Five algebra problems numbered 26 to 30 requiring students to add and subtract rational expressions with variable denominators.

Problem 26: Simplify the expression

$$
\frac{4}{4x-1} + \frac{8x-15}{4x^2 + 11x - 3}
$$

#### Step 1: Factor the denominator of the second fraction
The denominator of the second fraction is $4x^2 + 11x - 3$. We need to factor it:
$$
4x^2 + 11x - 3
$$
We look for two numbers that multiply to $4 \cdot (-3) = -12$ and add to $11$. These numbers are $12$ and $-1$. Thus, we can rewrite the quadratic as:
$$
4x^2 + 12x - x - 3 = 4x(x + 3) - 1(x + 3) = (4x - 1)(x + 3)
$$
So, the factored form of the denominator is:
$$
4x^2 + 11x - 3 = (4x - 1)(x + 3)
$$

#### Step 2: Rewrite the expression with a common denominator
The expression becomes:
$$
\frac{4}{4x-1} + \frac{8x-15}{(4x-1)(x+3)}
$$
The common denominator is $(4x-1)(x+3)$. Rewrite the first fraction with this common denominator:
$$
\frac{4}{4x-1} = \frac{4(x+3)}{(4x-1)(x+3)}
$$
So the expression is:
$$
\frac{4(x+3)}{(4x-1)(x+3)} + \frac{8x-15}{(4x-1)(x+3)}
$$

#### Step 3: Combine the fractions
Since the denominators are the same, we can combine the numerators:
$$
\frac{4(x+3) + (8x-15)}{(4x-1)(x+3)}
$$
Simplify the numerator:
$$
4(x+3) + (8x-15) = 4x + 12 + 8x - 15 = 12x - 3
$$
So the expression becomes:
$$
\frac{12x - 3}{(4x-1)(x+3)}
$$

#### Step 4: Factor the numerator
The numerator $12x - 3$ can be factored as:
$$
12x - 3 = 3(4x - 1)
$$
So the expression is:
$$
\frac{3(4x-1)}{(4x-1)(x+3)}
$$

#### Step 5: Simplify the fraction
Cancel the common factor $(4x-1)$ in the numerator and the denominator:
$$
\frac{3(4x-1)}{(4x-1)(x+3)} = \frac{3}{x+3}
$$

#### Final Answer:
$$
\boxed{\frac{3}{x+3}}
$$

---

Problem 27: Simplify the expression

$$
\frac{9}{2y+3} - \frac{5}{4y}
$$

#### Step 1: Find the least common denominator (LCD)
The denominators are $2y+3$ and $4y$. The LCD is:
$$
4y(2y+3)
$$

#### Step 2: Rewrite each fraction with the LCD
For the first fraction:
$$
\frac{9}{2y+3} = \frac{9 \cdot 4y}{(2y+3) \cdot 4y} = \frac{36y}{4y(2y+3)}
$$
For the second fraction:
$$
\frac{5}{4y} = \frac{5 \cdot (2y+3)}{4y \cdot (2y+3)} = \frac{5(2y+3)}{4y(2y+3)}
$$
So the expression becomes:
$$
\frac{36y}{4y(2y+3)} - \frac{5(2y+3)}{4y(2y+3)}
$$

#### Step 3: Combine the fractions
Since the denominators are the same, we can combine the numerators:
$$
\frac{36y - 5(2y+3)}{4y(2y+3)}
$$
Simplify the numerator:
$$
36y - 5(2y+3) = 36y - 10y - 15 = 26y - 15
$$
So the expression becomes:
$$
\frac{26y - 15}{4y(2y+3)}
$$

#### Final Answer:
$$
\boxed{\frac{26y-15}{4y(2y+3)}}
$$

---

Problem 28: Simplify the expression

$$
\frac{4}{5x+7} + \frac{7}{6x}
$$

#### Step 1: Find the least common denominator (LCD)
The denominators are $5x+7$ and $6x$. The LCD is:
$$
6x(5x+7)
$$

#### Step 2: Rewrite each fraction with the LCD
For the first fraction:
$$
\frac{4}{5x+7} = \frac{4 \cdot 6x}{(5x+7) \cdot 6x} = \frac{24x}{6x(5x+7)}
$$
For the second fraction:
$$
\frac{7}{6x} = \frac{7 \cdot (5x+7)}{6x \cdot (5x+7)} = \frac{7(5x+7)}{6x(5x+7)}
$$
So the expression becomes:
$$
\frac{24x}{6x(5x+7)} + \frac{7(5x+7)}{6x(5x+7)}
$$

