Worksheet on adding and subtracting algebraic fractions with various expressions.

To solve the given problems involving algebraic fractions, we need to follow these steps:

1. Find a common denominator for the fractions.
2. Rewrite each fraction with the common denominator.
3. Combine the numerators and simplify the resulting expression.

Let's solve each problem step by step.

---

Problem (a):

$$
\frac{1}{f} + \frac{2}{3f}
$$

#### Step 1: Identify the denominators
The denominators are \( f \) and \( 3f \). The least common denominator (LCD) is \( 3f \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{1}{f} \), multiply numerator and denominator by 3:
$$
\frac{1}{f} = \frac{1 \cdot 3}{f \cdot 3} = \frac{3}{3f}
$$
- For \( \frac{2}{3f} \), it already has the denominator \( 3f \):
$$
\frac{2}{3f} = \frac{2}{3f}
$$

#### Step 3: Combine the fractions
$$
\frac{1}{f} + \frac{2}{3f} = \frac{3}{3f} + \frac{2}{3f} = \frac{3 + 2}{3f} = \frac{5}{3f}
$$

#### Final Answer:
$$
\boxed{\frac{5}{3f}}
$$

---

Problem (b):

$$
\frac{h-1}{2} - \frac{h-2}{5}
$$

#### Step 1: Identify the denominators
The denominators are \( 2 \) and \( 5 \). The least common denominator (LCD) is \( 10 \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{h-1}{2} \), multiply numerator and denominator by 5:
$$
\frac{h-1}{2} = \frac{(h-1) \cdot 5}{2 \cdot 5} = \frac{5(h-1)}{10} = \frac{5h - 5}{10}
$$
- For \( \frac{h-2}{5} \), multiply numerator and denominator by 2:
$$
\frac{h-2}{5} = \frac{(h-2) \cdot 2}{5 \cdot 2} = \frac{2(h-2)}{10} = \frac{2h - 4}{10}
$$

#### Step 3: Combine the fractions
$$
\frac{h-1}{2} - \frac{h-2}{5} = \frac{5h - 5}{10} - \frac{2h - 4}{10} = \frac{(5h - 5) - (2h - 4)}{10}
$$
Simplify the numerator:
$$
(5h - 5) - (2h - 4) = 5h - 5 - 2h + 4 = 3h - 1
$$
So the expression becomes:
$$
\frac{3h - 1}{10}
$$

#### Final Answer:
$$
\boxed{\frac{3h - 1}{10}}
$$

---

Problem (c):

$$
\frac{4}{2d} - \frac{2r}{4d}
$$

#### Step 1: Simplify each fraction
- Simplify \( \frac{4}{2d} \):
$$
\frac{4}{2d} = \frac{2}{d}
$$
- Simplify \( \frac{2r}{4d} \):
$$
\frac{2r}{4d} = \frac{r}{2d}
$$

#### Step 2: Identify the denominators
The denominators are \( d \) and \( 2d \). The least common denominator (LCD) is \( 2d \).

#### Step 3: Rewrite each fraction with the LCD
- For \( \frac{2}{d} \), multiply numerator and denominator by 2:
$$
\frac{2}{d} = \frac{2 \cdot 2}{d \cdot 2} = \frac{4}{2d}
$$
- For \( \frac{r}{2d} \), it already has the denominator \( 2d \):
$$
\frac{r}{2d} = \frac{r}{2d}
$$

#### Step 4: Combine the fractions
$$
\frac{4}{2d} - \frac{r}{2d} = \frac{4 - r}{2d}
$$

#### Final Answer:
$$
\boxed{\frac{4 - r}{2d}}
$$

---

Problem (d):

$$
\frac{2}{c} - \frac{y+2}{4}
$$

#### Step 1: Identify the denominators
The denominators are \( c \) and \( 4 \). The least common denominator (LCD) is \( 4c \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{2}{c} \), multiply numerator and denominator by 4:
$$
\frac{2}{c} = \frac{2 \cdot 4}{c \cdot 4} = \frac{8}{4c}
$$
- For \( \frac{y+2}{4} \), multiply numerator and denominator by \( c \):
$$
\frac{y+2}{4} = \frac{(y+2) \cdot c}{4 \cdot c} = \frac{c(y+2)}{4c} = \frac{cy + 2c}{4c}
$$

#### Step 3: Combine the fractions
$$
\frac{2}{c} - \frac{y+2}{4} = \frac{8}{4c} - \frac{cy + 2c}{4c} = \frac{8 - (cy + 2c)}{4c}
$$
Simplify the numerator:
$$
8 - (cy + 2c) = 8 - cy - 2c
$$
So the expression becomes:
$$
\frac{8 - cy - 2c}{4c}
$$

#### Final Answer:
$$
\boxed{\frac{8 - cy - 2c}{4c}}
$$

---

Problem (e):

$$
\frac{w}{x} - \frac{1}{y}
$$

#### Step 1: Identify the denominators
The denominators are \( x \) and \( y \). The least common denominator (LCD) is \( xy \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{w}{x} \), multiply numerator and denominator by \( y \):
$$
\frac{w}{x} = \frac{w \cdot y}{x \cdot y} = \frac{wy}{xy}
$$
- For \( \frac{1}{y} \), multiply numerator and denominator by \( x \):
$$
\frac{1}{y} = \frac{1 \cdot x}{y \cdot x} = \frac{x}{xy}
$$

