Question

10 which expression is equivalent to \\(\frac{x + 7}{x^2 + 4x - 21} \div \frac{x + 5}{x^2 + 8x + 15}\\) when x is restricted so that the expressions are defined?\
a \\(\frac{x + 3}{x - 3}\\) \
c 1\
b \\(\frac{x - 3}{x + 3}\\) \
d 1\
\\(\bigcirc\\) b\
\\(\bigcirc\\) c\
\\(\bigcirc\\) a\
\\(\bigcirc\\) d

Explanation:

Step1: Factor Denominators

Factor \(x^2 + 4x - 21\): Find two numbers that multiply to \(-21\) and add to \(4\), which are \(7\) and \(-3\). So, \(x^2 + 4x - 21=(x + 7)(x - 3)\).
Factor \(x^2 + 8x + 15\): Find two numbers that multiply to \(15\) and add to \(8\), which are \(3\) and \(5\). So, \(x^2 + 8x + 15=(x + 3)(x + 5)\).

The expression becomes \(\frac{x + 7}{(x + 7)(x - 3)} \div \frac{x + 5}{(x + 3)(x + 5)}\).

Step2: Rewrite Division as Multiplication

Dividing by a fraction is multiplying by its reciprocal:
\(\frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\).

Step3: Cancel Common Factors

Cancel \(x + 7\) (since \(x eq -7\) for the expression to be defined), \(x + 5\) (since \(x eq -5\) for the expression to be defined):
\(\frac{\cancel{x + 7}}{\cancel{(x + 7)}(x - 3)} \times \frac{(x + 3)\cancel{(x + 5)}}{\cancel{x + 5}}=\frac{x + 3}{x - 3}\)? Wait, no—wait, let's check again. Wait, the original problem: Wait, maybe I misread the operation. Wait, the problem is \(\frac{x + 7}{x^2 + 4x - 21} \div \frac{x + 5}{x^2 + 8x + 15}\). Wait, no, maybe the operation is multiplication? Wait, no, the user's image: Wait, maybe a typo, but assuming it's division. Wait, no, let's re-express:

Wait, after factoring, \(\frac{x + 7}{(x + 7)(x - 3)} \div \frac{x + 5}{(x + 3)(x + 5)}=\frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\). Cancel \(x + 7\), \(x + 5\): \(\frac{1}{x - 3} \times (x + 3)=\frac{x + 3}{x - 3}\)? But that's not an option. Wait, maybe the operation is multiplication? Wait, maybe the original problem has a multiplication sign? Let me check the options. Wait, the options include \(\frac{x + 3}{x - 3}\) (option A), \(\frac{x - 3}{x + 3}\) (option B), 1 (C, D). Wait, maybe I made a mistake in factoring. Wait, \(x^2 + 4x - 21\): \(x^2 + 4x - 21=(x + 7)(x - 3)\) (correct, because \(7 \times (-3)=-21\), \(7 + (-3)=4\)). \(x^2 + 8x + 15=(x + 3)(x + 5)\) (correct, \(3 \times 5=15\), \(3 + 5=8\)).

Wait, maybe the operation is multiplication? Let's assume it's multiplication: \(\frac{x + 7}{x^2 + 4x - 21} \times \frac{x + 5}{x^2 + 8x + 15}\). Then:

\(\frac{x + 7}{(x + 7)(x - 3)} \times \frac{x + 5}{(x + 3)(x + 5)}=\frac{1}{x - 3} \times \frac{1}{x + 3}=\frac{1}{(x - 3)(x + 3)}\), which is not an option. So probably the operation is division. Wait, maybe the original problem has a typo, but looking at the options, option A is \(\frac{x + 3}{x - 3}\), which matches our result. Wait, but let's check again. Wait, the user's image: the problem is \(\frac{x + 7}{x^2 + 4x - 21} \div \frac{x + 5}{x^2 + 8x + 15}\). So division. So after canceling, we have \(\frac{x + 3}{x - 3}\), which is option A? Wait, no, option A is \(\frac{x + 3}{x - 3}\), option B is \(\frac{x - 3}{x + 3}\). Wait, maybe I flipped the reciprocal. Wait, division: \(\frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \times \frac{d}{c}\). So \(\frac{x + 7}{(x + 7)(x - 3)} \div \frac{x + 5}{(x + 3)(x + 5)}=\frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\). Cancel \(x + 7\), \(x + 5\): \(\frac{1}{x - 3} \times (x + 3)=\frac{x + 3}{x - 3}\), which is option A. But wait, the options: A is \(\frac{x + 3}{x - 3}\), B is \(\frac{x - 3}{x + 3}\), C and D are 1. Wait, maybe the problem was multiplication? No, division. So the correct answer should be A. But let's check again. Wait, maybe the original problem's denominators are different. Wait, \(x^2 + 4x - 21=(x + 7)(x - 3)\), correct. \(x^2 + 8x + 15=(x + 3)(x + 5)\), correct. So reciprocal of \(\frac{x + 5}{(x + 3)(x + 5)}\) is \(\frac{(x + 3)(x + 5)}{x + 5}\). So multiplying: \(\frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}=\frac{x + 3}{x - 3}\), which is option A.

Answer:

A. \(\frac{x + 3}{x - 3}\)