Simplify the given algebraic expressions in Gamma Exercise 1.

Let's solve each of these algebraic expressions step by step, simplifying them as much as possible. We'll work through each one carefully.

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1)

$$
\frac{x+4}{x-3} - \frac{x-7}{x+1}
$$

We need a common denominator: $(x - 3)(x + 1)$

Rewrite both fractions:

$$
= \frac{(x+4)(x+1)}{(x-3)(x+1)} - \frac{(x-7)(x-3)}{(x+1)(x-3)}
$$

Now subtract the numerators:

$$
= \frac{(x+4)(x+1) - (x-7)(x-3)}{(x-3)(x+1)}
$$

Expand both:

- $(x+4)(x+1) = x^2 + x + 4x + 4 = x^2 + 5x + 4$
- $(x-7)(x-3) = x^2 - 3x - 7x + 21 = x^2 - 10x + 21$

Subtract:

$$
x^2 + 5x + 4 - (x^2 - 10x + 21) = x^2 + 5x + 4 - x^2 + 10x - 21 = 15x - 17
$$

So the expression becomes:

$$
\boxed{\frac{15x - 17}{(x-3)(x+1)}}
$$

✔ Answer: $\boxed{\frac{15x - 17}{(x-3)(x+1)}}$

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2)

$$
\frac{2w+4}{w-5} - \frac{w-7}{3w+1}
$$

Factor numerator of first fraction:

$$
= \frac{2(w+2)}{w-5} - \frac{w-7}{3w+1}
$$

Common denominator: $(w - 5)(3w + 1)$

$$
= \frac{2(w+2)(3w+1) - (w-7)(w-5)}{(w-5)(3w+1)}
$$

Expand numerator:

First term: $2(w+2)(3w+1)$

- $(w+2)(3w+1) = 3w^2 + w + 6w + 2 = 3w^2 + 7w + 2$
- Multiply by 2: $6w^2 + 14w + 4$

Second term: $(w-7)(w-5) = w^2 - 5w - 7w + 35 = w^2 - 12w + 35$

Now subtract:

$$
6w^2 + 14w + 4 - (w^2 - 12w + 35) = 6w^2 + 14w + 4 - w^2 + 12w - 35 = 5w^2 + 26w - 31
$$

So:

$$
\boxed{\frac{5w^2 + 26w - 31}{(w-5)(3w+1)}}
$$

✔ Answer: $\boxed{\frac{5w^2 + 26w - 31}{(w-5)(3w+1)}}$

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3)

$$
\frac{3(q+4)}{5-q} - \frac{2(q-7)}{q+4}
$$

Note: $5 - q = -(q - 5)$, but let’s keep it as is for now.

We can write $5 - q = -(q - 5)$, but perhaps better to just use common denominator.

Denominators: $5 - q$ and $q + 4$

Common denominator: $(5 - q)(q + 4)$

But note: $5 - q = -(q - 5)$, so we can write:

$$
= \frac{3(q+4)}{5-q} - \frac{2(q-7)}{q+4}
= \frac{3(q+4)}{-(q-5)} - \frac{2(q-7)}{q+4}
= -\frac{3(q+4)}{q-5} - \frac{2(q-7)}{q+4}
$$

But maybe easier to proceed directly.

Use common denominator: $(5 - q)(q + 4)$

So:

$$
= \frac{3(q+4)(q+4) - 2(q-7)(5-q)}{(5-q)(q+4)}
$$

Wait — careful: first term is $\frac{3(q+4)}{5-q}$, so numerator is $3(q+4)$ times $(q+4)$? No!

Actually:

$$
\frac{A}{B} - \frac{C}{D} = \frac{A \cdot D - C \cdot B}{B \cdot D}
$$

So:

$$
= \frac{3(q+4)(q+4) - 2(q-7)(5-q)}{(5-q)(q+4)}
$$

Wait — no! First fraction: numerator is $3(q+4)$, denominator $5 - q$

Second: numerator $2(q-7)$, denominator $q+4$

So:

$$
= \frac{3(q+4) \cdot (q+4) - 2(q-7) \cdot (5 - q)}{(5 - q)(q + 4)}
$$

Yes:

$$
= \frac{3(q+4)^2 - 2(q-7)(5 - q)}{(5 - q)(q + 4)}
$$

Now expand:

- $(q+4)^2 = q^2 + 8q + 16$, so $3(q^2 + 8q + 16) = 3q^2 + 24q + 48$

- $(q-7)(5 - q) = q(5 - q) - 7(5 - q) = 5q - q^2 - 35 + 7q = -q^2 + 12q - 35$

Then multiply by 2: $-2q^2 + 24q - 70$

Now subtract: $3q^2 + 24q + 48 - (-2q^2 + 24q - 70)$

Wait — in the formula: it's $A D - C B$, so:

$$
3(q+4)^2 - 2(q-7)(5 - q)
= (3q^2 + 24q + 48) - (-2q^2 + 24q - 70)
= 3q^2 + 24q + 48 + 2q^2 - 24q + 70
= 5q^2 + 118
$$

