Worksheet-3 on simple equations with 15 problems to solve, including rational expressions and instructions to check for extraneous solutions.

Problem: Solve each equation and check for extraneous solutions.

We will solve the equations step by step. Let's start with the first equation and proceed systematically.

---

#### 1. Solve the equation:
\[
\frac{5}{4n} + \frac{1}{4} = \frac{1}{4n}
\]

Step 1: Eliminate the denominators by finding a common denominator, which is \(4n\):
\[
\frac{5}{4n} + \frac{n}{4n} = \frac{1}{4n}
\]

Step 2: Combine the terms on the left-hand side:
\[
\frac{5 + n}{4n} = \frac{1}{4n}
\]

Step 3: Since the denominators are the same, equate the numerators:
\[
5 + n = 1
\]

Step 4: Solve for \(n\):
\[
n = 1 - 5 \implies n = -4
\]

Step 5: Check for extraneous solutions. Substitute \(n = -4\) back into the original equation:
\[
\frac{5}{4(-4)} + \frac{1}{4} = \frac{1}{4(-4)}
\]
\[
\frac{5}{-16} + \frac{1}{4} = \frac{1}{-16}
\]
\[
-\frac{5}{16} + \frac{4}{16} = -\frac{1}{16}
\]
\[
-\frac{1}{16} = -\frac{1}{16}
\]
The solution is valid.

Final Answer:
\[
\boxed{n = -4}
\]

---

#### 2. Solve the equation:
\[
\frac{1}{6x^2} = \frac{1}{6x} + \frac{1}{3x^2}
\]

Step 1: Find a common denominator, which is \(6x^2\):
\[
\frac{1}{6x^2} = \frac{x}{6x^2} + \frac{2}{6x^2}
\]

Step 2: Combine the terms on the right-hand side:
\[
\frac{1}{6x^2} = \frac{x + 2}{6x^2}
\]

Step 3: Since the denominators are the same, equate the numerators:
\[
1 = x + 2
\]

Step 4: Solve for \(x\):
\[
x = 1 - 2 \implies x = -1
\]

Step 5: Check for extraneous solutions. Substitute \(x = -1\) back into the original equation:
\[
\frac{1}{6(-1)^2} = \frac{1}{6(-1)} + \frac{1}{3(-1)^2}
\]
\[
\frac{1}{6} = -\frac{1}{6} + \frac{1}{3}
\]
\[
\frac{1}{6} = -\frac{1}{6} + \frac{2}{6}
\]
\[
\frac{1}{6} = \frac{1}{6}
\]
The solution is valid.

Final Answer:
\[
\boxed{x = -1}
\]

---

#### 3. Solve the equation:
\[
\frac{1}{6b^2} + \frac{b - 4}{6b^2} = \frac{1}{b^2}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{1 + b - 4}{6b^2} = \frac{1}{b^2}
\]
\[
\frac{b - 3}{6b^2} = \frac{1}{b^2}
\]

Step 2: Eliminate the denominators by multiplying through by \(6b^2\):
\[
b - 3 = 6
\]

Step 3: Solve for \(b\):
\[
b = 6 + 3 \implies b = 9
\]

Step 4: Check for extraneous solutions. Substitute \(b = 9\) back into the original equation:
\[
\frac{1}{6(9)^2} + \frac{9 - 4}{6(9)^2} = \frac{1}{(9)^2}
\]
\[
\frac{1}{486} + \frac{5}{486} = \frac{1}{81}
\]
\[
\frac{6}{486} = \frac{1}{81}
\]
\[
\frac{1}{81} = \frac{1}{81}
\]
The solution is valid.

