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 \\)
\\( \circ \\) a
\\( \circ \\) d
\\( \circ \\) b
\\( \circ \\) c

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, \(x^2 + 8x + 15=(x + 3)(x + 5)\).

Step2: Rewrite the Expression

Substitute the factored forms into the original expression:
\(\frac{x + 7}{(x + 7)(x - 3)} \cdot \frac{x + 5}{(x + 3)(x + 5)}\)

Step3: Cancel Common Factors

Cancel \((x + 7)\) from the numerator and denominator of the first fraction, and cancel \((x + 5)\) from the numerator and denominator of the second fraction:
\(\frac{1}{x - 3} \cdot \frac{1}{x + 3}\)? Wait, no—wait, the original is multiplication? Wait, no, the problem is \(\frac{x + 7}{x^2 + 4x - 21} \cdot \frac{x + 5}{x^2 + 8x + 15}\)? Wait, no, maybe it's division? Wait, the user's image: let me recheck. Wait, the original expression: maybe it's \(\frac{x + 7}{x^2 + 4x - 21} \div \frac{x + 5}{x^2 + 8x + 15}\)? Because the options don't match multiplication. Oh, likely a typo, division. Let's assume division (since multiplication would not give the options). So, dividing by a fraction is multiplying by its reciprocal.

So, correct step: If it's division, \(\frac{x + 7}{x^2 + 4x - 21} \div \frac{x + 5}{x^2 + 8x + 15} = \frac{x + 7}{(x + 7)(x - 3)} \cdot \frac{(x + 3)(x + 5)}{x + 5}\).

Now cancel: \((x + 7)\) cancels, \((x + 5)\) cancels. So we get \(\frac{1}{x - 3} \cdot (x + 3) = \frac{x + 3}{x - 3}\)? Wait, no: Wait, \(\frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\). Cancel \((x + 7)\) and \((x + 5)\): left with \(\frac{1}{x - 3} \times (x + 3) = \frac{x + 3}{x - 3}\), which is option A. Wait, but let's re-express:

Original denominators:
\(x^2 + 4x - 21 = (x + 7)(x - 3)\)
\(x^2 + 8x + 15 = (x + 3)(x + 5)\)

So, if the operation is division: \(\frac{x + 7}{(x + 7)(x - 3)} \div \frac{x + 5}{(x + 3)(x + 5)} = \frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\)

Cancel \((x + 7)\) and \((x + 5)\):
\(\frac{1}{x - 3} \times (x + 3) = \frac{x + 3}{x - 3}\), which is option A.

Wait, but let's check the options. Option A is \(\frac{x + 3}{x - 3}\), option B is \(\frac{x - 3}{x + 3}\), C is 1, D is -1. So if the operation is division, then A is correct.

So, step-by-step:

Step1: Factor Denominators

\(x^2 + 4x - 21 = (x + 7)(x - 3)\) (since \(7 \times -3 = -21\), \(7 + (-3) = 4\))
\(x^2 + 8x + 15 = (x + 3)(x + 5)\) (since \(3 \times 5 = 15\), \(3 + 5 = 8\))

Step2: Rewrite Division as Multiplication (Reciprocal)

\(\frac{x + 7}{(x + 7)(x - 3)} \div \frac{x + 5}{(x + 3)(x + 5)} = \frac{x + 7}{(x + 7)(x - 3)} \times \frac{(x + 3)(x + 5)}{x + 5}\)

Step3: Cancel Common Factors

Cancel \((x + 7)\) (numerator of first fraction and denominator of first fraction), cancel \((x + 5)\) (numerator of second fraction and denominator of second fraction):
\(\frac{1}{x - 3} \times (x + 3) = \frac{x + 3}{x - 3}\)

Answer:

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