Question

Divide x over x+3 divided by x over x-4

Answer 1

Ok bestie im gonna need you to size the explanation down for me thank you

Answer 2

I need like 4 senteces at most please

Answer 3

subract x over x+1 minus 8 over x-7

Answer 4

2 things
1. I need the explantion trimmed down to like 4 sentences
2. can you do the expalnaton where any time theres fractions you say like how I said it in my question with saying x over x-1
Also can you not add any double parenthesis or {} these things or backwards slash

Answer 5

I said remove the double parenthesis

Answer 6

Bestie thats not an answer but theres one similar its
x^2-15x-8 over x^2-6x-7

Answer 7

To divide the expression \(\frac{x}{x+3}\) by \(\frac{x}{x-4}\), you can follow these steps:

1. **Write the division as a multiplication by the reciprocal**:
\[
\frac{x}{x+3} \div \frac{x}{x-4} = \frac{x}{x+3} \times \frac{x-4}{x}
\]

2. **Multiply the fractions**:
\[
\frac{x \cdot (x-4)}{(x+3) \cdot x}
\]

3. **Cancel out the common factors** (if \(x \neq 0\)):
\[
\frac{x-4}{x+3}
\]

So the simplified result is:
\[
\frac{x-4}{x+3}
\]

This assumes \(x \neq 0\), \(x \neq -3\), and \(x \neq 4\) to avoid division by zero.

Answer 8

Sure! Here’s the concise version:

To divide \(\frac{x}{x+3}\) by \(\frac{x}{x-4}\):

1. Rewrite it as a multiplication:
\[
\frac{x}{x+3} \times \frac{x-4}{x}
\]

2. The \(x\) cancels out (assuming \(x \neq 0\)):
\[
\frac{x-4}{x+3}
\]

So, the answer is:
\[
\frac{x-4}{x+3}
\]

(Note: \(x \neq 0, -3, 4\) to avoid division by zero.)

Answer 9

To divide \(\frac{x}{x+3}\) by \(\frac{x}{x-4}\), rewrite it as \(\frac{x}{x+3} \times \frac{x-4}{x}\). The \(x\) cancels out, giving you \(\frac{x-4}{x+3}\). The result is \(\frac{x-4}{x+3}\), assuming \(x \neq 0, -3, 4\).

Answer 10

To subtract \(\frac{x}{x+1}\) and \(\frac{8}{x-7}\), follow these steps:

1. Find a common denominator, which is \((x+1)(x-7)\).
2. Rewrite the fractions:
\[
\frac{x}{x+1} = \frac{x(x-7)}{(x+1)(x-7)}
\]
\[
\frac{8}{x-7} = \frac{8(x+1)}{(x+1)(x-7)}
\]

3. Now subtract:
\[
\frac{x(x-7) - 8(x+1)}{(x+1)(x-7)}
\]

4. Simplify the numerator:
\[
x^2 - 7x - 8x - 8 = x^2 - 15x - 8
\]

5. The final result is:
\[
\frac{x^2 - 15x - 8}{(x+1)(x-7)}
\]

So,
\[
\frac{x^2 - 15x - 8}{(x+1)(x-7)}
\] is the simplified result.

Answer 11

To subtract x over x+1 and 8 over x-7, find a common denominator, which is (x+1)(x-7). Rewrite the fractions as (x(x-7) over (x+1)(x-7)) and (8(x+1) over (x+1)(x-7)). Now subtract: (x(x-7) - 8(x+1)) over (x+1)(x-7). Simplifying gives you (x^2 - 15x - 8) over (x+1)(x-7).

Answer 12

Got it! Here’s the explanation without double parentheses:

To subtract x over x+1 and 8 over x-7, find a common denominator, which is x+1 times x-7. Rewrite the fractions as x times (x-7) over that denominator and 8 times (x+1) over the same denominator. Now subtract: (x times (x-7) minus 8 times (x+1)) over x+1 times x-7. Simplifying gives you (x^2 - 15x - 8) over x+1 times x-7.

Answer 13

I apologize for the confusion! Let's adjust the final expression accordingly:

To subtract x over x+1 and 8 over x-7, find a common denominator, which is x+1 times x-7. Rewrite the fractions as x times (x-7) over that denominator and 8 times (x+1) over the same denominator. Now subtract: (x times (x-7) minus 8 times (x+1)) over x+1 times x-7. Simplifying gives you (x^2 - 15x - 8) over (x^2 - 6x - 7).