A4f - Simplifying, multiplying and dividing algebraic fractions ... - Free Printable

To solve the given problems, we need to simplify each expression step by step. The key concept here is dividing fractions, which involves multiplying by the reciprocal of the divisor. Let's go through each problem in detail.

---

Problem 1:

\[
\frac{x}{4} \div 5
\]

#### Solution:
1. Dividing by a number is the same as multiplying by its reciprocal.
\[
\frac{x}{4} \div 5 = \frac{x}{4} \cdot \frac{1}{5}
\]
2. Multiply the numerators and the denominators:
\[
\frac{x}{4} \cdot \frac{1}{5} = \frac{x \cdot 1}{4 \cdot 5} = \frac{x}{20}
\]

#### Final Answer:
\[
\boxed{\frac{x}{20}}
\]

---

Problem 2:

\[
\frac{g}{4h} \div \frac{3j}{2h}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{g}{4h} \div \frac{3j}{2h} = \frac{g}{4h} \cdot \frac{2h}{3j}
\]
2. Multiply the numerators and the denominators:
\[
\frac{g}{4h} \cdot \frac{2h}{3j} = \frac{g \cdot 2h}{4h \cdot 3j} = \frac{2gh}{12hj}
\]
3. Simplify by canceling common factors:
- \( h \) in the numerator and denominator cancels out.
- Simplify \( \frac{2}{12} \) to \( \frac{1}{6} \):
\[
\frac{2gh}{12hj} = \frac{g}{6j}
\]

#### Final Answer:
\[
\boxed{\frac{g}{6j}}
\]

---

Problem 3:

\[
\frac{y}{y+7} \div \frac{y^4}{y-1}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{y}{y+7} \div \frac{y^4}{y-1} = \frac{y}{y+7} \cdot \frac{y-1}{y^4}
\]
2. Multiply the numerators and the denominators:
\[
\frac{y}{y+7} \cdot \frac{y-1}{y^4} = \frac{y \cdot (y-1)}{(y+7) \cdot y^4} = \frac{y(y-1)}{y^4(y+7)}
\]
3. Simplify by canceling common factors:
- \( y \) in the numerator and denominator cancels out:
\[
\frac{y(y-1)}{y^4(y+7)} = \frac{y-1}{y^3(y+7)}
\]

#### Final Answer:
\[
\boxed{\frac{y-1}{y^3(y+7)}}
\]

---

Problem 4:

\[
\frac{x-4}{x+3} \div \frac{x-1}{x+3}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{x-4}{x+3} \div \frac{x-1}{x+3} = \frac{x-4}{x+3} \cdot \frac{x+3}{x-1}
\]
2. Multiply the numerators and the denominators:
\[
\frac{x-4}{x+3} \cdot \frac{x+3}{x-1} = \frac{(x-4)(x+3)}{(x+3)(x-1)}
\]
3. Simplify by canceling common factors:
- \( x+3 \) in the numerator and denominator cancels out:
\[
\frac{(x-4)(x+3)}{(x+3)(x-1)} = \frac{x-4}{x-1}
\]

#### Final Answer:
\[
\boxed{\frac{x-4}{x-1}}
\]

---

Problem 5:

\[
\frac{(x+4)(x+7)}{x+5} \div \frac{(x+5)(x+7)}{(x-4)(x-7)}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{(x+4)(x+7)}{x+5} \div \frac{(x+5)(x+7)}{(x-4)(x-7)} = \frac{(x+4)(x+7)}{x+5} \cdot \frac{(x-4)(x-7)}{(x+5)(x+7)}
\]
2. Multiply the numerators and the denominators:
\[
\frac{(x+4)(x+7)}{x+5} \cdot \frac{(x-4)(x-7)}{(x+5)(x+7)} = \frac{(x+4)(x+7)(x-4)(x-7)}{(x+5)(x+5)(x+7)}
\]
3. Simplify by canceling common factors:
- \( x+7 \) in the numerator and denominator cancels out:
\[
\frac{(x+4)(x+7)(x-4)(x-7)}{(x+5)(x+5)(x+7)} = \frac{(x+4)(x-4)(x-7)}{(x+5)^2}
\]
- Note that \( (x+4)(x-4) = x^2 - 16 \):
\[
\frac{(x+4)(x-4)(x-7)}{(x+5)^2} = \frac{(x^2 - 16)(x-7)}{(x+5)^2}
\]

