Question

Subtract the rational expressions x22x−12−18x−6 . Simplify the answer.(1 point)
Responses

x2−362(x−6)
Start Fraction x squared minus 36 over 2 left parenthesis x minus 6 right parenthesis End Fraction

x2−18x−6
Start Fraction x squared minus 18 over x minus 6 End Fraction

x+62
Start Fraction x plus 6 over 2 End Fraction

x−62

Answer 1

What is the LCD in its factored form of the following rational expression subtraction?

x2+2xx2+7x−8−xx−1

(1 point)
Responses

(x−1)(x+8)
left parenthesis x minus 1 right parenthesis left parenthesis x plus 8 right parenthesis

(x−1)2(x+8)
left parenthesis x minus 1 right parenthesis squared left parenthesis x plus 8 right parenthesis

x(x+2)
x left parenthesis x plus 2 right parenthesis

(x−1)(x2+7x−8)

Answer 2

What value of x causes the rational equation 3x+5=1x+5+4
to be undefined?(1 point)

Answer 3

To subtract the rational expressions \(\frac{x^2}{2x-12} - \frac{18}{x-6}\), we first need to simplify the denominators and find a common denominator.

1. **Factor the denominators:**
- The first expression \(\frac{x^2}{2x-12}\) can be factored as:
\[
2x - 12 = 2(x - 6)
\]
- The second denominator \(x - 6\) is already factored.

2. **Write the expressions with common denominators:**
The common denominator of the two expressions will be \(2(x - 6)\).

Rewrite each expression with that common denominator:
\[
\frac{x^2}{2(x-6)} - \frac{18 \cdot 2}{2(x-6)} = \frac{x^2}{2(x-6)} - \frac{36}{2(x-6)}
\]

3. **Combine the two fractions:**
\[
\frac{x^2 - 36}{2(x - 6)}
\]

4. **Factor the numerator:**
The numerator \(x^2 - 36\) is a difference of squares and can be factored as:
\[
x^2 - 36 = (x - 6)(x + 6)
\]

5. **Substitute back into the expression:**
\[
\frac{(x - 6)(x + 6)}{2(x - 6)}
\]

Since \(x - 6\) is a common factor in both the numerator and the denominator, it can be canceled (as long as \(x \neq 6\)):
\[
\frac{x + 6}{2}
\]

Thus, the simplified answer is:
\[
\frac{x + 6}{2}
\]

Based on the provided options, the correct response is:
\[
\text{Start Fraction x plus 6 over 2 End Fraction}
\]

Answer 4

To find the least common denominator (LCD) of the rational expressions involved in the subtraction \(\frac{x^2 + 2x}{x^2 + 7x - 8} - \frac{x}{x - 1}\), we need to factor the denominators of the rational expressions.

### Step 1: Factor each denominator.

1. **First denominator:** \(x^2 + 7x - 8\)
- We can factor this quadratic expression. We need two numbers that multiply to \(-8\) and add to \(7\). These numbers are \(8\) and \(-1\).
- Thus, we can factor it as:
\[
x^2 + 7x - 8 = (x + 8)(x - 1)
\]

2. **Second denominator:** \(x - 1\)
- The second denominator is already in factored form.

### Step 2: Identify the unique factors.

Now that we have factored both denominators, we have:
- From the first expression: \((x + 8)(x - 1)\)
- From the second expression: \((x - 1)\)

### Step 3: Construct the LCD.

The LCD is the product of the highest power of each unique factor present in any of the denominators.
- The factor \((x - 1)\) appears in both denominators, but only once is needed for the LCD.
- The factor \((x + 8)\) appears only in the first denominator.

Thus, the LCD, in its factored form, is:
\[
(x - 1)(x + 8)
\]

### Conclusion

From the provided responses, the correct LCD in its factored form is:
\[
(x - 1)(x + 8)
\]

Answer 5

To determine the value of \( x \) that causes the rational equation

\[
\frac{3x + 5}{1} = \frac{x + 5}{1} + 4
\]

to be undefined, we need to look for any values of \( x \) that would make the denominator of any rational expression equal to zero. However, in this equation, all denominators are \( 1 \), which means they will never be zero and will not cause the equation to be undefined.

If we were considering a more typical rational equation with the form where one or more denominators contain variables, we would set those denominators equal to zero and solve for \( x \).

For example, if your equation looked like:

\[
\frac{3x + 5}{x - 1} = \frac{x + 5}{x + 2} + 4
\]

You would then set \( x - 1 = 0 \) or \( x + 2 = 0 \) to find the values that would make the equation undefined.

In your specific equation, however, since all terms are defined and we don't have any variables in the denominator, there isn't a value of \( x \) that will make the given rational equation undefined.

Thus, there are no values of \( x \) that cause the equation \( 3x + 5 = x + 5 + 4 \) to be undefined. If you have a specific form that includes denominators, please provide it for further examination.