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

Let's solve each of these rational expression division problems step by step. The general rule for dividing fractions (or rational expressions) is:

> $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$

We'll apply this rule and simplify where possible.

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1) $\frac{x}{4} \div 5$

Convert 5 to a fraction: $5 = \frac{5}{1}$

$$
\frac{x}{4} \div 5 = \frac{x}{4} \cdot \frac{1}{5} = \frac{x}{20}
$$

✔ Answer: $\boxed{\frac{x}{20}}$

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2) $\frac{g}{4h} \div \frac{3j}{2h}$

Multiply by the reciprocal:

$$
\frac{g}{4h} \cdot \frac{2h}{3j}
$$

Simplify:
- $h$ in numerator and denominator cancels.
- $2$ and $4$: reduce $2/4 = 1/2$

$$
= \frac{g}{4h} \cdot \frac{2h}{3j} = \frac{g \cdot 2h}{4h \cdot 3j} = \frac{2gh}{12hj} = \frac{g}{6j}
$$

✔ Answer: $\boxed{\frac{g}{6j}}$

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3) $\frac{y}{y+7} \div \frac{y^4}{y-1}$

Multiply by reciprocal:

$$
\frac{y}{y+7} \cdot \frac{y-1}{y^4}
$$

Simplify:
- $y$ in numerator and $y^4$ → $y / y^4 = 1/y^3$

$$
= \frac{y(y-1)}{(y+7)(y^4)} = \frac{y-1}{(y+7)y^3}
$$

✔ Answer: $\boxed{\frac{y-1}{y^3(y+7)}}$

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4) $\frac{x-4}{x+3} \div \frac{x-1}{x+3}$

Multiply by reciprocal:

$$
\frac{x-4}{x+3} \cdot \frac{x+3}{x-1}
$$

Cancel $x+3$:

$$
= \frac{x-4}{1} \cdot \frac{1}{x-1} = \frac{x-4}{x-1}
$$

✔ Answer: $\boxed{\frac{x-4}{x-1}}$

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

Multiply by reciprocal:

$$
\frac{(x+4)(x+7)}{x+5} \cdot \frac{(x-4)(x-7)}{(x+5)(x+7)}
$$

Now cancel common factors:
- $(x+7)$: top and bottom
- $(x+5)$: one from denominator, one from numerator

$$
= \frac{(x+4)}{1} \cdot \frac{(x-4)(x-7)}{(x+5)} = \frac{(x+4)(x-4)(x-7)}{x+5}
$$

Note: $(x+4)(x-4) = x^2 - 16$, so:

$$
= \frac{(x^2 - 16)(x - 7)}{x + 5}
$$

✔ Answer: $\boxed{\frac{(x^2 - 16)(x - 7)}{x + 5}}$

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6) $\frac{p+1}{(p-3)^2} \div \frac{q-3}{(p-3)(p+1)}$

Multiply by reciprocal:

$$
\frac{p+1}{(p-3)^2} \cdot \frac{(p-3)(p+1)}{q-3}
$$

Simplify:
- One $(p-3)$ cancels from denominator and numerator
- $(p+1)$ appears twice: $(p+1) \cdot (p+1) = (p+1)^2$

$$
= \frac{p+1}{(p-3)} \cdot \frac{p+1}{q-3} = \frac{(p+1)^2}{(p-3)(q-3)}
$$

✔ Answer: $\boxed{\frac{(p+1)^2}{(p-3)(q-3)}}$

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7) $\frac{x^2 + 2x - 3}{x+2} \div \frac{4(x-4)}{5(x+2)}$

First, factor numerator:
$x^2 + 2x - 3 = (x+3)(x-1)$

So expression becomes:

$$
\frac{(x+3)(x-1)}{x+2} \div \frac{4(x-4)}{5(x+2)}
$$

Multiply by reciprocal:

$$
\frac{(x+3)(x-1)}{x+2} \cdot \frac{5(x+2)}{4(x-4)}
$$

Cancel $x+2$:

$$
= \frac{(x+3)(x-1)}{1} \cdot \frac{5}{4(x-4)} = \frac{5(x+3)(x-1)}{4(x-4)}
$$

✔ Answer: $\boxed{\frac{5(x+3)(x-1)}{4(x-4)}}$

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8) $\frac{x^2 + 4x - 5}{x^2 - 2x - 3} \div \frac{x^2 - 4x + 3}{x^2 + 6x + 5}$

Factor all quadratics:

- $x^2 + 4x - 5 = (x+5)(x-1)$
- $x^2 - 2x - 3 = (x-3)(x+1)$
- $x^2 - 4x + 3 = (x-3)(x-1)$
- $x^2 + 6x + 5 = (x+5)(x+1)$

Now substitute:

$$
\frac{(x+5)(x-1)}{(x-3)(x+1)} \div \frac{(x-3)(x-1)}{(x+5)(x+1)}
$$

Multiply by reciprocal:

$$
\frac{(x+5)(x-1)}{(x-3)(x+1)} \cdot \frac{(x+5)(x+1)}{(x-3)(x-1)}
$$

Now cancel:
- $(x-1)$: top and bottom
- $(x+1)$: top and bottom
- $(x+5)$: one from each side
- $(x-3)$: one from each side

After cancellation:

$$
= \frac{(x+5)}{(x-3)} \cdot \frac{(x+5)}{(x-3)} = \frac{(x+5)^2}{(x-3)^2}
$$

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

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✔ Final Answers Summary:

1) $\boxed{\frac{x}{20}}$
2) $\boxed{\frac{g}{6j}}$
3) $\boxed{\frac{y-1}{y^3(y+7)}}$
4) $\boxed{\frac{x-4}{x-1}}$
5) $\boxed{\frac{(x^2 - 16)(x - 7)}{x + 5}}$
6) $\boxed{\frac{(p+1)^2}{(p-3)(q-3)}}$
7) $\boxed{\frac{5(x+3)(x-1)}{4(x-4)}}$
8) $\boxed{\frac{(x+5)^2}{(x-3)^2}}$

Let me know if you'd like any explanation visualized or further simplified! 🦆