Collection of algebraic equations with fractions and variables for solving.

Let's solve each of these rational equations step by step. We'll go through them one at a time, showing the algebraic steps and identifying any restrictions (values that make denominators zero). We'll also check for extraneous solutions.

---

a) $\frac{1}{x+1} = \frac{1}{2x}$

Step 1: Cross-multiply:
$$
1 \cdot 2x = 1 \cdot (x + 1)
\Rightarrow 2x = x + 1
$$

Step 2: Solve:
$$
2x - x = 1 \Rightarrow x = 1
$$

Check restrictions: $x \ne -1$, $x \ne 0$ → $x=1$ is valid.

✔ Answer: $x = 1$

---

b) $\frac{x-4}{x-8} = 5$

Step 1: Multiply both sides by $(x - 8)$:
$$
x - 4 = 5(x - 8)
\Rightarrow x - 4 = 5x - 40
$$

Step 2: Solve:
$$
-4 + 40 = 5x - x \Rightarrow 36 = 4x \Rightarrow x = 9
$$

Check restriction: $x \ne 8$ → $x = 9$ is valid.

✔ Answer: $x = 9$

---

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

Step 1: Cross-multiply:
$$
2(3x - 4) = 1(4x - 3)
\Rightarrow 6x - 8 = 4x - 3
$$

Step 2: Solve:
$$
6x - 4x = -3 + 8 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}
$$

Check restriction: $x \ne \frac{3}{4}$ → $x = \frac{5}{2}$ is valid.

✔ Answer: $x = \frac{5}{2}$

---

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

Note: $5 - x = -(x - 5)$, so rewrite:

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

Now combine:
$$
\frac{3}{x - 5} + 2 = -\frac{5}{x - 5}
\Rightarrow \frac{3 + 5}{x - 5} = -2
\Rightarrow \frac{8}{x - 5} = -2
$$

Multiply both sides:
$$
8 = -2(x - 5) \Rightarrow 8 = -2x + 10 \Rightarrow -2 = -2x \Rightarrow x = 1
$$

Check restriction: $x \ne 5$ → $x = 1$ is valid.

✔ Answer: $x = 1$

---

e) $\frac{2t + 3}{3t + 1} - \frac{t + 5}{3t + 1} = \frac{1}{4}$

Same denominator:
$$
\frac{(2t + 3) - (t + 5)}{3t + 1} = \frac{1}{4}
\Rightarrow \frac{2t + 3 - t - 5}{3t + 1} = \frac{1}{4}
\Rightarrow \frac{t - 2}{3t + 1} = \frac{1}{4}
$$

Cross-multiply:
$$
4(t - 2) = 1(3t + 1) \Rightarrow 4t - 8 = 3t + 1 \Rightarrow t = 9
$$

Check restriction: $t \ne -\frac{1}{3}$ → $t = 9$ is valid.

✔ Answer: $t = 9$

---

f) $\frac{1}{x - 3} + \frac{2}{x - 4} = \frac{5}{(x - 3)(x - 4)}$

Common denominator: $(x - 3)(x - 4)$

Multiply both sides:
$$
(x - 4) + 2(x - 3) = 5
\Rightarrow x - 4 + 2x - 6 = 5
\Rightarrow 3x - 10 = 5 \Rightarrow 3x = 15 \Rightarrow x = 5
$$

Restrictions: $x \ne 3, 4$ → $x = 5$ is valid.

✔ Answer: $x = 5$

---

g) $\frac{1}{3} - \frac{23 - x}{3x} = \frac{7}{12} - \frac{1}{4x} - \frac{7}{x}$

Let’s simplify both sides.

Left side:
$$
\frac{1}{3} - \frac{23 - x}{3x} = \frac{1}{3} - \left(\frac{23}{3x} - \frac{x}{3x}\right) = \frac{1}{3} - \frac{23}{3x} + \frac{1}{3}
= \frac{2}{3} - \frac{23}{3x}
$$

Right side:
$$
\frac{7}{12} - \frac{1}{4x} - \frac{7}{x} = \frac{7}{12} - \left(\frac{1}{4x} + \frac{28}{4x}\right) = \frac{7}{12} - \frac{29}{4x}
$$

Now set equal:
$$
\frac{2}{3} - \frac{23}{3x} = \frac{7}{12} - \frac{29}{4x}
$$

Multiply all terms by LCM of denominators: $12x$

$$
12x \cdot \left(\frac{2}{3}\right) - 12x \cdot \left(\frac{23}{3x}\right) = 12x \cdot \left(\frac{7}{12}\right) - 12x \cdot \left(\frac{29}{4x}\right)
\Rightarrow 8x - 92 = 7x - 87
$$

Solve:
$$
8x - 7x = -87 + 92 \Rightarrow x = 5
$$

Check restrictions: $x \ne 0$ → valid.

