Solving Rational Equations - Free Printable

Let's solve each of these rational equations step by step. We'll follow the standard procedure:

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

---

1) $\frac{a+1}{5a} - \frac{1}{a} = 1$

LCD: $5a$

Multiply both sides:
$$
5a\left(\frac{a+1}{5a} - \frac{1}{a}\right) = 5a(1)
\Rightarrow (a+1) - 5 = 5a
\Rightarrow a + 1 - 5 = 5a
\Rightarrow a - 4 = 5a
\Rightarrow -4 = 4a
\Rightarrow a = -1
$$

Check: $a = -1$, no denominator is zero → valid.

✔ Solution: $a = -1$

---

2) $\frac{6v - 6}{v^2} + \frac{2}{v^2} = \frac{1}{v^2}$

All terms have same denominator:

$$
\frac{6v - 6 + 2}{v^2} = \frac{1}{v^2}
\Rightarrow \frac{6v - 4}{v^2} = \frac{1}{v^2}
\Rightarrow 6v - 4 = 1
\Rightarrow 6v = 5
\Rightarrow v = \frac{5}{6}
$$

Check: $v = \frac{5}{6} \neq 0$, so no division by zero.

✔ Solution: $v = \frac{5}{6}$

---

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

LCD: $n^2$

Multiply both sides:
$$
n^2\left(\frac{1}{n^2} + \frac{4}{n}\right) = n^2\left(\frac{3}{n^2}\right)
\Rightarrow 1 + 4n = 3
\Rightarrow 4n = 2
\Rightarrow n = \frac{1}{2}
$$

Check: $n = \frac{1}{2} \neq 0$, valid.

✔ Solution: $n = \frac{1}{2}$

---

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

LCD: $5x^2$

Multiply:
$$
5x^2\left(\frac{4}{x^2} + \frac{1}{5x}\right) = 5x^2\left(\frac{1}{5x^2}\right)
\Rightarrow 20 + x = 1
\Rightarrow x = -19
$$

Check: $x = -19 \neq 0$, valid.

✔ Solution: $x = -19$

---

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

LCD: $3k^2$

Multiply:
$$
3k^2\left(\frac{1}{k^2}\right) = 3k^2\left(\frac{1}{3k^2} + \frac{k+5}{3k^2}\right)
\Rightarrow 3 = 1 + (k+5)
\Rightarrow 3 = k + 6
\Rightarrow k = -3
$$

Check: $k = -3 \neq 0$, valid.

✔ Solution: $k = -3$

---

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

LCD: $x^2$

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

Check: $x \neq 0$, valid.

✔ Solution: $x = -\frac{5}{4}$

---

7) $\frac{6}{k} - \frac{1}{k^2 + 6k} = \frac{1}{k}$

Note: $k^2 + 6k = k(k + 6)$

LCD: $k(k+6)$

Multiply:
$$
k(k+6)\left(\frac{6}{k} - \frac{1}{k(k+6)}\right) = k(k+6)\left(\frac{1}{k}\right)
\Rightarrow 6(k+6) - 1 = k+6
\Rightarrow 6k + 36 - 1 = k + 6
\Rightarrow 6k + 35 = k + 6
\Rightarrow 5k = -29
\Rightarrow k = -\frac{29}{5}
$$

Check: $k \neq 0, -6$, yes → valid.

✔ Solution: $k = -\frac{29}{5}$

---

8) $\frac{4}{n+1} + \frac{1}{n^2 - 5n - 6} = \frac{-1}{n-6}$

Factor: $n^2 - 5n - 6 = (n - 6)(n + 1)$

So LCD: $(n+1)(n-6)$

Multiply:
$$
(n+1)(n-6)\left(\frac{4}{n+1} + \frac{1}{(n-6)(n+1)}\right) = (n+1)(n-6)\left(\frac{-1}{n-6}\right)
\Rightarrow 4(n-6) + 1 = -(n+1)
\Rightarrow 4n - 24 + 1 = -n - 1
\Rightarrow 4n - 23 = -n - 1
\Rightarrow 5n = 22
\Rightarrow n = \frac{22}{5}
$$

Check: $n \neq -1, 6$, yes → valid.

✔ Solution: $n = \frac{22}{5}$

---

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

Note: $x^2 + 5x = x(x+5)$

LCD: $x(x+5)$

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

Check: $x = 5$, not $0$ or $-5$, valid.

