Question27. $\frac{3}{4(x+1)}+\frac{x}{x+1}$
High School of American Studies atgebraic Fractions Unit 4; Alge
28. $\frac{2 x}{x-1}+\frac{3}{2(x-1)}$
29. $\frac{4 n-3}{15}-\frac{3(n-2)}{10}$
Write each expression as a fraction in simplest form.
9. $6+\frac{1}{x}$
10. $2+\frac{5}{a}$
11. $3-\frac{2}{x}$
12. $5-\frac{3}{n}$
Write each expression as a fraction in simplest form.
17. $2+\frac{x}{y}$
18. $3-\frac{2}{x+1}$
19. $7+\frac{y}{y-2}$
Write each expression as a fraction in simplest form. $\begin{array}{ll}\text { 29. } \frac{x}{x-1}+\frac{x-1}{x}-2 & \text { 30. } x+\frac{2 x+1}{x-2}\end{array}$
31. $\frac{3}{x+1}+\frac{x}{x+1}-1$

Studdy Solution
The fraction $\frac{3 + 4x}{4(x+1)}$ is already in its simplest form.
### Problem 28: $\frac{2x}{x-1}+\frac{3}{2(x-1)}$
STEP_1: Identify the least common denominator (LCD) for the fractions $\frac{2x}{x-1}$ and $\frac{3}{2(x-1)}$. The LCD is $2(x-1)$.
STEP_2: Rewrite each fraction with the common denominator:
$\frac{2x}{x-1} = \frac{4x}{2(x-1)}$
$\frac{3}{2(x-1)} = \frac{3}{2(x-1)}$
STEP_3: Combine the fractions:
$\frac{4x + 3}{2(x-1)}$
STEP_4: The fraction $\frac{4x + 3}{2(x-1)}$ is already in its simplest form.
### Problem 29: $\frac{4n-3}{15}-\frac{3(n-2)}{10}$
STEP_1: Identify the least common denominator (LCD) for the fractions $\frac{4n-3}{15}$ and $\frac{3(n-2)}{10}$. The LCD is $30$.
STEP_2: Rewrite each fraction with the common denominator:
$\frac{4n-3}{15} = \frac{2(4n-3)}{30} = \frac{8n-6}{30}$
$\frac{3(n-2)}{10} = \frac{3(n-2) \cdot 3}{30} = \frac{9(n-2)}{30} = \frac{9n-18}{30}$
STEP_3: Combine the fractions:
$\frac{8n-6 - (9n-18)}{30} = \frac{8n-6 - 9n + 18}{30} = \frac{-n + 12}{30}$
STEP_4: The fraction $\frac{-n + 12}{30}$ is already in its simplest form.
### Problem 31: $\frac{3}{x+1}+\frac{x}{x+1}-1$
STEP_1: Identify the least common denominator (LCD) for the fractions $\frac{3}{x+1}$ and $\frac{x}{x+1}$. The LCD is $x+1$.
STEP_2: Rewrite each fraction with the common denominator and include the whole number:
$\frac{3}{x+1} + \frac{x}{x+1} - 1 = \frac{3 + x}{x+1} - \frac{x+1}{x+1}$
STEP_3: Combine the fractions:
$\frac{3 + x - (x+1)}{x+1} = \frac{3 + x - x - 1}{x+1} = \frac{2}{x+1}$
STEP_4: The fraction $\frac{2}{x+1}$ is already in its simplest form.