Question

divide and simplify.\\(
\frac{5x + 5}{2x} \div \frac{x^2 - 9x}{x^2 - 8x - 9}\\)
\\(\square\\), \\(x \
eq -1, 0, 9\\)
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Explanation:

Step1: Rewrite division as multiplication

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. So, $\frac{5x + 5}{2x} \div \frac{x^2 - 9x}{x^2 - 8x - 9}$ becomes $\frac{5x + 5}{2x} \times \frac{x^2 - 8x - 9}{x^2 - 9x}$.

Step2: Factor each expression

- Factor $5x + 5$: We can factor out a 5, so $5x + 5 = 5(x + 1)$.
- Factor $x^2 - 8x - 9$: We need two numbers that multiply to -9 and add to -8. Those numbers are -9 and 1. So, $x^2 - 8x - 9=(x - 9)(x + 1)$.
- Factor $x^2 - 9x$: We can factor out an $x$, so $x^2 - 9x = x(x - 9)$.

Substituting these factored forms back into the expression, we get: $\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$ (Wait, no, wait. Wait, the second fraction's numerator is $x^2 - 8x - 9=(x - 9)(x + 1)$ and denominator is $x^2 - 9x = x(x - 9)$. So the expression is $\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$? Wait, no, no, I made a mistake. Wait, the reciprocal of $\frac{x^2 - 9x}{x^2 - 8x - 9}$ is $\frac{x^2 - 8x - 9}{x^2 - 9x}$, so the correct factoring is:

$5x + 5 = 5(x + 1)$

$x^2 - 8x - 9=(x - 9)(x + 1)$ (since -9*1=-9 and -9+1=-8)

$x^2 - 9x = x(x - 9)$

So substituting back, we have:

$\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$? Wait, no, wait, the second fraction's numerator is $x^2 - 8x - 9=(x - 9)(x + 1)$ and denominator is $x^2 - 9x = x(x - 9)$. So the multiplication is $\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$? Wait, no, that can't be. Wait, no, the original division is $\frac{5x + 5}{2x} \div \frac{x^2 - 9x}{x^2 - 8x - 9}$, so reciprocal of the second fraction is $\frac{x^2 - 8x - 9}{x^2 - 9x}$, so the expression is $\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$? Wait, no, I think I messed up the factoring of $x^2 - 8x - 9$. Wait, let's do it again. $x^2 - 8x - 9$. Let's use the quadratic formula: $x=\frac{8\pm\sqrt{64 + 36}}{2}=\frac{8\pm\sqrt{100}}{2}=\frac{8\pm10}{2}$. So $x=\frac{18}{2}=9$ or $x=\frac{-2}{2}=-1$. So yes, $x^2 - 8x - 9=(x - 9)(x + 1)$. And $x^2 - 9x = x(x - 9)$. And $5x + 5 = 5(x + 1)$. So now, substituting back, the expression is:

$\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$? Wait, no, wait, the second fraction's numerator is $x^2 - 8x - 9=(x - 9)(x + 1)$ and denominator is $x^2 - 9x = x(x - 9)$. So when we multiply, we have:

Numerator: $5(x + 1) \times (x - 9)(x + 1)$

Denominator: $2x \times x(x - 9)$

But now we can cancel out common factors. We have $(x - 9)$ in the numerator and denominator, so we can cancel that. We also have $(x + 1)$? Wait, no, wait, no, wait, I think I made a mistake in the factoring. Wait, no, the first fraction's numerator is $5(x + 1)$, the second fraction's numerator is $(x - 9)(x + 1)$, the first fraction's denominator is $2x$, the second fraction's denominator is $x(x - 9)$. So when we multiply, the $(x - 9)$ terms cancel, and we have:

$\frac{5(x + 1) \times (x + 1)}{2x \times x}$? Wait, no, that's not right. Wait, no, the second fraction's numerator is $(x - 9)(x + 1)$ and denominator is $x(x - 9)$, so when we multiply by the reciprocal, it's $\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$. Wait, no, that would be $\frac{5(x + 1)(x - 9)(x + 1)}{2x \times x(x - 9)}$. Then we can cancel $(x - 9)$ from numerator and denominator, leaving $\frac{5(x + 1)^2}{2x^2}$. But that seems wrong. Wait, no, wait, maybe I factored the wrong quadratic. Wait, no, let's check again.

Wait, the original problem is $\frac{5x + 5}{2x} \div \frac{x^2 - 9x}{x^2 - 8x - 9}$. So reciprocal of the second fraction is $\frac{x^2 - 8x - 9}{x^2 - 9x}$. So let's factor each part:

- $5x + 5 = 5(x + 1)$ (correct)
- $x^2 - 8x - 9$: Let's factor this. We need two numbers that multiply to -9 and add to -8. So -9 and +1. So $(x - 9)(x + 1)$ (correct, because $(x - 9)(x + 1)=x^2 + x - 9x - 9 = x^2 - 8x - 9$)
- $x^2 - 9x = x(x - 9)$ (correct)

So now, substituting into the multiplication:

$\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$

Now, we can cancel the $(x - 9)$ terms:

$\frac{5(x + 1)}{2x} \times \frac{(x + 1)}{x}$

Because the $(x - 9)$ in the numerator and denominator cancels. Then we multiply the numerators and denominators:

Numerator: $5(x + 1)(x + 1) = 5(x + 1)^2$

Denominator: $2x \times x = 2x^2$

Wait, but that would be $\frac{5(x + 1)^2}{2x^2}$, but that seems incorrect. Wait, maybe I made a mistake in the reciprocal. Wait, no, dividing by a fraction is multiplying by its reciprocal. So $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$. So in this case, $a = 5x + 5$, $b = 2x$, $c = x^2 - 9x$, $d = x^2 - 8x - 9$. So it's $\frac{5x + 5}{2x} \times \frac{x^2 - 8x - 9}{x^2 - 9x}$. So that's correct. Then factoring:

$5x + 5 = 5(x + 1)$

$x^2 - 8x - 9 = (x - 9)(x + 1)$

$x^2 - 9x = x(x - 9)$

So substituting, we have:

$\frac{5(x + 1)}{2x} \times \frac{(x - 9)(x + 1)}{x(x - 9)}$

Now, cancel $(x - 9)$ from numerator and denominator:

$\frac{5(x + 1)}{2x} \times \frac{(x + 1)}{x}$

Now, multiply the numerators: $5(x + 1)(x + 1) = 5(x + 1)^2$

Multiply the denominators: $2x \times x = 2x^2$

Wait, but that seems to give $\frac{5(x + 1)^2}{2x^2}$, but let's check with a value. Let's pick $x = 1$ (which is not -1, 0, or 9). Original expression:

First fraction: $\frac{5(1) + 5}{2(1)} = \frac{10}{2} = 5$

Second fraction: $\frac{1^2 - 9(1)}{1^2 - 8(1) - 9} = \frac{1 - 9}{1 - 8 - 9} = \frac{-8}{-16} = \frac{1}{2}$

So dividing 5 by $\frac{1}{2}$ is 10. Now, let's plug $x = 1$ into our simplified expression: $\frac{5(1 + 1)^2}{2(1)^2} = \frac{5(4)}{2} = 10$. Okay, that works. Wait, but let's check another value. Let's take $x = 2$.

Original expression:

First fraction: $\frac{5(2) + 5}{2(2)} = \frac{15}{4}$

Second fraction: $\frac{2^2 - 9(2)}{2^2 - 8(2) - 9} = \frac{4 - 18}{4 - 16 - 9} = \frac{-14}{-21} = \frac{2}{3}$

Dividing $\frac{15}{4}$ by $\frac{2}{3}$ is $\frac{15}{4} \times \frac{3}{2} = \frac{45}{8}$. Now, our simplified expression: $\frac{5(2 + 1)^2}{2(2)^2} = \frac{5(9)}{8} = \frac{45}{8}$. Okay, that works too. So maybe that's correct. Wait, but let's check the factoring again. Wait, the second fraction's numerator is $x^2 - 9x$ and denominator is $x^2 - 8x - 9$, so reciprocal is $\frac{x^2 - 8x - 9}{x^2 - 9x}$. So factoring $x^2 - 8x - 9$ as $(x - 9)(x + 1)$ and $x^2 - 9x$ as $x(x - 9)$. Then canceling $(x - 9)$ gives $\frac{(x + 1)}{x}$ in the second fraction's part. Then multiplying by $\frac{5(x + 1)}{2x}$ gives $\frac{5(x + 1)^2}{2x^2}$. So that seems correct.

Wait, but let's check the initial factoring again. Wait, $x^2 - 8x - 9$: when we factor, it's (x - 9)(x + 1) because -9*1=-9 and -9+1=-8. Correct. $x^2 - 9x = x(x - 9)$. Correct. $5x + 5 = 5(x + 1)$. Correct. So the cancellation of (x - 9) is valid because $x
eq 9$, so we don't have division by zero. Then we have $\frac{5(x + 1)}{2x} \times \frac{(x + 1)}{x} = \frac{5(x + 1)^2}{2x^2}$. So that's the simplified form.

Answer:

$\frac{5(x + 1)^2}{2x^2}$ (or expanded, $\frac{5(x^2 + 2x + 1)}{2x^2} = \frac{5x^2 + 10x + 5}{2x^2}$, but the factored form is simpler)