Simplifying Expressions Worksheets - Math Monks - Free Printable

Let's solve each expression step by step and simplify them.

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(1) $\frac{1}{x - 1} \times \frac{8x - 8}{8}$

Step 1: Factor numerator of second fraction:
$$
8x - 8 = 8(x - 1)
$$

So the expression becomes:
$$
\frac{1}{x - 1} \times \frac{8(x - 1)}{8}
$$

Step 2: Cancel common terms:
- $8$ in numerator and denominator cancel.
- $(x - 1)$ in numerator and denominator cancel.

$$
= 1
$$

✔ Answer: $ \boxed{1} $

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(2) $\frac{2y - 5}{(y + 2)(y - 3)} \times \frac{y - 3}{4y - 10}$

Step 1: Factor $4y - 10$:
$$
4y - 10 = 2(2y - 5)
$$

Now rewrite:
$$
\frac{2y - 5}{(y + 2)(y - 3)} \times \frac{y - 3}{2(2y - 5)}
$$

Step 2: Cancel common factors:
- $2y - 5$ cancels.
- $y - 3$ cancels.

Left with:
$$
\frac{1}{(y + 2)} \times \frac{1}{2} = \frac{1}{2(y + 2)}
$$

✔ Answer: $ \boxed{\frac{1}{2(y + 2)}} $

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(3) $\frac{3x - 9y}{x^2 - xy} \div \frac{x^2 - 9y^2}{x^2 - y^2}$

Step 1: Rewrite division as multiplication by reciprocal:
$$
\frac{3x - 9y}{x^2 - xy} \times \frac{x^2 - y^2}{x^2 - 9y^2}
$$

Step 2: Factor all expressions:

- $3x - 9y = 3(x - 3y)$
- $x^2 - xy = x(x - y)$
- $x^2 - y^2 = (x - y)(x + y)$
- $x^2 - 9y^2 = (x - 3y)(x + 3y)$

Now substitute:
$$
\frac{3(x - 3y)}{x(x - y)} \times \frac{(x - y)(x + y)}{(x - 3y)(x + 3y)}
$$

Step 3: Cancel common factors:
- $(x - 3y)$ cancels.
- $(x - y)$ cancels.

Remaining:
$$
\frac{3}{x} \times \frac{x + y}{x + 3y} = \frac{3(x + y)}{x(x + 3y)}
$$

✔ Answer: $ \boxed{\frac{3(x + y)}{x(x + 3y)}} $

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(4) $\frac{r - 2}{8r + 16} \times \frac{r + 2}{r^2 - 2r}$

Step 1: Factor all expressions:

- $8r + 16 = 8(r + 2)$
- $r^2 - 2r = r(r - 2)$

So:
$$
\frac{r - 2}{8(r + 2)} \times \frac{r + 2}{r(r - 2)}
$$

Step 2: Cancel common factors:
- $r - 2$ cancels.
- $r + 2$ cancels.

Left with:
$$
\frac{1}{8} \times \frac{1}{r} = \frac{1}{8r}
$$

✔ Answer: $ \boxed{\frac{1}{8r}} $

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(5) $\frac{1}{p - 4} \div \frac{3p}{4p - 16}$

Step 1: Rewrite as multiplication:
$$
\frac{1}{p - 4} \times \frac{4p - 16}{3p}
$$

Step 2: Factor $4p - 16 = 4(p - 4)$

So:
$$
\frac{1}{p - 4} \times \frac{4(p - 4)}{3p}
$$

Step 3: Cancel $p - 4$:
$$
1 \times \frac{4}{3p} = \frac{4}{3p}
$$

✔ Answer: $ \boxed{\frac{4}{3p}} $

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(6) $\frac{x^3 + 2x^2}{y^3 - y} \times \frac{y^2 - 1}{x^2 - 4}$

Step 1: Factor all expressions:

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

Now substitute:
$$
\frac{x^2(x + 2)}{y(y - 1)(y + 1)} \times \frac{(y - 1)(y + 1)}{(x - 2)(x + 2)}
$$

Step 2: Cancel common factors:
- $(y - 1)$ cancels.
- $(y + 1)$ cancels.
- $(x + 2)$ cancels.

Left with:
$$
\frac{x^2}{y} \times \frac{1}{x - 2} = \frac{x^2}{y(x - 2)}
$$

✔ Answer: $ \boxed{\frac{x^2}{y(x - 2)}} $

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(7) $\frac{5b^2c^2}{10bc} \div \frac{6b^2c}{2c^2}$

Step 1: Rewrite as multiplication:
$$
\frac{5b^2c^2}{10bc} \times \frac{2c^2}{6b^2c}
$$

Step 2: Simplify each fraction first:

- $\frac{5b^2c^2}{10bc} = \frac{1}{2} \cdot \frac{b^2c^2}{bc} = \frac{1}{2} \cdot bc = \frac{bc}{2}$

But better to do it directly:

Simplify both fractions:

First fraction:
$$
\frac{5b^2c^2}{10bc} = \frac{5}{10} \cdot \frac{b^2}{b} \cdot \frac{c^2}{c} = \frac{1}{2} \cdot b \cdot c = \frac{bc}{2}
$$

Second fraction:
$$
\frac{2c^2}{6b^2c} = \frac{2}{6} \cdot \frac{c^2}{c} \cdot \frac{1}{b^2} = \frac{1}{3} \cdot c \cdot \frac{1}{b^2} = \frac{c}{3b^2}
$$

Wait — we are multiplying, so let's go back:

Actually, we have:
$$
\frac{5b^2c^2}{10bc} \times \frac{2c^2}{6b^2c}
$$

Multiply numerators and denominators:

Numerator: $5b^2c^2 \cdot 2c^2 = 10b^2c^4$

Denominator: $10bc \cdot 6b^2c = 60b^3c^2$

So:
$$
\frac{10b^2c^4}{60b^3c^2} = \frac{10}{60} \cdot \frac{b^2}{b^3} \cdot \frac{c^4}{c^2} = \frac{1}{6} \cdot \frac{1}{b} \cdot c^2 = \frac{c^2}{6b}
$$

✔ Answer: $ \boxed{\frac{c^2}{6b}} $

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(8) $\frac{2p}{q} \div \frac{4p}{q^2}$

Step 1: Rewrite as multiplication:
$$
\frac{2p}{q} \times \frac{q^2}{4p}
$$

Step 2: Cancel common factors:
- $2p$ and $4p$: $2p / 4p = 1/2$
- $q$ and $q^2$: $q^2 / q = q$

So:
$$
\frac{2p}{q} \times \frac{q^2}{4p} = \frac{2p \cdot q^2}{q \cdot 4p} = \frac{2pq^2}{4pq} = \frac{2}{4} \cdot \frac{q^2}{q} = \frac{1}{2} \cdot q = \frac{q}{2}
$$

✔ Answer: $ \boxed{\frac{q}{2}} $

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✔ Final Answers Summary:

1. $ \boxed{1} $
2. $ \boxed{\frac{1}{2(y + 2)}} $
3. $ \boxed{\frac{3(x + y)}{x(x + 3y)}} $
4. $ \boxed{\frac{1}{8r}} $
5. $ \boxed{\frac{4}{3p}} $
6. $ \boxed{\frac{x^2}{y(x - 2)}} $
7. $ \boxed{\frac{c^2}{6b}} $
8. $ \boxed{\frac{q}{2}} $

Let me know if you'd like a printable version or explanation for any specific one!