So numerator: $5q^2 + 118$

Denominator: $(5 - q)(q + 4)$

So:

$$
\boxed{\frac{5q^2 + 118}{(5 - q)(q + 4)}}
$$

Alternatively, write $5 - q = -(q - 5)$, so:

$$
= \frac{5q^2 + 118}{-(q - 5)(q + 4)} = -\frac{5q^2 + 118}{(q - 5)(q + 4)}
$$

But unless factoring helps, leave as:

✔ Answer: $\boxed{\frac{5q^2 + 118}{(5 - q)(q + 4)}}$

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4)

$$
\frac{2(1 - x)}{x + 2} + \frac{2x}{x + 7}
$$

Simplify first: $2(1 - x) = 2 - 2x$, so:

$$
= \frac{2 - 2x}{x + 2} + \frac{2x}{x + 7}
$$

Factor numerator: $2(1 - x)$

But $1 - x = -(x - 1)$, so:

$$
= \frac{-2(x - 1)}{x + 2} + \frac{2x}{x + 7}
$$

Now common denominator: $(x + 2)(x + 7)$

$$
= \frac{-2(x - 1)(x + 7) + 2x(x + 2)}{(x + 2)(x + 7)}
$$

Expand numerator:

First: $-2(x - 1)(x + 7)$

- $(x - 1)(x + 7) = x^2 + 7x - x - 7 = x^2 + 6x - 7$
- Multiply by -2: $-2x^2 - 12x + 14$

Second: $2x(x + 2) = 2x^2 + 4x$

Add together:

$$
(-2x^2 - 12x + 14) + (2x^2 + 4x) = (-2x^2 + 2x^2) + (-12x + 4x) + 14 = -8x + 14
$$

So:

$$
\boxed{\frac{-8x + 14}{(x + 2)(x + 7)}}
$$

Factor numerator: $-2(4x - 7)$ or $2(-4x + 7)$

So:

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

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

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5)

$$
\frac{5 - x}{y + 3} + \frac{2y}{x}
$$

Different variables — no common factors.

Common denominator: $x(y + 3)$

$$
= \frac{(5 - x)x + 2y(y + 3)}{x(y + 3)}
$$

Expand numerator:

- $(5 - x)x = 5x - x^2$
- $2y(y + 3) = 2y^2 + 6y$

Add: $5x - x^2 + 2y^2 + 6y$

Arrange: $-x^2 + 5x + 2y^2 + 6y$

So:

$$
\boxed{\frac{-x^2 + 5x + 2y^2 + 6y}{x(y + 3)}}
$$

No further simplification possible.

✔ Answer: $\boxed{\frac{-x^2 + 5x + 2y^2 + 6y}{x(y + 3)}}$

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6)

$$
\frac{3(a+4)}{b+2} - \frac{2(a-7)}{b+4}
$$

Different denominators: $b+2$, $b+4$

Common denominator: $(b+2)(b+4)$

$$
= \frac{3(a+4)(b+4) - 2(a-7)(b+2)}{(b+2)(b+4)}
$$

Expand numerator:

First: $3(a+4)(b+4)$

- $(a+4)(b+4) = ab + 4a + 4b + 16$
- Multiply by 3: $3ab + 12a + 12b + 48$

Second: $2(a-7)(b+2)$

- $(a-7)(b+2) = ab + 2a - 7b - 14$
- Multiply by 2: $2ab + 4a - 14b - 28$

Now subtract:

$$
(3ab + 12a + 12b + 48) - (2ab + 4a - 14b - 28)
= 3ab - 2ab + 12a - 4a + 12b + 14b + 48 + 28
= ab + 8a + 26b + 76
$$

So:

$$
\boxed{\frac{ab + 8a + 26b + 76}{(b+2)(b+4)}}
$$

Can this be factored? Let's see:

Numerator: $ab + 8a + 26b + 76$

Try grouping:

= $a(b + 8) + 2(13b + 38)$ — doesn't help.

Or: $a(b + 8) + 26b + 76$

Not factorable easily.

So leave as is.

✔ Answer: $\boxed{\frac{ab + 8a + 26b + 76}{(b+2)(b+4)}}$

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✔ Final Answers Summary:

1) $\boxed{\frac{15x - 17}{(x-3)(x+1)}}$
2) $\boxed{\frac{5w^2 + 26w - 31}{(w-5)(3w+1)}}$
3) $\boxed{\frac{5q^2 + 118}{(5 - q)(q + 4)}}$
4) $\boxed{\frac{2(7 - 4x)}{(x + 2)(x + 7)}}$
5) $\boxed{\frac{-x^2 + 5x + 2y^2 + 6y}{x(y + 3)}}$
6) $\boxed{\frac{ab + 8a + 26b + 76}{(b+2)(b+4)}}$

Let me know if you'd like any of them checked further!