✔ Solution: $x = 5$

---

10) $\frac{5}{p+6} - \frac{1}{p^2 + 6p} = \frac{2}{p^2 + 6p}$

Note: $p^2 + 6p = p(p+6)$

LCD: $p(p+6)$

Multiply:
$$
p(p+6)\left(\frac{5}{p+6} - \frac{1}{p(p+6)}\right) = p(p+6)\left(\frac{2}{p(p+6)}\right)
\Rightarrow 5p - 1 = 2
\Rightarrow 5p = 3
\Rightarrow p = \frac{3}{5}
$$

Check: $p \neq 0, -6$, valid.

✔ Solution: $p = \frac{3}{5}$

---

11) $\frac{1}{2v} - \frac{5v + 15}{v^2 - 6v} = \frac{v + 6}{2v^2 - 12v}$

Factor:
- $v^2 - 6v = v(v - 6)$
- $2v^2 - 12v = 2v(v - 6)$

LCD: $2v(v - 6)$

Multiply:
$$
2v(v-6)\left(\frac{1}{2v} - \frac{5v+15}{v(v-6)}\right) = 2v(v-6)\left(\frac{v+6}{2v(v-6)}\right)
\Rightarrow (v - 6) - 2(5v + 15) = v + 6
\Rightarrow v - 6 - 10v - 30 = v + 6
\Rightarrow -9v - 36 = v + 6
\Rightarrow -10v = 42
\Rightarrow v = -\frac{21}{5}
$$

Check: $v \neq 0, 6$, yes → valid.

✔ Solution: $v = -\frac{21}{5}$

---

12) $\frac{5}{x+1} = \frac{6}{x^2 - 2x - 3} + \frac{1}{x - 3}$

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

LCD: $(x+1)(x-3)$

Multiply:
$$
(x+1)(x-3)\left(\frac{5}{x+1}\right) = (x+1)(x-3)\left(\frac{6}{(x-3)(x+1)} + \frac{1}{x-3}\right)
\Rightarrow 5(x - 3) = 6 + (x + 1)
\Rightarrow 5x - 15 = 6 + x + 1
\Rightarrow 5x - 15 = x + 7
\Rightarrow 4x = 22
\Rightarrow x = \frac{11}{2}
$$

Check: $x \neq -1, 3$, valid.

✔ Solution: $x = \frac{11}{2}$

---

13) $\frac{n^2 + 7n + 6}{n^2} = \frac{1}{6} - \frac{1}{6n^2}$

Left side: factor numerator: $n^2 + 7n + 6 = (n+1)(n+6)$

LCD: $6n^2$

Multiply:
$$
6n^2\left(\frac{(n+1)(n+6)}{n^2}\right) = 6n^2\left(\frac{1}{6} - \frac{1}{6n^2}\right)
\Rightarrow 6(n+1)(n+6) = n^2 - 1
$$

Expand left:
$$
6(n^2 + 7n + 6) = n^2 - 1
\Rightarrow 6n^2 + 42n + 36 = n^2 - 1
\Rightarrow 5n^2 + 42n + 37 = 0
$$

Use quadratic formula:
$$
n = \frac{-42 \pm \sqrt{42^2 - 4(5)(37)}}{2(5)} = \frac{-42 \pm \sqrt{1764 - 740}}{10} = \frac{-42 \pm \sqrt{1024}}{10}
= \frac{-42 \pm 32}{10}
$$

So:
- $n = \frac{-10}{10} = -1$
- $n = \frac{-74}{10} = -7.4 = -\frac{37}{5}$

Check:
- $n = -1$: denominator $n^2 = 1$, okay? But check original: $n^2 \neq 0$, fine. But in original, $n^2$ is in denominator — no problem.
But wait: let's plug back into original.

Wait — but check if it makes any denominator zero: only $n = 0$ is invalid. So both are okay?

Wait — but we must check whether they satisfy original equation.

Try $n = -1$:
LHS: $\frac{(-1)^2 + 7(-1) + 6}{(-1)^2} = \frac{1 - 7 + 6}{1} = 0$

RHS: $\frac{1}{6} - \frac{1}{6(1)} = \frac{1}{6} - \frac{1}{6} = 0$ → OK

Now $n = -\frac{37}{5}$: messy, but algebraically derived, and no denominator zero → acceptable.

✔ Solutions: $n = -1$, $n = -\frac{37}{5}$

---

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

Right side: write as single fraction.