Final Answer:
\[
\boxed{b = 9}
\]

---

#### 4. Solve the equation:
\[
\frac{4}{x^2} = \frac{5}{3x} - \frac{1}{x^2}
\]

Step 1: Combine the terms involving \(x^2\) on one side:
\[
\frac{4}{x^2} + \frac{1}{x^2} = \frac{5}{3x}
\]
\[
\frac{5}{x^2} = \frac{5}{3x}
\]

Step 2: Eliminate the denominators by multiplying through by \(3x^2\):
\[
3x \cdot 5 = 5x^2
\]
\[
15x = 5x^2
\]

Step 3: Rearrange the equation:
\[
5x^2 - 15x = 0
\]
\[
5x(x - 3) = 0
\]

Step 4: Solve for \(x\):
\[
x = 0 \quad \text{or} \quad x = 3
\]

Step 5: Check for extraneous solutions. \(x = 0\) is not valid because it makes the denominators zero in the original equation. Substitute \(x = 3\):
\[
\frac{4}{(3)^2} = \frac{5}{3(3)} - \frac{1}{(3)^2}
\]
\[
\frac{4}{9} = \frac{5}{9} - \frac{1}{9}
\]
\[
\frac{4}{9} = \frac{4}{9}
\]
The solution is valid.

Final Answer:
\[
\boxed{x = 3}
\]

---

#### 5. Solve the equation:
\[
\frac{4r + 16}{r^2} + \frac{1}{r} = \frac{6}{r}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{4r + 16}{r^2} + \frac{r}{r^2} = \frac{6}{r}
\]
\[
\frac{4r + 16 + r}{r^2} = \frac{6}{r}
\]
\[
\frac{5r + 16}{r^2} = \frac{6}{r}
\]

Step 2: Eliminate the denominators by multiplying through by \(r^2\):
\[
5r + 16 = 6r
\]

Step 3: Solve for \(r\):
\[
16 = 6r - 5r \implies r = 16
\]

Step 4: Check for extraneous solutions. Substitute \(r = 16\) back into the original equation:
\[
\frac{4(16) + 16}{(16)^2} + \frac{1}{16} = \frac{6}{16}
\]
\[
\frac{64 + 16}{256} + \frac{1}{16} = \frac{6}{16}
\]
\[
\frac{80}{256} + \frac{1}{16} = \frac{6}{16}
\]
\[
\frac{5}{16} + \frac{1}{16} = \frac{6}{16}
\]
\[
\frac{6}{16} = \frac{6}{16}
\]
The solution is valid.

Final Answer:
\[
\boxed{r = 16}
\]

---

#### 6. Solve the equation:
\[
\frac{1}{n^2} = \frac{n + 5}{3n^2} + \frac{1}{3n^2}
\]

Step 1: Combine the terms on the right-hand side:
\[
\frac{1}{n^2} = \frac{n + 5 + 1}{3n^2}
\]
\[
\frac{1}{n^2} = \frac{n + 6}{3n^2}
\]

Step 2: Eliminate the denominators by multiplying through by \(3n^2\):
\[
3 = n + 6
\]

Step 3: Solve for \(n\):
\[
n = 3 - 6 \implies n = -3
\]

Step 4: Check for extraneous solutions. Substitute \(n = -3\) back into the original equation:
\[
\frac{1}{(-3)^2} = \frac{-3 + 5}{3(-3)^2} + \frac{1}{3(-3)^2}
\]
\[
\frac{1}{9} = \frac{2}{27} + \frac{1}{27}
\]
\[
\frac{1}{9} = \frac{3}{27}
\]
\[
\frac{1}{9} = \frac{1}{9}
\]
The solution is valid.

Final Answer:
\[
\boxed{n = -3}
\]

---

#### 7. Solve the equation:
\[
\frac{1}{3b} = \frac{b + 2}{6b^2} + \frac{1}{3b^2}
\]

Step 1: Find a common denominator, which is \(6b^2\):
\[
\frac{2b}{6b^2} = \frac{b + 2}{6b^2} + \frac{2}{6b^2}
\]

Step 2: Combine the terms on the right-hand side:
\[
\frac{2b}{6b^2} = \frac{b + 2 + 2}{6b^2}
\]
\[
\frac{2b}{6b^2} = \frac{b + 4}{6b^2}
\]

Step 3: Since the denominators are the same, equate the numerators:
\[
2b = b + 4
\]

Step 4: Solve for \(b\):
\[
2b - b = 4 \implies b = 4
\]

Step 5: Check for extraneous solutions. Substitute \(b = 4\) back into the original equation:
\[
\frac{1}{3(4)} = \frac{4 + 2}{6(4)^2} + \frac{1}{3(4)^2}
\]
\[
\frac{1}{12} = \frac{6}{96} + \frac{1}{48}
\]
\[
\frac{1}{12} = \frac{1}{16} + \frac{1}{48}
\]
\[
\frac{1}{12} = \frac{3}{48} + \frac{1}{48}
\]
\[
\frac{1}{12} = \frac{4}{48}
\]
\[
\frac{1}{12} = \frac{1}{12}
\]
The solution is valid.

