301 Moved Permanently - Free Printable

Let's solve each equation from Worksheet-3 step by step. These are rational equations, so we'll:

1. Find the least common denominator (LCD).
2. Eliminate denominators by multiplying both sides by the LCD.
3. Solve the resulting equation.
4. Check for extraneous solutions (values that make any denominator zero).

---

1) $\frac{5}{4n} + \frac{1}{4} = \frac{1}{4n}$

Multiply both sides by $4n$ (LCD):

$$
4n \left( \frac{5}{4n} + \frac{1}{4} \right) = 4n \left( \frac{1}{4n} \right)
\Rightarrow 5 + n = 1
\Rightarrow n = -4
$$

✔ Check: $n = -4$, no denominator is zero → valid.

Answer: $n = -4$

---

2) $\frac{1}{6x^2} = \frac{1}{6x} + \frac{1}{3x^2}$

LCD = $6x^2$

Multiply both sides:

$$
6x^2 \cdot \frac{1}{6x^2} = 6x^2 \left( \frac{1}{6x} + \frac{1}{3x^2} \right)
\Rightarrow 1 = x + 2
\Rightarrow x = -1
$$

Check: $x = -1$: denominators are non-zero → valid.

Answer: $x = -1$

---

3) $\frac{1}{6b^2} + \frac{b - 4}{6b^2} = \frac{1}{b^2}$

Left side: combine numerators:

$$
\frac{1 + b - 4}{6b^2} = \frac{b - 3}{6b^2}
$$

Set equal to right side:

$$
\frac{b - 3}{6b^2} = \frac{1}{b^2}
$$

Multiply both sides by $6b^2$:

$$
b - 3 = 6 \Rightarrow b = 9
$$

Check: $b = 9$, $b^2 \neq 0$ → valid.

Answer: $b = 9$

---

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

LCD = $3x^2$

Multiply both sides:

$$
3x^2 \cdot \frac{4}{x^2} = 3x^2 \left( \frac{5}{3x} - \frac{1}{x^2} \right)
\Rightarrow 12 = 5x - 3
\Rightarrow 5x = 15 \Rightarrow x = 3
$$

Check: $x = 3$, no division by zero → valid.

Answer: $x = 3$

---

5) $\frac{4r + 16}{r^2} + \frac{1}{r} = \frac{6}{r}$

LCD = $r^2$

Multiply both sides:

$$
r^2 \left( \frac{4r + 16}{r^2} + \frac{1}{r} \right) = r^2 \cdot \frac{6}{r}
\Rightarrow (4r + 16) + r = 6r
\Rightarrow 5r + 16 = 6r
\Rightarrow r = 16
$$

Check: $r = 16$, not zero → valid.

Answer: $r = 16$

---

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

Right side: combine:

$$
\frac{n + 5 + 1}{3n^2} = \frac{n + 6}{3n^2}
$$

So:

$$
\frac{1}{n^2} = \frac{n + 6}{3n^2}
$$

Multiply both sides by $3n^2$:

$$
3 = n + 6 \Rightarrow n = -3
$$

Check: $n = -3$, $n^2 \neq 0$ → valid.

Answer: $n = -3$

---

7) $\frac{1}{3b} = \frac{b + 2}{6b^2} + \frac{1}{3b^2}$

LCD = $6b^2$

Multiply both sides:

$$
6b^2 \cdot \frac{1}{3b} = 6b^2 \left( \frac{b + 2}{6b^2} + \frac{1}{3b^2} \right)
\Rightarrow 2b = (b + 2) + 2
\Rightarrow 2b = b + 4
\Rightarrow b = 4
$$

Check: $b = 4$, no zero denominators → valid.

Answer: $b = 4$

---

8) $\frac{v + 4}{v^2} + \frac{1}{v} = \frac{3v + 9}{v^2}$

LCD = $v^2$

Multiply both sides:

$$
v^2 \left( \frac{v + 4}{v^2} + \frac{1}{v} \right) = v^2 \cdot \frac{3v + 9}{v^2}
\Rightarrow (v + 4) + v = 3v + 9
\Rightarrow 2v + 4 = 3v + 9
\Rightarrow -v = 5 \Rightarrow v = -5
$$

Check: $v = -5$, $v \neq 0$ → valid.

Answer: $v = -5$

---

9) $\frac{6}{5x} - \frac{x + 4}{x^2} = \frac{1}{x}$

LCD = $5x^2$

Multiply both sides:

$$
5x^2 \left( \frac{6}{5x} - \frac{x + 4}{x^2} \right) = 5x^2 \cdot \frac{1}{x}
\Rightarrow 6x - 5(x + 4) = 5x
\Rightarrow 6x - 5x - 20 = 5x
\Rightarrow x - 20 = 5x
\Rightarrow -4x = 20 \Rightarrow x = -5
$$

Check: $x = -5$, not zero → valid.