First, simplify right side:
$$
1 - \frac{k^2 - 3k - 4}{4k} = \frac{4k}{4k} - \frac{k^2 - 3k - 4}{4k} = \frac{4k - (k^2 - 3k - 4)}{4k} = \frac{4k - k^2 + 3k + 4}{4k} = \frac{-k^2 + 7k + 4}{4k}
$$

Now equation:
$$
\frac{k+1}{k} = \frac{-k^2 + 7k + 4}{4k}
$$

Multiply both sides by $4k$:
$$
4(k+1) = -k^2 + 7k + 4
\Rightarrow 4k + 4 = -k^2 + 7k + 4
\Rightarrow k^2 - 3k = 0
\Rightarrow k(k - 3) = 0
\Rightarrow k = 0 \text{ or } k = 3
$$

Check: $k = 0$ → denominator zero → extraneous

$k = 3$: check
LHS: $\frac{3+1}{3} = \frac{4}{3}$

RHS: $1 - \frac{9 - 9 - 4}{12} = 1 - \frac{-4}{12} = 1 + \frac{1}{3} = \frac{4}{3}$ → OK

✔ Solution: $k = 3$

---

15) $1 = \frac{2}{r^2} - \frac{1}{r}$

Multiply by $r^2$:
$$
r^2 = 2 - r
\Rightarrow r^2 + r - 2 = 0
\Rightarrow (r + 2)(r - 1) = 0
\Rightarrow r = -2, 1
$$

Check: $r \neq 0$, both valid.

Check $r = -2$: LHS = 1, RHS = $2/4 - (-1/2) = 0.5 + 0.5 = 1$ → OK
$r = 1$: $2/1 - 1/1 = 2 - 1 = 1$ → OK

✔ Solutions: $r = -2$, $r = 1$

---

16) $\frac{2n^2 - 8n - 10}{5n} - 1 = \frac{n + 6}{5n}$

Multiply both sides by $5n$:
$$
2n^2 - 8n - 10 - 5n = n + 6
\Rightarrow 2n^2 - 13n - 10 = n + 6
\Rightarrow 2n^2 - 14n - 16 = 0
\Rightarrow n^2 - 7n - 8 = 0
\Rightarrow (n - 8)(n + 1) = 0
\Rightarrow n = 8, -1
$$

Check: $n \neq 0$, both valid.

Check:
- $n = 8$: LHS: $\frac{2(64) - 64 - 10}{40} - 1 = \frac{128 - 64 - 10}{40} - 1 = \frac{54}{40} - 1 = 1.35 - 1 = 0.35$
RHS: $\frac{8+6}{40} = 14/40 = 0.35$ → OK

- $n = -1$: LHS: $\frac{2(1) + 8 - 10}{-5} - 1 = \frac{2 + 8 - 10}{-5} - 1 = 0 - 1 = -1$
RHS: $\frac{-1 + 6}{-5} = \frac{5}{-5} = -1$ → OK

✔ Solutions: $n = -1$, $n = 8$

---

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

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

LCD: $x^2(x - 1)$

Multiply:
$$
x^2(x-1)\left(\frac{(x-4)(x+1)}{x(x-1)} - \frac{1}{x^2}\right) = x^2(x-1)\left(\frac{x-2}{x}\right)
$$

Left:
- First term: $x^2(x-1) \cdot \frac{(x-4)(x+1)}{x(x-1)} = x(x-4)(x+1)$
- Second: $x^2(x-1) \cdot \frac{1}{x^2} = x - 1$

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

Compute left:
$x(x-4)(x+1) = x[(x)(x+1) - 4(x+1)] = x[x^2 + x - 4x - 4] = x(x^2 - 3x - 4) = x^3 - 3x^2 - 4x$

Then subtract $(x - 1)$: $x^3 - 3x^2 - 4x - x + 1 = x^3 - 3x^2 - 5x + 1$

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

Set equal:
$$
x^3 - 3x^2 - 5x + 1 = x^3 - 3x^2 + 2x
\Rightarrow -5x + 1 = 2x
\Rightarrow 1 = 7x
\Rightarrow x = \frac{1}{7}
$$

Check: $x \neq 0, 1$, yes → valid.

✔ Solution: $x = \frac{1}{7}$

---

18) $1 = \frac{n - 2}{n - 1} + \frac{3}{n^2 + 3n - 4}$

Factor: $n^2 + 3n - 4 = (n + 4)(n - 1)$

LCD: $(n - 1)(n + 4)$

Multiply:
$$
(n - 1)(n + 4)\left(1\right) = (n - 1)(n + 4)\left(\frac{n - 2}{n - 1} + \frac{3}{(n - 1)(n + 4)}\right)
\Rightarrow (n - 1)(n + 4) = (n - 2)(n + 4) + 3
$$

Left: $n^2 + 3n - 4$

Right: $(n - 2)(n + 4) + 3 = n^2 + 4n - 2n - 8 + 3 = n^2 + 2n - 5$

Set equal:
$$
n^2 + 3n - 4 = n^2 + 2n - 5
\Rightarrow 3n - 4 = 2n - 5
\Rightarrow n = -1
$$

Check: $n = -1$, not $1$ or $-4$ → valid.

