6.3 complex fractions
directions: use either method to simplify each complex fraction. show all work.
1.
$\dfrac{\dfrac{5}{x}}{\dfrac{9}{x + 2}}$
2.
$\dfrac{6 + \dfrac{1}{x}}{7 - \dfrac{3}{x}}$
3.
$\dfrac{\dfrac{8}{x + 7}}{\dfrac{24}{x^2 - 49}}$
4.
$\dfrac{4 + \dfrac{2}{x}}{\dfrac{x}{3} + \dfrac{1}{6}}$
5.
$\dfrac{x + \dfrac{x - 3}{3}}{\dfrac{4}{3} + \dfrac{2}{3x}}$

Question

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Recall that dividing by a fraction is multiplying by its reciprocal. So, $\frac{\frac{5}{x}}{\frac{9}{x + 2}}=\frac{5}{x}\times\frac{x + 2}{9}$
## Step2: Multiply the numerators and denominators: $\frac{5(x + 2)}{9x}$
# Answer: $\frac{5(x + 2)}{9x}$

### Problem 2:
# Explanation:
## Step1: First, simplify the numerator and denominator separately. For the numerator: $6+\frac{1}{x}=\frac{6x + 1}{x}$. For the denominator: $7-\frac{3}{x}=\frac{7x-3}{x}$
## Step2: Now, divide the two fractions: $\frac{\frac{6x + 1}{x}}{\frac{7x-3}{x}}=\frac{6x + 1}{x}\times\frac{x}{7x-3}$
## Step3: Cancel out the $x$ terms: $\frac{6x + 1}{7x-3}$
# Answer: $\frac{6x + 1}{7x-3}$

### Problem 3:
# Explanation:
## Step1: Recall that $x^{2}-49=(x + 7)(x - 7)$ (difference of squares). Now, rewrite the complex fraction as $\frac{\frac{8}{x + 7}}{\frac{24}{(x + 7)(x - 7)}}$
## Step2: Divide by multiplying by the reciprocal: $\frac{8}{x + 7}\times\frac{(x + 7)(x - 7)}{24}$
## Step3: Cancel out the $x + 7$ terms and simplify $\frac{8}{24}$ to $\frac{1}{3}$: $\frac{1\times(x - 7)}{3}=\frac{x - 7}{3}$
# Answer: $\frac{x - 7}{3}$

### Problem 4:
# Explanation:
## Step1: Simplify the numerator and denominator separately. Numerator: $4+\frac{2}{x}=\frac{4x + 2}{x}$. Denominator: $\frac{x}{3}+\frac{1}{6}=\frac{2x+1}{6}$ (by finding a common denominator of 6)
## Step2: Divide the two fractions: $\frac{\frac{4x + 2}{x}}{\frac{2x + 1}{6}}=\frac{4x + 2}{x}\times\frac{6}{2x + 1}$
## Step3: Factor out 2 from $4x + 2$: $4x + 2 = 2(2x + 1)$. Now, substitute: $\frac{2(2x + 1)}{x}\times\frac{6}{2x + 1}$
## Step4: Cancel out the $2x + 1$ terms: $\frac{2\times6}{x}=\frac{12}{x}$
# Answer: $\frac{12}{x}$

### Problem 5:
# Explanation:
## Step1: Simplify the numerator and denominator separately. For the numerator: $x+\frac{x - 3}{3}=\frac{3x+x - 3}{3}=\frac{4x-3}{3}$. For the denominator: $\frac{4}{3}+\frac{2}{3x}=\frac{4x + 2}{3x}$ (common denominator of $3x$)
## Step2: Now, divide the two fractions: $\frac{\frac{4x-3}{3}}{\frac{4x + 2}{3x}}=\frac{4x-3}{3}\times\frac{3x}{4x + 2}$
## Step3: Cancel out the 3 terms: $\frac{(4x-3)x}{4x + 2}=\frac{4x^{2}-3x}{4x + 2}$ (we can also factor numerator and denominator, but this is simplified)
# Answer: $\frac{4x^{2}-3x}{4x + 2}$ (or factored as $\frac{x(4x - 3)}{2(2x+1)}$)

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Recall that dividing by a fraction is multiplying by its reciprocal. So, \(\frac{\frac{5}{x}}{\frac{9}{x + 2}}=\frac{5}{x}\times\frac{x + 2}{9}\)

Step2: Multiply the numerators and denominators: \(\frac{5(x + 2)}{9x}\)

Step1: First, simplify the numerator and denominator separately. For the numerator: \(6+\frac{1}{x}=\frac{6x + 1}{x}\). For the denominator: \(7-\frac{3}{x}=\frac{7x-3}{x}\)

Step2: Now, divide the two fractions: \(\frac{\frac{6x + 1}{x}}{\frac{7x-3}{x}}=\frac{6x + 1}{x}\times\frac{x}{7x-3}\)

Step3: Cancel out the \(x\) terms: \(\frac{6x + 1}{7x-3}\)

Step1: Recall that \(x^{2}-49=(x + 7)(x - 7)\) (difference of squares). Now, rewrite the complex fraction as \(\frac{\frac{8}{x + 7}}{\frac{24}{(x + 7)(x - 7)}}\)

Step2: Divide by multiplying by the reciprocal: \(\frac{8}{x + 7}\times\frac{(x + 7)(x - 7)}{24}\)

Step3: Cancel out the \(x + 7\) terms and simplify \(\frac{8}{24}\) to \(\frac{1}{3}\): \(\frac{1\times(x - 7)}{3}=\frac{x - 7}{3}\)

Answer:

Problem 2: