Question

Add together the rational expressions and write your answer in simplest form.
3x-4/x^2 -9 + 2x-1/x+3=

Answer 1

To add the rational expressions \(\frac{3x-4}{x^2 - 9} + \frac{2x-1}{x+3}\), we need to first simplify the denominators and then find a common denominator.

### Step 1: Factor the denominators

The first denominator \(x^2 - 9\) is a difference of squares and can be factored as:

\[
x^2 - 9 = (x - 3)(x + 3)
\]

So now we have:

\[
\frac{3x-4}{(x - 3)(x + 3)} + \frac{2x - 1}{x + 3}
\]

### Step 2: Find a common denominator

The common denominator for these two fractions will be \((x - 3)(x + 3)\). We need to rewrite the second fraction to have this common denominator:

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

### Step 3: Rewrite the addition

Now, we can rewrite the addition as:

\[
\frac{3x - 4}{(x - 3)(x + 3)} + \frac{(2x - 1)(x - 3)}{(x - 3)(x + 3)} = \frac{3x - 4 + (2x - 1)(x - 3)}{(x - 3)(x + 3)}
\]

### Step 4: Expand the numerator

Next, we will expand the numerator \((2x - 1)(x - 3)\):

\[
(2x - 1)(x - 3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3
\]

Now, substitute back into the numerator:

\[
3x - 4 + (2x^2 - 7x + 3) = 2x^2 - 4x - 1
\]

### Step 5: Write the complete expression

Combine everything to get:

\[
\frac{2x^2 - 4x - 1}{(x - 3)(x + 3)}
\]

### Step 6: Simplify the numerator if possible

Now, let's check if the numerator \(2x^2 - 4x - 1\) can be factored. To factor \(2x^2 - 4x - 1\), we can use the quadratic formula; however, let's check if it factors easily:

The coefficients are \(a = 2\), \(b = -4\), and \(c = -1\).

The discriminant \(D = b^2 - 4ac = (-4)^2 - 4(2)(-1) = 16 + 8 = 24\), which is not a perfect square. So it won't factor nicely.

Thus, we keep it as is. There are no common factors in the numerator and denominator, so the expression is already in simplest form.

### Final Answer

\[
\frac{2x^2 - 4x - 1}{(x - 3)(x + 3)}
\]