✔ Answer: $x = 5$

---

h) $\frac{x + 7}{2x + 2} = 1 + \frac{x + 4}{4x + 4}$

Factor denominators:
$$
\frac{x + 7}{2(x + 1)} = 1 + \frac{x + 4}{4(x + 1)}
$$

Multiply both sides by $4(x + 1)$ to eliminate denominators:

Left:
$$
4(x + 1) \cdot \frac{x + 7}{2(x + 1)} = 2(x + 7) = 2x + 14
$$

Right:
$$
4(x + 1) \cdot 1 + 4(x + 1) \cdot \frac{x + 4}{4(x + 1)} = 4(x + 1) + (x + 4) = 4x + 4 + x + 4 = 5x + 8
$$

So:
$$
2x + 14 = 5x + 8 \Rightarrow 14 - 8 = 5x - 2x \Rightarrow 6 = 3x \Rightarrow x = 2
$$

Check restriction: $x \ne -1$ → $x = 2$ is valid.

✔ Answer: $x = 2$

---

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

Move all terms to left:
$$
\frac{1 + x}{x - 1} - \frac{3 + x}{x + 1} - \frac{4}{x + 1} = 0
\Rightarrow \frac{1 + x}{x - 1} - \left(\frac{3 + x + 4}{x + 1}\right) = 0
\Rightarrow \frac{1 + x}{x - 1} - \frac{x + 7}{x + 1} = 0
$$

So:
$$
\frac{1 + x}{x - 1} = \frac{x + 7}{x + 1}
$$

Cross-multiply:
$$
(1 + x)(x + 1) = (x + 7)(x - 1)
$$

Left: $(x + 1)^2 = x^2 + 2x + 1$

Right: $x^2 - x + 7x - 7 = x^2 + 6x - 7$

Set equal:
$$
x^2 + 2x + 1 = x^2 + 6x - 7 \Rightarrow 2x + 1 = 6x - 7 \Rightarrow 8 = 4x \Rightarrow x = 2
$$

Check restrictions: $x \ne 1, -1$ → $x = 2$ is valid.

✔ Answer: $x = 2$

---

j) $\frac{12 - 7y}{y - 1} = \frac{4}{y + 1} - 7$

Move $-7$ to left:
$$
\frac{12 - 7y}{y - 1} + 7 = \frac{4}{y + 1}
$$

Write $7$ as $\frac{7(y - 1)}{y - 1}$:
$$
\frac{12 - 7y + 7(y - 1)}{y - 1} = \frac{4}{y + 1}
\Rightarrow \frac{12 - 7y + 7y - 7}{y - 1} = \frac{4}{y + 1}
\Rightarrow \frac{5}{y - 1} = \frac{4}{y + 1}
$$

Cross-multiply:
$$
5(y + 1) = 4(y - 1) \Rightarrow 5y + 5 = 4y - 4 \Rightarrow y = -9
$$

Check restrictions: $y \ne 1, -1$ → $y = -9$ is valid.

✔ Answer: $y = -9$

---

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

Find common denominator: $(x - 3)(x - 2)(x - 4)$

Multiply both sides by this:

Left:
$$
4(x - 2)(x - 4) - 3(x - 3)(x - 4) = (x - 3)(x - 2)
$$

Compute each term:

First term: $4(x^2 - 6x + 8) = 4x^2 - 24x + 32$

Second term: $3(x^2 - 7x + 12) = 3x^2 - 21x + 36$

So left side:
$$
(4x^2 - 24x + 32) - (3x^2 - 21x + 36) = x^2 - 3x - 4
$$

Right side: $(x - 3)(x - 2) = x^2 - 5x + 6$

Set equal:
$$
x^2 - 3x - 4 = x^2 - 5x + 6 \Rightarrow -3x - 4 = -5x + 6 \Rightarrow 2x = 10 \Rightarrow x = 5
$$

Check restrictions: $x \ne 3, 2, 4$ → $x = 5$ is valid.

