Examples of simplifying algebraic expressions involving fractions and variables.

Let's solve each of these algebraic expressions as simply as possible, meaning we'll simplify them by combining like terms, finding common denominators, factoring where necessary, and reducing fractions.

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1) $\frac{x}{3} + 2$

This is already simple, but we can write it with a common denominator:

$$
\frac{x}{3} + \frac{6}{3} = \frac{x + 6}{3}
$$

✔ Answer: $\boxed{\frac{x + 6}{3}}$

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2) $\frac{x}{3} - \frac{2x}{7}$

Find the least common denominator (LCD) of 3 and 7: $21$

$$
= \frac{7x}{21} - \frac{6x}{21} = \frac{x}{21}
$$

✔ Answer: $\boxed{\frac{x}{21}}$

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3) $\frac{2}{x} + \frac{3x}{5}$

These are not like terms, and the denominators are different. We can combine them over a common denominator:

LCD of $x$ and $5$ is $5x$

$$
= \frac{2 \cdot 5}{5x} + \frac{3x \cdot x}{5x} = \frac{10 + 3x^2}{5x}
$$

✔ Answer: $\boxed{\frac{3x^2 + 10}{5x}}$

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

LCD is $(x - 1)(x + 4)$

$$
= \frac{3(x + 4)}{(x - 1)(x + 4)} - \frac{2(x - 1)}{(x - 1)(x + 4)}
= \frac{3x + 12 - (2x - 2)}{(x - 1)(x + 4)}
= \frac{3x + 12 - 2x + 2}{(x - 1)(x + 4)}
= \frac{x + 14}{(x - 1)(x + 4)}
$$

✔ Answer: $\boxed{\frac{x + 14}{(x - 1)(x + 4)}}$

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

Note that both have $(x+2)$ in the denominator. LCD is $(x+2)(x-5)$

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

✔ Answer: $\boxed{\frac{x^2 - 4x + 7}{(x+2)(x - 5)}}$

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6) $\frac{y+7}{(y-5)^2} + \frac{y}{(y-5)(y+5)}$

Note:
- First denominator: $(y - 5)^2$
- Second: $(y - 5)(y + 5)$

LCD is $(y - 5)^2(y + 5)$

Now rewrite both:

First term:
$$
\frac{(y+7)(y+5)}{(y-5)^2(y+5)}
$$

Second term:
$$
\frac{y(y - 5)}{(y - 5)^2(y + 5)}
$$

Add:
$$
\frac{(y+7)(y+5) + y(y - 5)}{(y - 5)^2(y + 5)}
$$

Expand numerator:
- $(y+7)(y+5) = y^2 + 5y + 7y + 35 = y^2 + 12y + 35$
- $y(y - 5) = y^2 - 5y$

Sum: $y^2 + 12y + 35 + y^2 - 5y = 2y^2 + 7y + 35$

So:
$$
\frac{2y^2 + 7y + 35}{(y - 5)^2(y + 5)}
$$

✔ Answer: $\boxed{\frac{2y^2 + 7y + 35}{(y - 5)^2(y + 5)}}$

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7) $\frac{3q + 1}{q^2 - 8q + 15} - \frac{q + 7}{q^2 - 4q + 3}$

Factor denominators:

- $q^2 - 8q + 15 = (q - 3)(q - 5)$
- $q^2 - 4q + 3 = (q - 1)(q - 3)$

So LCD is $(q - 1)(q - 3)(q - 5)$

Now rewrite both:

First term:
$$
\frac{(3q + 1)(q - 1)}{(q - 1)(q - 3)(q - 5)}
$$

Second term:
$$
\frac{(q + 7)(q - 5)}{(q - 1)(q - 3)(q - 5)}
$$

Subtract:
$$
\frac{(3q + 1)(q - 1) - (q + 7)(q - 5)}{(q - 1)(q - 3)(q - 5)}
$$

Now expand numerator:

1. $(3q + 1)(q - 1) = 3q(q) - 3q(1) + 1(q) - 1(1) = 3q^2 - 3q + q - 1 = 3q^2 - 2q - 1$

2. $(q + 7)(q - 5) = q^2 - 5q + 7q - 35 = q^2 + 2q - 35$

Now subtract:
$$
(3q^2 - 2q - 1) - (q^2 + 2q - 35) = 3q^2 - 2q - 1 - q^2 - 2q + 35 = 2q^2 - 4q + 34
$$

So:
$$
\frac{2q^2 - 4q + 34}{(q - 1)(q - 3)(q - 5)}
$$

We can factor out 2 from numerator:
$$
= \frac{2(q^2 - 2q + 17)}{(q - 1)(q - 3)(q - 5)}
$$

No further simplification possible.

✔ Answer: $\boxed{\frac{2(q^2 - 2q + 17)}{(q - 1)(q - 3)(q - 5)}}$

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8) $\frac{b}{a + 6} + \frac{3b}{2(a + 1)}$

Different denominators: $a+6$ and $2(a+1)$

LCD is $2(a+6)(a+1)$

Rewrite:

First term:
$$
\frac{b \cdot 2(a+1)}{2(a+6)(a+1)} = \frac{2b(a+1)}{2(a+6)(a+1)}
$$

Second term:
$$
\frac{3b(a+6)}{2(a+6)(a+1)}
$$

Add:
$$
\frac{2b(a+1) + 3b(a+6)}{2(a+6)(a+1)}
$$

Expand numerator:
- $2b(a+1) = 2ab + 2b$
- $3b(a+6) = 3ab + 18b$

Sum: $2ab + 2b + 3ab + 18b = 5ab + 20b = 5b(a + 4)$

So:
$$
\frac{5b(a + 4)}{2(a+6)(a+1)}
$$

✔ Answer: $\boxed{\frac{5b(a + 4)}{2(a+6)(a+1)}}$

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✔ Final Answers Summary:

1) $\boxed{\frac{x + 6}{3}}$
2) $\boxed{\frac{x}{21}}$
3) $\boxed{\frac{3x^2 + 10}{5x}}$
4) $\boxed{\frac{x + 14}{(x - 1)(x + 4)}}$
5) $\boxed{\frac{x^2 - 4x + 7}{(x+2)(x - 5)}}$
6) $\boxed{\frac{2y^2 + 7y + 35}{(y - 5)^2(y + 5)}}$
7) $\boxed{\frac{2(q^2 - 2q + 17)}{(q - 1)(q - 3)(q - 5)}}$
8) $\boxed{\frac{5b(a + 4)}{2(a+6)(a+1)}}$

Let me know if you'd like any step explained more clearly! 🦆🦆🦆