Rational Expressions Worksheets - Math Monks - Free Printable

Problem: Simplify each rational expression and state the excluded values where applicable.

#### Part 1: Simplify each expression

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Problem 1:

$$
\frac{5p^2 - 5p}{1 - p}
$$

Step 1: Factor the numerator.
$$
5p^2 - 5p = 5p(p - 1)
$$

So the expression becomes:
$$
\frac{5p(p - 1)}{1 - p}
$$

Step 2: Notice that \(1 - p\) can be rewritten as \(-(p - 1)\):
$$
1 - p = -(p - 1)
$$

Thus, the expression becomes:
$$
\frac{5p(p - 1)}{-(p - 1)}
$$

Step 3: Cancel the common factor \((p - 1)\) in the numerator and denominator. Note that \(p \neq 1\) (since division by zero is undefined).

After canceling:
$$
\frac{5p(p - 1)}{-(p - 1)} = \frac{5p}{-1} = -5p
$$

Final Answer for Problem 1:
$$
\boxed{-5p}
$$

---

Problem 2:

$$
\frac{45}{10q - 10}
$$

Step 1: Factor the denominator.
$$
10q - 10 = 10(q - 1)
$$

So the expression becomes:
$$
\frac{45}{10(q - 1)}
$$

Step 2: Simplify the fraction by dividing the numerator and denominator by their greatest common divisor (GCD), which is 5:
$$
\frac{45}{10(q - 1)} = \frac{9}{2(q - 1)}
$$

Final Answer for Problem 2:
$$
\boxed{\frac{9}{2(q - 1)}}
$$

---

Problem 3:

$$
\frac{15y - 3}{24}
$$

Step 1: Factor the numerator.
$$
15y - 3 = 3(5y - 1)
$$

So the expression becomes:
$$
\frac{3(5y - 1)}{24}
$$

Step 2: Simplify the fraction by dividing the numerator and denominator by their GCD, which is 3:
$$
\frac{3(5y - 1)}{24} = \frac{5y - 1}{8}
$$

Final Answer for Problem 3:
$$
\boxed{\frac{5y - 1}{8}}
$$

---

Problem 4:

$$
\frac{4s - 4}{6s - 20}
$$

Step 1: Factor both the numerator and the denominator.
- Numerator: \(4s - 4 = 4(s - 1)\)
- Denominator: \(6s - 20 = 2(3s - 10)\)

So the expression becomes:
$$
\frac{4(s - 1)}{2(3s - 10)}
$$

Step 2: Simplify the fraction by dividing the numerator and denominator by their GCD, which is 2:
$$
\frac{4(s - 1)}{2(3s - 10)} = \frac{2(s - 1)}{3s - 10}
$$

Final Answer for Problem 4:
$$
\boxed{\frac{2(s - 1)}{3s - 10}}
$$

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#### Part 2: Simplify each expression and state the excluded values

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Problem 5:

$$
\frac{10x - 6}{10x - 6}
$$

Step 1: Notice that the numerator and denominator are identical. Therefore, the expression simplifies to:
$$
\frac{10x - 6}{10x - 6} = 1
$$

Step 2: Determine the excluded value. The expression is undefined when the denominator is zero:
$$
10x - 6 = 0 \implies x = \frac{6}{10} = \frac{3}{5}
$$

So the excluded value is \(x = \frac{3}{5}\).

Final Answer for Problem 5:
$$
\boxed{1 \text{ (excluded value: } x = \frac{3}{5})}
$$

---

Problem 6:

$$
\frac{2x^2 + 10x}{3x^2 + 15x}
$$

Step 1: Factor both the numerator and the denominator.
- Numerator: \(2x^2 + 10x = 2x(x + 5)\)
- Denominator: \(3x^2 + 15x = 3x(x + 5)\)

So the expression becomes:
$$
\frac{2x(x + 5)}{3x(x + 5)}
$$

Step 2: Cancel the common factors \(x\) and \((x + 5)\). Note that \(x \neq 0\) and \(x \neq -5\) (to avoid division by zero).

After canceling:
$$
\frac{2x(x + 5)}{3x(x + 5)} = \frac{2}{3}
$$

Step 3: The excluded values are \(x = 0\) and \(x = -5\).

Final Answer for Problem 6:
$$
\boxed{\frac{2}{3} \text{ (excluded values: } x = 0, x = -5)}
$$

---

Problem 7:

$$
\frac{p^2 - 3p - 10}{p^2 + p - 2}
$$

Step 1: Factor both the numerator and the denominator.
- Numerator: \(p^2 - 3p - 10 = (p - 5)(p + 2)\)
- Denominator: \(p^2 + p - 2 = (p + 2)(p - 1)\)

So the expression becomes:
$$
\frac{(p - 5)(p + 2)}{(p + 2)(p - 1)}
$$

Step 2: Cancel the common factor \((p + 2)\). Note that \(p \neq -2\) (to avoid division by zero).

After canceling:
$$
\frac{(p - 5)(p + 2)}{(p + 2)(p - 1)} = \frac{p - 5}{p - 1}
$$

Step 3: The excluded values are \(p = -2\) and \(p = 1\) (since the original denominator was zero at these points).

Final Answer for Problem 7:
$$
\boxed{\frac{p - 5}{p - 1} \text{ (excluded values: } p = -2, p = 1)}
$$

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Problem 8:

$$
\frac{l^2 + l - 6}{l^2 + 8l + 15}
$$

Step 1: Factor both the numerator and the denominator.
- Numerator: \(l^2 + l - 6 = (l + 3)(l - 2)\)
- Denominator: \(l^2 + 8l + 15 = (l + 3)(l + 5)\)

So the expression becomes:
$$
\frac{(l + 3)(l - 2)}{(l + 3)(l + 5)}
$$

Step 2: Cancel the common factor \((l + 3)\). Note that \(l \neq -3\) (to avoid division by zero).

After canceling:
$$
\frac{(l + 3)(l - 2)}{(l + 3)(l + 5)} = \frac{l - 2}{l + 5}
$$

Step 3: The excluded values are \(l = -3\) and \(l = -5\) (since the original denominator was zero at these points).

Final Answer for Problem 8:
$$
\boxed{\frac{l - 2}{l + 5} \text{ (excluded values: } l = -3, l = -5)}
$$

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Final Answers:

1. \(\boxed{-5p}\)
2. \(\boxed{\frac{9}{2(q - 1)}}\)
3. \(\boxed{\frac{5y - 1}{8}}\)
4. \(\boxed{\frac{2(s - 1)}{3s - 10}}\)
5. \(\boxed{1 \text{ (excluded value: } x = \frac{3}{5})}\)
6. \(\boxed{\frac{2}{3} \text{ (excluded values: } x = 0, x = -5)}\)
7. \(\boxed{\frac{p - 5}{p - 1} \text{ (excluded values: } p = -2, p = 1)}\)
8. \(\boxed{\frac{l - 2}{l + 5} \text{ (excluded values: } l = -3, l = -5)}\)