✔ Answer: $x = 5$

---

l) $2 - \frac{4x + 3}{x + x^2} = \frac{2x - 5}{x + 1} - \frac{5}{x}$

Note: $x + x^2 = x(x + 1)$

So:
$$
2 - \frac{4x + 3}{x(x + 1)} = \frac{2x - 5}{x + 1} - \frac{5}{x}
$$

Multiply both sides by $x(x + 1)$:

Left:
$$
2x(x + 1) - (4x + 3) = 2x^2 + 2x - 4x - 3 = 2x^2 - 2x - 3
$$

Right:
$$
x(2x - 5) - 5(x + 1) = 2x^2 - 5x - 5x - 5 = 2x^2 - 10x - 5
$$

Set equal:
$$
2x^2 - 2x - 3 = 2x^2 - 10x - 5 \Rightarrow -2x - 3 = -10x - 5 \Rightarrow 8x = -2 \Rightarrow x = -\frac{1}{4}
$$

Check restrictions: $x \ne 0, -1$ → $x = -\frac{1}{4}$ is valid.

✔ Answer: $x = -\frac{1}{4}$

---

m) $\frac{y + 5}{y - 7} - 2 = \frac{y + 7}{y - 5}$

Move $-2$ to right:
$$
\frac{y + 5}{y - 7} = \frac{y + 7}{y - 5} + 2
$$

Write $2 = \frac{2(y - 5)}{y - 5}$:
$$
\frac{y + 5}{y - 7} = \frac{y + 7 + 2(y - 5)}{y - 5} = \frac{y + 7 + 2y - 10}{y - 5} = \frac{3y - 3}{y - 5}
$$

Now cross-multiply:
$$
(y + 5)(y - 5) = (3y - 3)(y - 7)
$$

Left: $y^2 - 25$

Right: $3y(y - 7) - 3(y - 7) = 3y^2 - 21y - 3y + 21 = 3y^2 - 24y + 21$

So:
$$
y^2 - 25 = 3y^2 - 24y + 21 \Rightarrow 0 = 2y^2 - 24y + 46
\Rightarrow y^2 - 12y + 23 = 0
$$

Use quadratic formula:
$$
y = \frac{12 \pm \sqrt{144 - 92}}{2} = \frac{12 \pm \sqrt{52}}{2} = \frac{12 \pm 2\sqrt{13}}{2} = 6 \pm \sqrt{13}
$$

Check restrictions: $y \ne 7, 5$ → both roots are valid.

✔ Answer: $y = 6 \pm \sqrt{13}$

---

n) $1 - \frac{5}{2y + 6} = \frac{3}{y + 3}$

Note: $2y + 6 = 2(y + 3)$

So:
$$
1 - \frac{5}{2(y + 3)} = \frac{3}{y + 3}
$$

Multiply both sides by $2(y + 3)$:

Left:
$$
2(y + 3) - 5 = 2y + 6 - 5 = 2y + 1
$$

Right:
$$
2(y + 3) \cdot \frac{3}{y + 3} = 6
$$

So:
$$
2y + 1 = 6 \Rightarrow 2y = 5 \Rightarrow y = \frac{5}{2}
$$

Check restriction: $y \ne -3$ → valid.

✔ Answer: $y = \frac{5}{2}$

---

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

Factor:
- $2x + 2 = 2(x + 1)$
- $4x + 4 = 4(x + 1)$

So:
$$
\frac{x + 2}{2(x + 1)} - \frac{1}{2} = -\frac{x + 4}{4(x + 1)}
$$

Multiply both sides by $4(x + 1)$:

Left:
$$
4(x + 1) \cdot \frac{x + 2}{2(x + 1)} = 2(x + 2) = 2x + 4
$$
$$
4(x + 1) \cdot \frac{1}{2} = 2(x + 1) = 2x + 2
$$
So total left: $2x + 4 - (2x + 2) = 2$

Right:
$$
- \frac{x + 4}{4(x + 1)} \cdot 4(x + 1) = -(x + 4)
$$

So:
$$
2 = -x - 4 \Rightarrow x = -6
$$

Check restriction: $x \ne -1$ → valid.

✔ Answer: $x = -6$

---

p) $\frac{z - 5}{z} + \frac{3z + 1}{z - 3} = 4$

Multiply both sides by $z(z - 3)$:

Left:
$$
(z - 5)(z - 3) + z(3z + 1) = 4z(z - 3)
$$

Expand:

First: $(z - 5)(z - 3) = z^2 - 8z + 15$

Second: $z(3z + 1) = 3z^2 + z$

Sum: $z^2 - 8z + 15 + 3z^2 + z = 4z^2 - 7z + 15$

Right: $4z^2 - 12z$

Set equal:
$$
4z^2 - 7z + 15 = 4z^2 - 12z \Rightarrow -7z + 15 = -12z \Rightarrow 5z = -15 \Rightarrow z = -3
$$

Check restrictions: $z \ne 0, 3$ → $z = -3$ is valid.