Final Answer:
\[
\boxed{b = 4}
\]

---

#### 8. Solve the equation:
\[
\frac{v + 4}{v^2} + \frac{1}{v} = \frac{3v + 9}{v^2}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{v + 4}{v^2} + \frac{v}{v^2} = \frac{3v + 9}{v^2}
\]
\[
\frac{v + 4 + v}{v^2} = \frac{3v + 9}{v^2}
\]
\[
\frac{2v + 4}{v^2} = \frac{3v + 9}{v^2}
\]

Step 2: Since the denominators are the same, equate the numerators:
\[
2v + 4 = 3v + 9
\]

Step 3: Solve for \(v\):
\[
2v - 3v = 9 - 4 \implies -v = 5 \implies v = -5
\]

Step 4: Check for extraneous solutions. Substitute \(v = -5\) back into the original equation:
\[
\frac{-5 + 4}{(-5)^2} + \frac{1}{-5} = \frac{3(-5) + 9}{(-5)^2}
\]
\[
\frac{-1}{25} + \frac{1}{-5} = \frac{-15 + 9}{25}
\]
\[
\frac{-1}{25} - \frac{5}{25} = \frac{-6}{25}
\]
\[
\frac{-6}{25} = \frac{-6}{25}
\]
The solution is valid.

Final Answer:
\[
\boxed{v = -5}
\]

---

#### 9. Solve the equation:
\[
\frac{6}{5x} - \frac{x + 4}{x^2} = \frac{1}{x}
\]

Step 1: Find a common denominator, which is \(5x^2\):
\[
\frac{6x}{5x^2} - \frac{5(x + 4)}{5x^2} = \frac{5x}{5x^2}
\]
\[
\frac{6x - 5(x + 4)}{5x^2} = \frac{5x}{5x^2}
\]
\[
\frac{6x - 5x - 20}{5x^2} = \frac{5x}{5x^2}
\]
\[
\frac{x - 20}{5x^2} = \frac{5x}{5x^2}
\]

Step 2: Since the denominators are the same, equate the numerators:
\[
x - 20 = 5x
\]

Step 3: Solve for \(x\):
\[
x - 5x = 20 \implies -4x = 20 \implies x = -5
\]

Step 4: Check for extraneous solutions. Substitute \(x = -5\) back into the original equation:
\[
\frac{6}{5(-5)} - \frac{-5 + 4}{(-5)^2} = \frac{1}{-5}
\]
\[
\frac{6}{-25} - \frac{-1}{25} = \frac{1}{-5}
\]
\[
-\frac{6}{25} + \frac{1}{25} = -\frac{1}{5}
\]
\[
-\frac{5}{25} = -\frac{1}{5}
\]
\[
-\frac{1}{5} = -\frac{1}{5}
\]
The solution is valid.

Final Answer:
\[
\boxed{x = -5}
\]

---

#### 10. Solve the equation:
\[
\frac{2x - 12}{5x^2} = \frac{1}{5x} - \frac{x - 6}{x^2}
\]

Step 1: Find a common denominator, which is \(5x^2\):
\[
\frac{2x - 12}{5x^2} = \frac{x}{5x^2} - \frac{5(x - 6)}{5x^2}
\]
\[
\frac{2x - 12}{5x^2} = \frac{x - 5(x - 6)}{5x^2}
\]
\[
\frac{2x - 12}{5x^2} = \frac{x - 5x + 30}{5x^2}
\]
\[
\frac{2x - 12}{5x^2} = \frac{-4x + 30}{5x^2}
\]

Step 2: Since the denominators are the same, equate the numerators:
\[
2x - 12 = -4x + 30
\]

Step 3: Solve for \(x\):
\[
2x + 4x = 30 + 12 \implies 6x = 42 \implies x = 7
\]

Step 4: Check for extraneous solutions. Substitute \(x = 7\) back into the original equation:
\[
\frac{2(7) - 12}{5(7)^2} = \frac{1}{5(7)} - \frac{7 - 6}{(7)^2}
\]
\[
\frac{14 - 12}{245} = \frac{1}{35} - \frac{1}{49}
\]
\[
\frac{2}{245} = \frac{7}{245} - \frac{5}{245}
\]
\[
\frac{2}{245} = \frac{2}{245}
\]
The solution is valid.

