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

Question

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

To solve the problem I first dissembled the equation on the denominator $ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$

I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this?

Answer 1

$$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{(x+3)\color{red}{(x+1)}-(x-2)\color{red}{(x-1)}}{(x-1)(x+1)^2}$$

Answer 2

$a=(x^2-1)=(x+1)(x-1)$

$b=(x^2+2x+1)=(x+1)^2=(x+1)(x+1)$

So, we have:

$$\frac{(x+3)}{a}-\frac{(x-2)}{b}$$

$$\frac{b(x+3)-a(x-2)}{ab}$$

$ab=(x+1)^3(x-1)$

$b(x+3)=(x+3)(x+1)(x+1)$

$a(x-2)=(x-2)(x+1)(x-1)$

Then,

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

$$\frac{{(x+1)}[(x+3)(x+1)-(x-2)(x-1)]}{{(x+1)^{3}}(x-1)}$$

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

Expanding:

$$\frac{7x+1}{x^3+x^2-x-1}$$

Answer 3

Let's start here and simplify the problem.

$$\frac{(x+3)}{(x-1)(x+1)} - \frac{(x-2)}{(x+1)(x+1)}$$

I'm going to call $(x+1)=a$ and $(x-1)=b$

$$\frac{(x+3)}{(b)(a)} - \frac{(x-2)}{(a)(a)}$$

So, on the right side we have (a)(a) in the denominator, and only one in the denominator on the right. So we have to multiply the term on the right by $\frac{a}{a}$.

The term on the left has (b) in the denominator, while the one on the right doesn't. So we're going to to multiply the one on the right by $\frac{b}{b}$.

Let's take those steps, and we get:

$$\frac{(x+3)(a)}{(b)(a)(a)} - \frac{(x-2)(b)}{(a)(a)(b)}$$

See how both denominators are identical!? Great! Now we can put both of the numerators on top of the denominator we worked so hard to get.

$$\frac{(x+3)(a)-(x-2)(b)}{(b)(a)(a)}$$

Now all you have to do is plug back in $a=(x+1)$ and $b=(x-1)$, open up the brackets and collect like terms.

Answer 4

The common factor can be taken to front as coefficient in the numerator as well as in the denominator.

$$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{1}{(x+1)} \cdot [\frac{x+3}{(x-1)} - \frac{x-2}{(x+1)}] $$

$$=\frac{1}{(x+1)} \cdot \frac{(x^2+4 x+ 3) - (x^2- 3 x + 2)} {(x^2-1)} $$

$$=\frac{1}{(x+1)} \frac{(7 x + 1)} {(x^2-1)}. $$