✔ Answer: $z = -3$

---

q) $\frac{2h + 1}{h} = \frac{6h}{3h - 1}$

Cross-multiply:
$$
(2h + 1)(3h - 1) = 6h \cdot h = 6h^2
$$

Left:
$$
6h^2 - 2h + 3h - 1 = 6h^2 + h - 1
$$

Set equal:
$$
6h^2 + h - 1 = 6h^2 \Rightarrow h - 1 = 0 \Rightarrow h = 1
$$

Check restrictions: $h \ne 0, \frac{1}{3}$ → $h = 1$ is valid.

✔ Answer: $h = 1$

---

r) $\frac{2x + 5}{6} + \frac{10}{x - 3} = \frac{2x - 3}{6}$

Subtract $\frac{2x + 5}{6}$ from both sides:
$$
\frac{10}{x - 3} = \frac{2x - 3 - (2x + 5)}{6} = \frac{-8}{6} = -\frac{4}{3}
$$

So:
$$
\frac{10}{x - 3} = -\frac{4}{3}
\Rightarrow 10 \cdot 3 = -4(x - 3) \Rightarrow 30 = -4x + 12 \Rightarrow 18 = -4x \Rightarrow x = -\frac{9}{2}
$$

Check restriction: $x \ne 3$ → valid.

✔ Answer: $x = -\frac{9}{2}$

---

s) $\frac{11}{x^2 - 5x} + \frac{6}{x^2 + 4x} + \frac{4}{(x + 4)(x - 5)} = 0$

Factor denominators:

- $x^2 - 5x = x(x - 5)$
- $x^2 + 4x = x(x + 4)$
- $(x + 4)(x - 5)$

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

Rewrite each term:

1. $\frac{11}{x(x - 5)} = \frac{11(x + 4)}{x(x + 4)(x - 5)}$
2. $\frac{6}{x(x + 4)} = \frac{6(x - 5)}{x(x + 4)(x - 5)}$
3. $\frac{4}{(x + 4)(x - 5)} = \frac{4x}{x(x + 4)(x - 5)}$

Add numerators:
$$
11(x + 4) + 6(x - 5) + 4x = 11x + 44 + 6x - 30 + 4x = 21x + 14
$$

Set numerator = 0:
$$
21x + 14 = 0 \Rightarrow x = -\frac{2}{3}
$$

Check restrictions: $x \ne 0, -4, 5$ → $x = -\frac{2}{3}$ is valid.

✔ Answer: $x = -\frac{2}{3}$

---

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

Note: $5 + x = x + 5$

So:
$$
\frac{x + 7}{x - 5} - 2 = \frac{x + 5}{x - 7}
$$

Move $-2$ to right:
$$
\frac{x + 7}{x - 5} = \frac{x + 5}{x - 7} + 2
$$

Write $2 = \frac{2(x - 7)}{x - 7}$:
$$
\frac{x + 7}{x - 5} = \frac{x + 5 + 2x - 14}{x - 7} = \frac{3x - 9}{x - 7}
$$

Cross-multiply:
$$
(x + 7)(x - 7) = (3x - 9)(x - 5)
$$

Left: $x^2 - 49$

Right: $3x(x - 5) - 9(x - 5) = 3x^2 - 15x - 9x + 45 = 3x^2 - 24x + 45$

Set equal:
$$
x^2 - 49 = 3x^2 - 24x + 45 \Rightarrow 0 = 2x^2 - 24x + 94
\Rightarrow x^2 - 12x + 47 = 0
$$

Discriminant: $144 - 188 = -44 < 0$ → no real solution.

✘ No real solution

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| a) | $x = 1$ |
| b) | $x = 9$ |
| c) | $x = \frac{5}{2}$ |
| d) | $x = 1$ |
| e) | $t = 9$ |
| f) | $x = 5$ |
| g) | $x = 5$ |
| h) | $x = 2$ |
| i) | $x = 2$ |
| j) | $y = -9$ |
| k) | $x = 5$ |
| l) | $x = -\frac{1}{4}$ |
| m) | $y = 6 \pm \sqrt{13}$ |
| n) | $y = \frac{5}{2}$ |
| o) | $x = -6$ |
| p) | $z = -3$ |
| q) | $h = 1$ |
| r) | $x = -\frac{9}{2}$ |
| s) | $x = -\frac{2}{3}$ |
| t) | No real solution |

Let me know if you'd like a visual explanation or graphing for any!