Final Answer:
\[
\boxed{x = 7}
\]

---

#### 11. Solve the equation:
\[
\frac{k + 3}{2k^2} + \frac{1}{2k^2} = \frac{1}{k^2}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{k + 3 + 1}{2k^2} = \frac{1}{k^2}
\]
\[
\frac{k + 4}{2k^2} = \frac{1}{k^2}
\]

Step 2: Eliminate the denominators by multiplying through by \(2k^2\):
\[
k + 4 = 2
\]

Step 3: Solve for \(k\):
\[
k = 2 - 4 \implies k = -2
\]

Step 4: Check for extraneous solutions. Substitute \(k = -2\) back into the original equation:
\[
\frac{-2 + 3}{2(-2)^2} + \frac{1}{2(-2)^2} = \frac{1}{(-2)^2}
\]
\[
\frac{1}{8} + \frac{1}{8} = \frac{1}{4}
\]
\[
\frac{2}{8} = \frac{1}{4}
\]
\[
\frac{1}{4} = \frac{1}{4}
\]
The solution is valid.

Final Answer:
\[
\boxed{k = -2}
\]

---

#### 12. Solve the equation:
\[
\frac{a + 4}{4a} + \frac{1}{4a} = \frac{a + 1}{a}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{a + 4 + 1}{4a} = \frac{a + 1}{a}
\]
\[
\frac{a + 5}{4a} = \frac{a + 1}{a}
\]

Step 2: Eliminate the denominators by multiplying through by \(4a\):
\[
a + 5 = 4(a + 1)
\]
\[
a + 5 = 4a + 4
\]

Step 3: Solve for \(a\):
\[
a - 4a = 4 - 5 \implies -3a = -1 \implies a = \frac{1}{3}
\]

Step 4: Check for extraneous solutions. Substitute \(a = \frac{1}{3}\) back into the original equation:
\[
\frac{\frac{1}{3} + 4}{4\left(\frac{1}{3}\right)} + \frac{1}{4\left(\frac{1}{3}\right)} = \frac{\frac{1}{3} + 1}{\frac{1}{3}}
\]
\[
\frac{\frac{1}{3} + \frac{12}{3}}{\frac{4}{3}} + \frac{1}{\frac{4}{3}} = \frac{\frac{1}{3} + \frac{3}{3}}{\frac{1}{3}}
\]
\[
\frac{\frac{13}{3}}{\frac{4}{3}} + \frac{3}{4} = \frac{\frac{4}{3}}{\frac{1}{3}}
\]
\[
\frac{13}{4} + \frac{3}{4} = 4
\]
\[
\frac{16}{4} = 4
\]
\[
4 = 4
\]
The solution is valid.

Final Answer:
\[
\boxed{a = \frac{1}{3}}
\]

---

#### 13. Solve the equation:
\[
\frac{x + 5}{x^2} - \frac{1}{x} = \frac{x + 1}{x^2}
\]

Step 1: Combine the terms on the left-hand side:
\[
\frac{x + 5}{x^2} - \frac{x}{x^2} = \frac{x + 1}{x^2}
\]
\[
\frac{x + 5 - x}{x^2} = \frac{x + 1}{x^2}
\]
\[
\frac{5}{x^2} = \frac{x + 1}{x^2}
\]

Step 2: Since the denominators are the same, equate the numerators:
\[
5 = x + 1
\]

Step 3: Solve for \(x\):
\[
x = 5 - 1 \implies x = 4
\]

Step 4: Check for extraneous solutions. Substitute \(x = 4\) back into the original equation:
\[
\frac{4 + 5}{(4)^2} - \frac{1}{4} = \frac{4 + 1}{(4)^2}
\]
\[
\frac{9}{16} - \frac{1}{4} = \frac{5}{16}
\]
\[
\frac{9}{16} - \frac{4}{16} = \frac{5}{16}
\]
\[
\frac{5}{16} = \frac{5}{16}
\]
The solution is valid.

