Examples of multiplying algebraic fractions with step-by-step problems.

Problem: Solve the given algebraic expressions and simplify them.

The task involves simplifying each of the given algebraic expressions. Let's solve them step by step.

---

#### Expression 1:
\[
\frac{x}{3} \times x
\]

1. Multiply the terms:
\[
\frac{x}{3} \times x = \frac{x \cdot x}{3} = \frac{x^2}{3}
\]

Answer:
\[
\boxed{\frac{x^2}{3}}
\]

---

#### Expression 2:
\[
\frac{m}{4b} \times \frac{2a}{5}
\]

1. Multiply the numerators and denominators:
\[
\frac{m}{4b} \times \frac{2a}{5} = \frac{m \cdot 2a}{4b \cdot 5} = \frac{2am}{20b}
\]

2. Simplify the fraction by dividing numerator and denominator by their greatest common divisor (GCD), which is 2:
\[
\frac{2am}{20b} = \frac{am}{10b}
\]

Answer:
\[
\boxed{\frac{am}{10b}}
\]

---

#### Expression 3:
\[
\frac{y}{y+3} \times \frac{y^4}{y-2}
\]

1. Multiply the numerators and denominators:
\[
\frac{y}{y+3} \times \frac{y^4}{y-2} = \frac{y \cdot y^4}{(y+3)(y-2)} = \frac{y^5}{(y+3)(y-2)}
\]

Answer:
\[
\boxed{\frac{y^5}{(y+3)(y-2)}}
\]

---

#### Expression 4:
\[
\frac{x-2}{x+3} \times \frac{x+1}{x-2}
\]

1. Multiply the numerators and denominators:
\[
\frac{x-2}{x+3} \times \frac{x+1}{x-2} = \frac{(x-2)(x+1)}{(x+3)(x-2)}
\]

2. Notice that \(x-2\) in the numerator and denominator cancel out (assuming \(x \neq 2\)):
\[
\frac{(x-2)(x+1)}{(x+3)(x-2)} = \frac{x+1}{x+3}
\]

Answer:
\[
\boxed{\frac{x+1}{x+3}}
\]

---

#### Expression 5:
\[
\frac{(x+4)(x+2)}{x+7} \times \frac{x+7}{(x-2)(x-3)}
\]

1. Multiply the numerators and denominators:
\[
\frac{(x+4)(x+2)}{x+7} \times \frac{x+7}{(x-2)(x-3)} = \frac{(x+4)(x+2)(x+7)}{(x+7)(x-2)(x-3)}
\]

2. Notice that \(x+7\) in the numerator and denominator cancel out (assuming \(x \neq -7\)):
\[
\frac{(x+4)(x+2)(x+7)}{(x+7)(x-2)(x-3)} = \frac{(x+4)(x+2)}{(x-2)(x-3)}
\]

Answer:
\[
\boxed{\frac{(x+4)(x+2)}{(x-2)(x-3)}}
\]

---

#### Expression 6:
\[
\frac{(p+1)(q-3)}{(p-2)^2} \times \frac{p-2}{(q-2)(p-3)}
\]

1. Multiply the numerators and denominators:
\[
\frac{(p+1)(q-3)}{(p-2)^2} \times \frac{p-2}{(q-2)(p-3)} = \frac{(p+1)(q-3)(p-2)}{(p-2)^2(q-2)(p-3)}
\]

2. Notice that one \(p-2\) in the numerator and denominator cancels out (assuming \(p \neq 2\)):
\[
\frac{(p+1)(q-3)(p-2)}{(p-2)^2(q-2)(p-3)} = \frac{(p+1)(q-3)}{(p-2)(q-2)(p-3)}
\]

Answer:
\[
\boxed{\frac{(p+1)(q-3)}{(p-2)(q-2)(p-3)}}
\]

---

#### Expression 7:
\[
\frac{x^2 + 3x - 4}{x+2} \times \frac{x+2}{6(x+4)}
\]

1. Factor the quadratic expression \(x^2 + 3x - 4\):
\[
x^2 + 3x - 4 = (x+4)(x-1)
\]
So the expression becomes:
\[
\frac{(x+4)(x-1)}{x+2} \times \frac{x+2}{6(x+4)}
\]

2. Multiply the numerators and denominators:
\[
\frac{(x+4)(x-1)}{x+2} \times \frac{x+2}{6(x+4)} = \frac{(x+4)(x-1)(x+2)}{(x+2) \cdot 6(x+4)}
\]

3. Cancel out \(x+2\) and \(x+4\) (assuming \(x \neq -2\) and \(x \neq -4\)):
\[
\frac{(x+4)(x-1)(x+2)}{(x+2) \cdot 6(x+4)} = \frac{x-1}{6}
\]

Answer:
\[
\boxed{\frac{x-1}{6}}
\]

---

#### Expression 8:
\[
\frac{x^2 + 4x - 5}{x^2 - 2x - 3} \times \frac{x^2 + 6x + 5}{x^2 - 4x + 3}
\]

1. Factor each quadratic expression:
- \(x^2 + 4x - 5 = (x+5)(x-1)\)
- \(x^2 - 2x - 3 = (x-3)(x+1)\)
- \(x^2 + 6x + 5 = (x+5)(x+1)\)
- \(x^2 - 4x + 3 = (x-3)(x-1)\)

So the expression becomes:
\[
\frac{(x+5)(x-1)}{(x-3)(x+1)} \times \frac{(x+5)(x+1)}{(x-3)(x-1)}
\]

2. Multiply the numerators and denominators:
\[
\frac{(x+5)(x-1)}{(x-3)(x+1)} \times \frac{(x+5)(x+1)}{(x-3)(x-1)} = \frac{(x+5)(x-1)(x+5)(x+1)}{(x-3)(x+1)(x-3)(x-1)}
\]

3. Cancel out common factors \(x+1\) and \(x-1\) (assuming \(x \neq -1\) and \(x \neq 1\)):
\[
\frac{(x+5)(x-1)(x+5)(x+1)}{(x-3)(x+1)(x-3)(x-1)} = \frac{(x+5)^2}{(x-3)^2}
\]

Answer:
\[
\boxed{\frac{(x+5)^2}{(x-3)^2}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \ \frac{x^2}{3} \\
2) & \ \frac{am}{10b} \\
3) & \ \frac{y^5}{(y+3)(y-2)} \\
4) & \ \frac{x+1}{x+3} \\
5) & \ \frac{(x+4)(x+2)}{(x-2)(x-3)} \\
6) & \ \frac{(p+1)(q-3)}{(p-2)(q-2)(p-3)} \\
7) & \ \frac{x-1}{6} \\
8) & \ \frac{(x+5)^2}{(x-3)^2}
\end{aligned}
}
\]