#### Step 3: Combine the fractions
$$
\frac{w}{x} - \frac{1}{y} = \frac{wy}{xy} - \frac{x}{xy} = \frac{wy - x}{xy}
$$

#### Final Answer:
$$
\boxed{\frac{wy - x}{xy}}
$$

---

Problem (f):

$$
\frac{4}{x-y} + \frac{5}{x+y}
$$

#### Step 1: Identify the denominators
The denominators are \( x-y \) and \( x+y \). The least common denominator (LCD) is \( (x-y)(x+y) \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{4}{x-y} \), multiply numerator and denominator by \( x+y \):
$$
\frac{4}{x-y} = \frac{4(x+y)}{(x-y)(x+y)} = \frac{4x + 4y}{(x-y)(x+y)}
$$
- For \( \frac{5}{x+y} \), multiply numerator and denominator by \( x-y \):
$$
\frac{5}{x+y} = \frac{5(x-y)}{(x+y)(x-y)} = \frac{5x - 5y}{(x+y)(x-y)}
$$

#### Step 3: Combine the fractions
$$
\frac{4}{x-y} + \frac{5}{x+y} = \frac{4x + 4y}{(x-y)(x+y)} + \frac{5x - 5y}{(x+y)(x-y)} = \frac{(4x + 4y) + (5x - 5y)}{(x-y)(x+y)}
$$
Simplify the numerator:
$$
(4x + 4y) + (5x - 5y) = 4x + 4y + 5x - 5y = 9x - y
$$
So the expression becomes:
$$
\frac{9x - y}{(x-y)(x+y)}
$$

#### Final Answer:
$$
\boxed{\frac{9x - y}{(x-y)(x+y)}}
$$

---

Problem (g):

$$
\frac{2}{a+c} - \frac{5}{a-c}
$$

#### Step 1: Identify the denominators
The denominators are \( a+c \) and \( a-c \). The least common denominator (LCD) is \( (a+c)(a-c) \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{2}{a+c} \), multiply numerator and denominator by \( a-c \):
$$
\frac{2}{a+c} = \frac{2(a-c)}{(a+c)(a-c)} = \frac{2a - 2c}{(a+c)(a-c)}
$$
- For \( \frac{5}{a-c} \), multiply numerator and denominator by \( a+c \):
$$
\frac{5}{a-c} = \frac{5(a+c)}{(a-c)(a+c)} = \frac{5a + 5c}{(a-c)(a+c)}
$$

#### Step 3: Combine the fractions
$$
\frac{2}{a+c} - \frac{5}{a-c} = \frac{2a - 2c}{(a+c)(a-c)} - \frac{5a + 5c}{(a-c)(a+c)} = \frac{(2a - 2c) - (5a + 5c)}{(a+c)(a-c)}
$$
Simplify the numerator:
$$
(2a - 2c) - (5a + 5c) = 2a - 2c - 5a - 5c = -3a - 7c
$$
So the expression becomes:
$$
\frac{-3a - 7c}{(a+c)(a-c)}
$$

#### Final Answer:
$$
\boxed{\frac{-3a - 7c}{(a+c)(a-c)}}
$$

---

Problem (h):

$$
\frac{2}{w^2-1} + \frac{3}{w+1}
$$

#### Step 1: Factor the denominators
- The denominator \( w^2 - 1 \) can be factored as \( (w-1)(w+1) \).
- The denominator \( w+1 \) is already factored.

So the denominators are \( (w-1)(w+1) \) and \( w+1 \). The least common denominator (LCD) is \( (w-1)(w+1) \).

#### Step 2: Rewrite each fraction with the LCD
- For \( \frac{2}{w^2-1} \), it already has the denominator \( (w-1)(w+1) \):
$$
\frac{2}{w^2-1} = \frac{2}{(w-1)(w+1)}
$$
- For \( \frac{3}{w+1} \), multiply numerator and denominator by \( w-1 \):
$$
\frac{3}{w+1} = \frac{3(w-1)}{(w+1)(w-1)} = \frac{3w - 3}{(w+1)(w-1)}
$$

#### Step 3: Combine the fractions
$$
\frac{2}{w^2-1} + \frac{3}{w+1} = \frac{2}{(w-1)(w+1)} + \frac{3w - 3}{(w+1)(w-1)} = \frac{2 + (3w - 3)}{(w-1)(w+1)}
$$
Simplify the numerator:
$$
2 + (3w - 3) = 2 + 3w - 3 = 3w - 1
$$
So the expression becomes:
$$
\frac{3w - 1}{(w-1)(w+1)}
$$

#### Final Answer:
$$
\boxed{\frac{3w - 1}{(w-1)(w+1)}}
$$

---

Final Answers:

$$
\boxed{
\begin{aligned}
&\text{(a)} \ \frac{5}{3f}, \\
&\text{(b)} \ \frac{3h - 1}{10}, \\
&\text{(c)} \ \frac{4 - r}{2d}, \\
&\text{(d)} \ \frac{8 - cy - 2c}{4c}, \\
&\text{(e)} \ \frac{wy - x}{xy}, \\
&\text{(f)} \ \frac{9x - y}{(x-y)(x+y)}, \\
&\text{(g)} \ \frac{-3a - 7c}{(a+c)(a-c)}, \\
&\text{(h)} \ \frac{3w - 1}{(w-1)(w+1)}.
\end{aligned}
}
$$