Final Answer:
\[
\boxed{x = 4}
\]

---

#### 14. Solve the equation:
\[
\frac{3n + 2}{3n^2} = \frac{1}{3n} + \frac{n - 6}{n^2}
\]

Step 1: Find a common denominator, which is \(3n^2\):
\[
\frac{3n + 2}{3n^2} = \frac{n}{3n^2} + \frac{3(n - 6)}{3n^2}
\]
\[
\frac{3n + 2}{3n^2} = \frac{n + 3(n - 6)}{3n^2}
\]
\[
\frac{3n + 2}{3n^2} = \frac{n + 3n - 18}{3n^2}
\]
\[
\frac{3n + 2}{3n^2} = \frac{4n - 18}{3n^2}
\]

Step 2: Since the denominators are the same, equate the numerators:
\[
3n + 2 = 4n - 18
\]

Step 3: Solve for \(n\):
\[
3n - 4n = -18 - 2 \implies -n = -20 \implies n = 20
\]

Step 4: Check for extraneous solutions. Substitute \(n = 20\) back into the original equation:
\[
\frac{3(20) + 2}{3(20)^2} = \frac{1}{3(20)} + \frac{20 - 6}{(20)^2}
\]
\[
\frac{60 + 2}{1200} = \frac{1}{60} + \frac{14}{400}
\]
\[
\frac{62}{1200} = \frac{20}{1200} + \frac{42}{1200}
\]
\[
\frac{62}{1200} = \frac{62}{1200}
\]
The solution is valid.

Final Answer:
\[
\boxed{n = 20}
\]

---

#### 15. Solve the equation:
\[
\frac{1}{2p} = \frac{3}{p} - \frac{p - 5}{p}
\]

Step 1: Combine the terms on the right-hand side:
\[
\frac{1}{2p} = \frac{3 - (p - 5)}{p}
\]
\[
\frac{1}{2p} = \frac{3 - p + 5}{p}
\]
\[
\frac{1}{2p} = \frac{8 - p}{p}
\]

Step 2: Eliminate the denominators by multiplying through by \(2p\):
\[
1 = 2(8 - p)
\]
\[
1 = 16 - 2p
\]

Step 3: Solve for \(p\):
\[
2p = 16 - 1 \implies 2p = 15 \implies p = \frac{15}{2}
\]

Step 4: Check for extraneous solutions. Substitute \(p = \frac{15}{2}\) back into the original equation:
\[
\frac{1}{2\left(\frac{15}{2}\right)} = \frac{3}{\frac{15}{2}} - \frac{\frac{15}{2} - 5}{\frac{15}{2}}
\]
\[
\frac{1}{15} = \frac{6}{15} - \frac{\frac{15}{2} - \frac{10}{2}}{\frac{15}{2}}
\]
\[
\frac{1}{15} = \frac{6}{15} - \frac{\frac{5}{2}}{\frac{15}{2}}
\]
\[
\frac{1}{15} = \frac{6}{15} - \frac{5}{15}
\]
\[
\frac{1}{15} = \frac{1}{15}
\]
The solution is valid.

Final Answer:
\[
\boxed{p = \frac{15}{2}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad n = -4 \\
2. & \quad x = -1 \\
3. & \quad b = 9 \\
4. & \quad x = 3 \\
5. & \quad r = 16 \\
6. & \quad n = -3 \\
7. & \quad b = 4 \\
8. & \quad v = -5 \\
9. & \quad x = -5 \\
10. & \quad x = 7 \\
11. & \quad k = -2 \\
12. & \quad a = \frac{1}{3} \\
13. & \quad x = 4 \\
14. & \quad n = 20 \\
15. & \quad p = \frac{15}{2}
\end{aligned}
}
\]