rational expressions - Free Printable

You've uploaded an image showing three complex rational expressions being simplified step-by-step. Let’s go through each one carefully, verify the steps, and explain the reasoning behind each simplification.

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## 🔹 Problem 1:

> Simplify:
> \[
\frac{\frac{x}{y} + \frac{1}{x}}{\frac{y}{x} + \frac{1}{y}}
\]

✔ Step-by-step Solution:

#### Step 1: Combine terms in numerator and denominator.

Numerator:
\[
\frac{x}{y} + \frac{1}{x} = \frac{x^2 + y}{xy}
\quad \text{(common denominator is } xy\text{)}
\]

Denominator:
\[
\frac{y}{x} + \frac{1}{y} = \frac{y^2 + x}{xy}
\quad \text{(common denominator is } xy\text{)}
\]

So the whole expression becomes:
\[
\frac{\frac{x^2 + y}{xy}}{\frac{y^2 + x}{xy}} = \frac{x^2 + y}{xy} \cdot \frac{xy}{y^2 + x}
\]

The \( xy \) cancels out:
\[
= \frac{x^2 + y}{y^2 + x}
\]

But wait — in your image, you have a different path:

You wrote:
\[
\frac{\frac{x^2}{xy} + \frac{y}{xy}}{\frac{y^2}{xy} + \frac{x}{xy}} = \frac{\frac{x^2 + y}{xy}}{\frac{y^2 + x}{xy}} = \frac{x^2 + y}{y^2 + x}
\]

✔ That’s correct so far.

Then you wrote:
\[
= \frac{x^2 + y}{xy} \cdot \frac{x}{y + 1}
\]

Wait — this is incorrect. Where did \( \frac{x}{y+1} \) come from? There's no \( y+1 \) in the original expression.

Looking again at your image:

You wrote:
\[
\frac{\frac{x^2 + y}{xy}}{\frac{y+1}{x}} = \frac{x^2 + y}{xy} \cdot \frac{x}{y+1} = \frac{(x^2 + y)x}{xy(y+1)} = \frac{(x^2 + y)}{y(y+1)}
\]

This implies you mistakenly thought the denominator was \( \frac{y+1}{x} \), but it’s actually \( \frac{y^2 + x}{xy} \).

✘ So this part is wrong.

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✔ Correct Simplification:

From earlier:

\[
\frac{\frac{x^2 + y}{xy}}{\frac{y^2 + x}{xy}} = \frac{x^2 + y}{y^2 + x}
\]

Note: \( x^2 + y \) and \( y^2 + x \) are not factorable in a way that cancels unless specific values are given. So this is the simplest form.

✔ Final Answer for Problem 1:
\[
\boxed{\frac{x^2 + y}{x + y^2}}
\]

*(Note: Order doesn't matter in addition, so \( y^2 + x = x + y^2 \))*

---

## 🔹 Problem 2:

> Simplify:
> \[
\frac{\frac{1}{x+2}}{1 + \frac{1}{x+2}}
\]

✔ Step-by-step Solution:

First, simplify the denominator:

\[
1 + \frac{1}{x+2} = \frac{x+2}{x+2} + \frac{1}{x+2} = \frac{x+3}{x+2}
\]

Now the whole expression becomes:

\[
\frac{\frac{1}{x+2}}{\frac{x+3}{x+2}} = \frac{1}{x+2} \cdot \frac{x+2}{x+3} = \frac{1}{x+3}
\]

✔ This matches your work exactly — correct!

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✔ Final Answer for Problem 2:

\[
\boxed{\frac{1}{x+3}}
\]

---

## 🔹 Problem 3:

> Simplify:
> \[
\frac{\frac{3}{x^2 y} + \frac{2}{x y^2}}{\frac{1}{x^2 y} + \frac{2}{x y^3}}
\]

✔ Step-by-step Solution:

#### Step 1: Combine numerator and denominator separately.

Numerator:
\[
\frac{3}{x^2 y} + \frac{2}{x y^2} = \frac{3y}{x^2 y^2} + \frac{2x}{x^2 y^2} = \frac{3y + 2x}{x^2 y^2}
\]

*(Common denominator: \( x^2 y^2 \))*

Denominator:
\[
\frac{1}{x^2 y} + \frac{2}{x y^3} = \frac{y^2}{x^2 y^3} + \frac{2x}{x^2 y^3} = \frac{y^2 + 2x}{x^2 y^3}
\]

*(Common denominator: \( x^2 y^3 \))*

Now the expression is:

\[
\frac{\frac{3y + 2x}{x^2 y^2}}{\frac{y^2 + 2x}{x^2 y^3}} = \frac{3y + 2x}{x^2 y^2} \cdot \frac{x^2 y^3}{y^2 + 2x}
\]

Cancel \( x^2 \):

\[
= \frac{3y + 2x}{y^2} \cdot \frac{y^3}{y^2 + 2x}
\]

Cancel \( y^2 \) with \( y^3 \):

\[
= (3y + 2x) \cdot \frac{y}{y^2 + 2x} = \frac{(3y + 2x)y}{y^2 + 2x}
\]

Note: \( 3y + 2x = 2x + 3y \), and \( y^2 + 2x = 2x + y^2 \), so we can write:

\[
= \frac{(2x + 3y)y}{2x + y^2}
\]

✔ Your image shows:
\[
\frac{(3x + 2y)y}{y^2 + 2x}
\]

Wait — you wrote 3x + 2y, but we have 3y + 2x.

That’s a mistake!

In the numerator, you had:
\[
\frac{3}{x^2 y} + \frac{2}{x y^2} → \frac{3y + 2x}{x^2 y^2}
\]

So it’s 3y + 2x, not 3x + 2y.

Therefore, your final answer has a typo: you swapped x and y in the numerator.

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✔ Correct Final Answer for Problem 3:

\[
\boxed{\frac{(2x + 3y)y}{2x + y^2}}
\]

Or equivalently:
\[
\boxed{\frac{y(2x + 3y)}{2x + y^2}}
\]

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## 📌 Summary of Corrections:

| Problem | Your Answer | Correct Answer | Notes |
|--------|-------------|----------------|-------|
| 1 | \( \frac{x^2 + y}{y(y+1)} \) | \( \frac{x^2 + y}{x + y^2} \) | Mistakenly introduced \( y+1 \); denominator was miscomputed |
| 2 | \( \frac{1}{x+3} \) | ✔ Correct | Perfect |
| 3 | \( \frac{(3x + 2y)y}{y^2 + 2x} \) | \( \frac{(2x + 3y)y}{2x + y^2} \) | Swapped x and y in numerator |

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## 💡 Tips for Avoiding Errors:

- Always double-check common denominators.
- When simplifying complex fractions, write every step clearly.
- Watch variable order — especially in multivariable expressions.

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✔ Final Correct Answers:

1. \( \boxed{\frac{x^2 + y}{x + y^2}} \)

2. \( \boxed{\frac{1}{x+3}} \)

3. \( \boxed{\frac{y(2x + 3y)}{2x + y^2}} \)

Let me know if you’d like to see graphical verification or domain restrictions too!