Answer: $x = -5$

---

10) $\frac{2x - 12}{5x^2} = \frac{1}{5x} - \frac{x - 6}{x^2}$

LCD = $5x^2$

Multiply both sides:

$$
5x^2 \cdot \frac{2x - 12}{5x^2} = 5x^2 \left( \frac{1}{5x} - \frac{x - 6}{x^2} \right)
\Rightarrow 2x - 12 = x - (5)(x - 6)
\Rightarrow 2x - 12 = x - 5x + 30
\Rightarrow 2x - 12 = -4x + 30
\Rightarrow 6x = 42 \Rightarrow x = 7
$$

Check: $x = 7$, not zero → valid.

Answer: $x = 7$

---

11) $\frac{k + 3}{2k^2} + \frac{1}{2k^2} = \frac{1}{k^2}$

Combine left side:

$$
\frac{k + 3 + 1}{2k^2} = \frac{k + 4}{2k^2}
$$

Set equal:

$$
\frac{k + 4}{2k^2} = \frac{1}{k^2}
$$

Multiply both sides by $2k^2$:

$$
k + 4 = 2 \Rightarrow k = -2
$$

Check: $k = -2$, $k^2 \neq 0$ → valid.

Answer: $k = -2$

---

12) $\frac{a + 4}{4a} + \frac{1}{4a} = \frac{a + 1}{a}$

Left side:

$$
\frac{a + 4 + 1}{4a} = \frac{a + 5}{4a}
$$

Set equal:

$$
\frac{a + 5}{4a} = \frac{a + 1}{a}
$$

Multiply both sides by $4a$:

$$
a + 5 = 4(a + 1) = 4a + 4
\Rightarrow a + 5 = 4a + 4
\Rightarrow -3a = -1 \Rightarrow a = \frac{1}{3}
$$

Check: $a = \frac{1}{3}$, not zero → valid.

Answer: $a = \frac{1}{3}$

---

13) $\frac{x + 5}{x^2} - \frac{1}{x} = \frac{x + 1}{x^2}$

LCD = $x^2$

Multiply both sides:

$$
x^2 \left( \frac{x + 5}{x^2} - \frac{1}{x} \right) = x^2 \cdot \frac{x + 1}{x^2}
\Rightarrow (x + 5) - x = x + 1
\Rightarrow 5 = x + 1 \Rightarrow x = 4
$$

Check: $x = 4$, not zero → valid.

Answer: $x = 4$

---

14) $\frac{3n + 2}{3n^2} = \frac{1}{3n} + \frac{n - 6}{n^2}$

LCD = $3n^2$

Multiply both sides:

$$
3n^2 \cdot \frac{3n + 2}{3n^2} = 3n^2 \left( \frac{1}{3n} + \frac{n - 6}{n^2} \right)
\Rightarrow 3n + 2 = n + 3(n - 6)
\Rightarrow 3n + 2 = n + 3n - 18
\Rightarrow 3n + 2 = 4n - 18
\Rightarrow -n = -20 \Rightarrow n = 20
$$

Check: $n = 20$, not zero → valid.

Answer: $n = 20$

---

15) $\frac{1}{2p} = \frac{3}{p} - \frac{p - 5}{p}$

Simplify right side:

$$
\frac{3 - (p - 5)}{p} = \frac{3 - p + 5}{p} = \frac{8 - p}{p}
$$

So:

$$
\frac{1}{2p} = \frac{8 - p}{p}
$$

Multiply both sides by $2p$:

$$
1 = 2(8 - p) = 16 - 2p
\Rightarrow 2p = 15 \Rightarrow p = \frac{15}{2}
$$

Check: $p = \frac{15}{2}$, not zero → valid.

Answer: $p = \frac{15}{2}$

---

## ✔ Final Answers Summary:

| Problem | Solution |
|--------|---------|
| 1 | $n = -4$ |
| 2 | $x = -1$ |
| 3 | $b = 9$ |
| 4 | $x = 3$ |
| 5 | $r = 16$ |
| 6 | $n = -3$ |
| 7 | $b = 4$ |
| 8 | $v = -5$ |
| 9 | $x = -5$ |
| 10 | $x = 7$ |
| 11 | $k = -2$ |
| 12 | $a = \frac{1}{3}$ |
| 13 | $x = 4$ |
| 14 | $n = 20$ |
| 15 | $p = \frac{15}{2}$ |

All solutions have been checked for extraneous values — none found.

Let me know if you'd like these steps explained in more detail!