#### Final Answer:
\[
\boxed{\frac{(x^2 - 16)(x-7)}{(x+5)^2}}
\]

---

Problem 6:

\[
\frac{p+1}{(p-3)^2} \div \frac{q-3}{(p-3)(p+1)}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{p+1}{(p-3)^2} \div \frac{q-3}{(p-3)(p+1)} = \frac{p+1}{(p-3)^2} \cdot \frac{(p-3)(p+1)}{q-3}
\]
2. Multiply the numerators and the denominators:
\[
\frac{p+1}{(p-3)^2} \cdot \frac{(p-3)(p+1)}{q-3} = \frac{(p+1)(p-3)(p+1)}{(p-3)^2(q-3)}
\]
3. Simplify by canceling common factors:
- One \( p-3 \) in the numerator and denominator cancels out:
\[
\frac{(p+1)(p-3)(p+1)}{(p-3)^2(q-3)} = \frac{(p+1)^2}{(p-3)(q-3)}
\]

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

---

Problem 7:

\[
\frac{x^2 + 2x - 3}{x+2} \div \frac{4(x-4)}{5(x+2)}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{x^2 + 2x - 3}{x+2} \div \frac{4(x-4)}{5(x+2)} = \frac{x^2 + 2x - 3}{x+2} \cdot \frac{5(x+2)}{4(x-4)}
\]
2. Multiply the numerators and the denominators:
\[
\frac{x^2 + 2x - 3}{x+2} \cdot \frac{5(x+2)}{4(x-4)} = \frac{(x^2 + 2x - 3) \cdot 5(x+2)}{(x+2) \cdot 4(x-4)}
\]
3. Simplify by canceling common factors:
- \( x+2 \) in the numerator and denominator cancels out:
\[
\frac{(x^2 + 2x - 3) \cdot 5(x+2)}{(x+2) \cdot 4(x-4)} = \frac{5(x^2 + 2x - 3)}{4(x-4)}
\]
4. Factor \( x^2 + 2x - 3 \):
\[
x^2 + 2x - 3 = (x+3)(x-1)
\]
So the expression becomes:
\[
\frac{5(x+3)(x-1)}{4(x-4)}
\]

#### Final Answer:
\[
\boxed{\frac{5(x+3)(x-1)}{4(x-4)}}
\]

---

Problem 8:

\[
\frac{x^2 + 4x - 5}{x^2 - 2x - 3} \div \frac{x^2 - 4x + 3}{x^2 + 6x + 5}
\]

#### Solution:
1. Dividing by a fraction means multiplying by its reciprocal:
\[
\frac{x^2 + 4x - 5}{x^2 - 2x - 3} \div \frac{x^2 - 4x + 3}{x^2 + 6x + 5} = \frac{x^2 + 4x - 5}{x^2 - 2x - 3} \cdot \frac{x^2 + 6x + 5}{x^2 - 4x + 3}
\]
2. Multiply the numerators and the denominators:
\[
\frac{x^2 + 4x - 5}{x^2 - 2x - 3} \cdot \frac{x^2 + 6x + 5}{x^2 - 4x + 3} = \frac{(x^2 + 4x - 5)(x^2 + 6x + 5)}{(x^2 - 2x - 3)(x^2 - 4x + 3)}
\]
3. Factor each quadratic expression:
- \( x^2 + 4x - 5 = (x+5)(x-1) \)
- \( x^2 + 6x + 5 = (x+5)(x+1) \)
- \( x^2 - 2x - 3 = (x-3)(x+1) \)
- \( x^2 - 4x + 3 = (x-3)(x-1) \)
Substituting these factors:
\[
\frac{(x+5)(x-1)(x+5)(x+1)}{(x-3)(x+1)(x-3)(x-1)}
\]
4. Simplify by canceling common factors:
- \( x+1 \) in the numerator and denominator cancels out.
- \( x-1 \) in the numerator and denominator cancels out.
- One \( x-3 \) in the numerator and denominator cancels out:
\[
\frac{(x+5)(x-1)(x+5)(x+1)}{(x-3)(x+1)(x-3)(x-1)} = \frac{(x+5)^2}{(x-3)^2}
\]

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

---

Final Answers:

\[
\boxed{
\begin{aligned}
1) & \ \frac{x}{20} \\
2) & \ \frac{g}{6j} \\
3) & \ \frac{y-1}{y^3(y+7)} \\
4) & \ \frac{x-4}{x-1} \\
5) & \ \frac{(x^2 - 16)(x-7)}{(x+5)^2} \\
6) & \ \frac{(p+1)^2}{(p-3)(q-3)} \\
7) & \ \frac{5(x+3)(x-1)}{4(x-4)} \\
8) & \ \frac{(x+5)^2}{(x-3)^2}
\end{aligned}
}
\]