Check:
LHS: 1
RHS: $\frac{-1 - 2}{-1 - 1} + \frac{3}{1 - 3 - 4} = \frac{-3}{-2} + \frac{3}{-6} = 1.5 - 0.5 = 1$ → OK

✔ Solution: $n = -1$

---

19) $\frac{v - 6}{2v^2 + 2v - 4} + \frac{v}{2v - 2} = \frac{1}{2}$

Factor:
- $2v^2 + 2v - 4 = 2(v^2 + v - 2) = 2(v + 2)(v - 1)$
- $2v - 2 = 2(v - 1)$

LCD: $2(v + 2)(v - 1)$

Multiply:
$$
2(v+2)(v-1)\left(\frac{v - 6}{2(v+2)(v-1)} + \frac{v}{2(v-1)}\right) = 2(v+2)(v-1)\left(\frac{1}{2}\right)
$$

Left:
- First: $v - 6$
- Second: $v(v + 2)$

Right: $(v+2)(v-1)$

So:
$$
(v - 6) + v(v + 2) = (v+2)(v-1)
\Rightarrow v - 6 + v^2 + 2v = v^2 + v - 2
\Rightarrow v^2 + 3v - 6 = v^2 + v - 2
\Rightarrow 2v = 4
\Rightarrow v = 2
$$

Check: $v = 2$, not $1$ or $-2$ → valid.

Check:
LHS: $\frac{2-6}{2(4)+4-4} + \frac{2}{4-2} = \frac{-4}{8+4-4} + \frac{2}{2} = \frac{-4}{8} + 1 = -0.5 + 1 = 0.5$ → OK

✔ Solution: $v = 2$

---

20) $\frac{x - 3}{2x + 10} + 2x - 12 = \frac{x^2 + 3x - 18}{2x + 10}$

Note: $2x + 10 = 2(x + 5)$

Also, $x^2 + 3x - 18 = (x + 6)(x - 3)$

LCD: $2x + 10$

Multiply both sides:
$$
(2x + 10)\left(\frac{x - 3}{2x + 10} + 2x - 12\right) = (2x + 10)\left(\frac{x^2 + 3x - 18}{2x + 10}\right)
\Rightarrow (x - 3) + (2x - 12)(2x + 10) = x^2 + 3x - 18
$$

Compute $(2x - 12)(2x + 10)$:
$$
= 4x^2 + 20x - 24x - 120 = 4x^2 - 4x - 120
$$

Add $x - 3$: $4x^2 - 4x - 120 + x - 3 = 4x^2 - 3x - 123$

Set equal:
$$
4x^2 - 3x - 123 = x^2 + 3x - 18
\Rightarrow 3x^2 - 6x - 105 = 0
\Rightarrow x^2 - 2x - 35 = 0
\Rightarrow (x - 7)(x + 5) = 0
\Rightarrow x = 7, -5
$$

Check: $x = -5$: denominator $2(-5) + 10 = 0$ → extraneous

$x = 7$: check

LHS: $\frac{7-3}{14+10} + 2(7) - 12 = \frac{4}{24} + 14 - 12 = \frac{1}{6} + 2 = 2.166...$

RHS: $\frac{49 + 21 - 18}{24} = \frac{52}{24} = \frac{13}{6} \approx 2.166...$ → OK

✔ Solution: $x = 7$

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## ✔ Final Answers:

| Problem | Solution |
|--------|---------|
| 1 | $a = -1$ |
| 2 | $v = \frac{5}{6}$ |
| 3 | $n = \frac{1}{2}$ |
| 4 | $x = -19$ |
| 5 | $k = -3$ |
| 6 | $x = -\frac{5}{4}$ |
| 7 | $k = -\frac{29}{5}$ |
| 8 | $n = \frac{22}{5}$ |
| 9 | $x = 5$ |
| 10 | $p = \frac{3}{5}$ |
| 11 | $v = -\frac{21}{5}$ |
| 12 | $x = \frac{11}{2}$ |
| 13 | $n = -1$, $n = -\frac{37}{5}$ |
| 14 | $k = 3$ |
| 15 | $r = -2$, $r = 1$ |
| 16 | $n = -1$, $n = 8$ |
| 17 | $x = \frac{1}{7}$ |
| 18 | $n = -1$ |
| 19 | $v = 2$ |
| 20 | $x = 7$ |

All solutions checked for extraneous roots. Let me know if you'd like this formatted as a PDF or need steps explained further!