#### Step 3: Combine the fractions
Since the denominators are the same, we can combine the numerators:
$$
\frac{24x + 7(5x+7)}{6x(5x+7)}
$$
Simplify the numerator:
$$
24x + 7(5x+7) = 24x + 35x + 49 = 59x + 49
$$
So the expression becomes:
$$
\frac{59x + 49}{6x(5x+7)}
$$

#### Final Answer:
$$
\boxed{\frac{59x+49}{6x(5x+7)}}
$$

---

Problem 29: Simplify the expression

$$
\frac{1}{3x^2 - 8x} + \frac{6}{8 - 3x}
$$

#### Step 1: Factor the denominators
The first denominator is $3x^2 - 8x$, which can be factored as:
$$
3x^2 - 8x = x(3x - 8)
$$
The second denominator is $8 - 3x$, which can be rewritten as:
$$
8 - 3x = -(3x - 8)
$$
So the expression becomes:
$$
\frac{1}{x(3x-8)} + \frac{6}{-(3x-8)} = \frac{1}{x(3x-8)} - \frac{6}{3x-8}
$$

#### Step 2: Find the least common denominator (LCD)
The LCD is:
$$
x(3x-8)
$$

#### Step 3: Rewrite each fraction with the LCD
For the first fraction:
$$
\frac{1}{x(3x-8)} = \frac{1}{x(3x-8)}
$$
For the second fraction:
$$
-\frac{6}{3x-8} = -\frac{6 \cdot x}{(3x-8) \cdot x} = -\frac{6x}{x(3x-8)}
$$
So the expression becomes:
$$
\frac{1}{x(3x-8)} - \frac{6x}{x(3x-8)}
$$

#### Step 4: Combine the fractions
Since the denominators are the same, we can combine the numerators:
$$
\frac{1 - 6x}{x(3x-8)}
$$

#### Final Answer:
$$
\boxed{\frac{1-6x}{x(3x-8)}}
$$

---

Problem 30: Simplify the expression

$$
\frac{x+4}{3x^2 - 4x} - \frac{6}{4-3x}
$$

#### Step 1: Factor the denominators
The first denominator is $3x^2 - 4x$, which can be factored as:
$$
3x^2 - 4x = x(3x - 4)
$$
The second denominator is $4 - 3x$, which can be rewritten as:
$$
4 - 3x = -(3x - 4)
$$
So the expression becomes:
$$
\frac{x+4}{x(3x-4)} - \frac{6}{-(3x-4)} = \frac{x+4}{x(3x-4)} + \frac{6}{3x-4}
$$

#### Step 2: Find the least common denominator (LCD)
The LCD is:
$$
x(3x-4)
$$

#### Step 3: Rewrite each fraction with the LCD
For the first fraction:
$$
\frac{x+4}{x(3x-4)} = \frac{x+4}{x(3x-4)}
$$
For the second fraction:
$$
\frac{6}{3x-4} = \frac{6 \cdot x}{(3x-4) \cdot x} = \frac{6x}{x(3x-4)}
$$
So the expression becomes:
$$
\frac{x+4}{x(3x-4)} + \frac{6x}{x(3x-4)}
$$

#### Step 4: Combine the fractions
Since the denominators are the same, we can combine the numerators:
$$
\frac{x+4 + 6x}{x(3x-4)} = \frac{7x+4}{x(3x-4)}
$$

#### Final Answer:
$$
\boxed{\frac{7x+4}{x(3x-4)}}
$$

---

Final Answers:

1. Problem 26: $\boxed{\frac{3}{x+3}}$
2. Problem 27: $\boxed{\frac{26y-15}{4y(2y+3)}}$
3. Problem 28: $\boxed{\frac{59x+49}{6x(5x+7)}}$
4. Problem 29: $\boxed{\frac{1-6x}{x(3x-8)}}$
5. Problem 30: $\boxed{\frac{7x+4}{x(3x-4)}}$