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

Question

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

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 2

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 3

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 4

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

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.