Simplifying Rational Expressions Worksheet with six algebraic problems to solve.

Let's solve each of the rational expressions step by step. We'll simplify each expression by factoring the numerator and denominator, canceling common factors, and then writing the simplified form.

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1.

$$
\frac{2x - 4}{x - 2}
$$

Step 1: Factor numerator:
$$
2x - 4 = 2(x - 2)
$$

So,
$$
\frac{2(x - 2)}{x - 2}
$$

Step 2: Cancel common factor $ x - 2 $ (as long as $ x \ne 2 $):
$$
= 2
$$

✔ Answer: $ \boxed{2} $

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2.

$$
\frac{x^2 + 3x - 28}{x^2 - 49}
$$

Step 1: Factor both numerator and denominator.

Numerator: $ x^2 + 3x - 28 $
Find two numbers that multiply to $-28$ and add to $3$: $7$ and $-4$
$$
x^2 + 3x - 28 = (x + 7)(x - 4)
$$

Denominator: $ x^2 - 49 = (x + 7)(x - 7) $ ← difference of squares

So,
$$
\frac{(x + 7)(x - 4)}{(x + 7)(x - 7)}
$$

Step 2: Cancel $ x + 7 $ (as long as $ x \ne -7 $):
$$
= \frac{x - 4}{x - 7}
$$

✔ Answer: $ \boxed{\frac{x - 4}{x - 7}} $

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3.

$$
\frac{x + 8}{x^2 - 64}
$$

Step 1: Denominator is a difference of squares:
$$
x^2 - 64 = (x + 8)(x - 8)
$$

So,
$$
\frac{x + 8}{(x + 8)(x - 8)}
$$

Step 2: Cancel $ x + 8 $ (as long as $ x \ne -8 $):
$$
= \frac{1}{x - 8}
$$

✔ Answer: $ \boxed{\frac{1}{x - 8}} $

---

4.

$$
\frac{x^2 - 1}{x^2 + 1}
$$

Step 1: Factor numerator:
$$
x^2 - 1 = (x + 1)(x - 1)
$$

Denominator: $ x^2 + 1 $ — this cannot be factored over real numbers.

So,
$$
\frac{(x + 1)(x - 1)}{x^2 + 1}
$$

No common factors.

✔ Answer: $ \boxed{\frac{x^2 - 1}{x^2 + 1}} $ ← already simplified

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5.

$$
\frac{2x^2 - 14}{5x - 25}
$$

Step 1: Factor numerator and denominator.

Numerator: $ 2x^2 - 14 = 2(x^2 - 7) $ → cannot factor further over integers.

Wait! Let's double-check: Is $ x^2 - 7 $ factorable? No — not over rationals.

But wait: Maybe we can factor out GCFs first.

Actually:
$$
2x^2 - 14 = 2(x^2 - 7)
$$

Denominator: $ 5x - 25 = 5(x - 5) $

So,
$$
\frac{2(x^2 - 7)}{5(x - 5)}
$$

No common factors between numerator and denominator.

So, it cannot be simplified further.

But let’s check if I made a mistake.

Wait — perhaps there was a typo in the problem?

Alternatively, maybe it's meant to be $ 2x^2 - 14x $? But no, it says $ 2x^2 - 14 $.

So unless $ x^2 - 7 $ shares a factor with $ x - 5 $, which it doesn’t, we’re stuck.

✔ Answer: $ \boxed{\frac{2(x^2 - 7)}{5(x - 5)}} $ ← or leave as $ \frac{2x^2 - 14}{5x - 25} $, but factored form is better.

But actually, since no cancellation is possible, this is simplified.

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

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6.

$$
\frac{x + 6}{x^2 + 5x - 6}
$$

Step 1: Factor denominator.

We need two numbers that multiply to $-6$ and add to $5$: $6$ and $-1$

So:
$$
x^2 + 5x - 6 = (x + 6)(x - 1)
$$

So,
$$
\frac{x + 6}{(x + 6)(x - 1)}
$$

Step 2: Cancel $ x + 6 $ (as long as $ x \ne -6 $):
$$
= \frac{1}{x - 1}
$$

✔ Answer: $ \boxed{\frac{1}{x - 1}} $

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7.

$$
\frac{2x^2 + 3x + 1}{3x^2 + 5x + 2}
$$

Step 1: Factor both numerator and denominator.

Numerator: $ 2x^2 + 3x + 1 $

Find two numbers that multiply to $2 \cdot 1 = 2$, add to $3$: $2$ and $1$

Split middle term:
$$
2x^2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1)
$$

Denominator: $ 3x^2 + 5x + 2 $

Multiply $3 \cdot 2 = 6$, find two numbers that multiply to 6 and add to 5: $3$ and $2$

Split:
$$
3x^2 + 3x + 2x + 2 = 3x(x + 1) + 2(x + 1) = (3x + 2)(x + 1)
$$

So,
$$
\frac{(2x + 1)(x + 1)}{(3x + 2)(x + 1)}
$$

Cancel $ x + 1 $ (as long as $ x \ne -1 $):
$$
= \frac{2x + 1}{3x + 2}
$$

✔ Answer: $ \boxed{\frac{2x + 1}{3x + 2}} $

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✔ Final Answers:

1. $ \boxed{2} $
2. $ \boxed{\frac{x - 4}{x - 7}} $
3. $ \boxed{\frac{1}{x - 8}} $
4. $ \boxed{\frac{x^2 - 1}{x^2 + 1}} $
5. $ \boxed{\frac{2(x^2 - 7)}{5(x - 5)}} $
6. $ \boxed{\frac{1}{x - 1}} $
7. $ \boxed{\frac{2x + 1}{3x + 2}} $

> Note: Always state restrictions on variables where denominators would be zero (e.g., $ x \ne 2 $ in #1), but since the problem only asks to simplify and highlight, we focus on the simplified expressions.

Let me know if you'd like these boxed in a formatted worksheet style!