algebra

8. Rational Expressions and Functions

Adding and Subtracting Rational Expressions with Different Denominators

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Multiple Choice

Add the following expressions and simplify if possible:
$\frac{2}{x^{2} - 1} + \frac{3}{x + 1}$

$\frac{3x+1}{\left(x+1\right)\left(x-1\right)}$

$\frac{3x-1}{\left(x+1\right)\left(x-1\right)}$

$\frac{3x-1}{x^2-1}$

$\frac{3x-1}{x^2+1}$

Verified step by step guidance

Identify the expressions to be added: $\frac{2}{x^2 - 1} + \frac{3}{x + 1}$.

Recognize that the denominator $x^2 - 1$ can be factored using the difference of squares formula: $x^2 - 1 = (x + 1)(x - 1)$.

Rewrite the first fraction with the factored denominator: $\frac{2}{(x + 1)(x - 1)}$.

Find a common denominator for both fractions, which is $(x + 1)(x - 1)$, and rewrite the second fraction accordingly: $\frac{3}{x + 1} = \frac{3(x - 1)}{(x + 1)(x - 1)}$.

Add the numerators over the common denominator: $\frac{2 + 3(x - 1)}{(x + 1)(x - 1)}$, then simplify the numerator by distributing and combining like terms.

6:23m

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