Solved 6.4/6.5 Rational equation-solving rational equations ... - Free Printable

Let's solve each of these rational equations step by step. We'll simplify, find common denominators, eliminate fractions, solve the resulting equations, and check for extraneous solutions (values that make the denominator zero).

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1.

$$
\frac{x}{x+3} + \frac{6}{x^2 + 8x + 15} = 2
$$

#### Step 1: Factor the denominator
Note that:
$$
x^2 + 8x + 15 = (x+3)(x+5)
$$

So the equation becomes:
$$
\frac{x}{x+3} + \frac{6}{(x+3)(x+5)} = 2
$$

#### Step 2: Find the LCD
LCD = $(x+3)(x+5)$

Multiply both sides by LCD to eliminate denominators:
$$
(x+3)(x+5) \left( \frac{x}{x+3} + \frac{6}{(x+3)(x+5)} \right) = 2(x+3)(x+5)
$$

Simplify:
$$
x(x+5) + 6 = 2(x+3)(x+5)
$$

Expand both sides:
- Left: $x^2 + 5x + 6$
- Right: $2(x^2 + 8x + 15) = 2x^2 + 16x + 30$

So:
$$
x^2 + 5x + 6 = 2x^2 + 16x + 30
$$

Bring all terms to one side:
$$
0 = x^2 + 11x + 24
$$

Factor:
$$
x^2 + 11x + 24 = (x+3)(x+8) = 0
$$

So $x = -3$ or $x = -8$

#### Step 3: Check for extraneous solutions
Original denominators: $x+3$, $x^2 + 8x + 15 = (x+3)(x+5)$

If $x = -3$: denominator becomes 0 → invalid

If $x = -8$: check:
- $x+3 = -5 \neq 0$
- $x+5 = -3 \neq 0$ → valid

✔ So only solution: $\boxed{x = -8}$

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2.

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

#### Step 1: LCD = $(x+4)(x+1)$

Multiply both sides:
$$
(x+4)(x+1)\left(\frac{2x}{x+4} - \frac{3}{x+1}\right) = (x+4)(x+1)(1)
$$

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

Expand:
- Left: $2x^2 + 2x - 3x - 12 = 2x^2 - x - 12$
- Right: $x^2 + 5x + 4$

Set equal:
$$
2x^2 - x - 12 = x^2 + 5x + 4
$$

Bring all terms to left:
$$
2x^2 - x - 12 - x^2 - 5x - 4 = 0 \Rightarrow x^2 - 6x - 16 = 0
$$

Factor:
$$
(x - 8)(x + 2) = 0
\Rightarrow x = 8 \text{ or } x = -2
$$

Check for restrictions: $x \ne -4$, $x \ne -1$

Both $x=8$ and $x=-2$ are valid.

✔ Solutions: $\boxed{x = 8,\ x = -2}$

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3.

$$
\frac{2x}{x+2} - \frac{20}{x^2 - x - 6} = 1
$$

#### Step 1: Factor denominator
$$
x^2 - x - 6 = (x - 3)(x + 2)
$$

So equation becomes:
$$
\frac{2x}{x+2} - \frac{20}{(x-3)(x+2)} = 1
$$

LCD = $(x+2)(x-3)$

Multiply both sides:
$$
(x+2)(x-3)\left(\frac{2x}{x+2} - \frac{20}{(x-3)(x+2)}\right) = (x+2)(x-3)(1)
$$

Simplify:
$$
2x(x-3) - 20 = (x+2)(x-3)
$$

Expand:
- Left: $2x^2 - 6x - 20$
- Right: $x^2 - x - 6$

Set equal:
$$
2x^2 - 6x - 20 = x^2 - x - 6
$$

Bring all to left:
$$
2x^2 - 6x - 20 - x^2 + x + 6 = 0 \Rightarrow x^2 - 5x - 14 = 0
$$

Factor:
$$
(x - 7)(x + 2) = 0 \Rightarrow x = 7,\ x = -2
$$

Check restrictions: $x \ne -2$, $x \ne 3$

$x = -2$ makes denominator zero → invalid

$x = 7$: valid

✔ Solution: $\boxed{x = 7}$

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4.

$$
\frac{r+5}{r^2 - 2r} - 1 = \frac{1}{r^2 - 2r}
$$

#### Step 1: Factor denominator
$$
r^2 - 2r = r(r - 2)
$$

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

Move all terms to one side:
$$
\frac{r+5}{r(r-2)} - \frac{1}{r(r-2)} - 1 = 0
\Rightarrow \frac{(r+5) - 1}{r(r-2)} - 1 = 0
\Rightarrow \frac{r+4}{r(r-2)} = 1
$$

Now solve:
$$
\frac{r+4}{r(r-2)} = 1
$$

Multiply both sides by $r(r-2)$ (≠ 0):
$$
r+4 = r(r-2) = r^2 - 2r
$$

Bring all to one side:
$$
0 = r^2 - 2r - r - 4 = r^2 - 3r - 4
$$

Factor:
$$
(r - 4)(r + 1) = 0 \Rightarrow r = 4,\ r = -1
$$

Check restrictions: $r \ne 0$, $r \ne 2$

Both $r = 4$ and $r = -1$ are valid.

✔ Solutions: $\boxed{r = 4,\ r = -1}$

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5.

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

#### Step 1: Factor denominators
$$
p^2 + p = p(p+1)
$$

So:
$$
\frac{p+5}{p(p+1)} = \frac{1}{p(p+1)} - \frac{p-6}{p+1}
$$

Get common denominator on right side:
$$
\frac{1}{p(p+1)} - \frac{p-6}{p+1} = \frac{1}{p(p+1)} - \frac{p(p-6)}{p(p+1)} = \frac{1 - p(p-6)}{p(p+1)}
$$

Simplify numerator:
$$
1 - p(p - 6) = 1 - (p^2 - 6p) = 1 - p^2 + 6p = -p^2 + 6p + 1
$$

So right side becomes:
$$
\frac{-p^2 + 6p + 1}{p(p+1)}
$$

Left side: $\frac{p+5}{p(p+1)}$

Now set equal:
$$
\frac{p+5}{p(p+1)} = \frac{-p^2 + 6p + 1}{p(p+1)}
$$

Since denominators are same and ≠ 0, equate numerators:
$$
p + 5 = -p^2 + 6p + 1
$$

Bring all to one side:
$$
p + 5 + p^2 - 6p - 1 = 0 \Rightarrow p^2 - 5p + 4 = 0
$$

Factor:
$$
(p - 1)(p - 4) = 0 \Rightarrow p = 1,\ p = 4
$$

Check restrictions: $p \ne 0$, $p \ne -1$

Both $p=1$ and $p=4$ are valid.

✔ Solutions: $\boxed{p = 1,\ p = 4}$

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✔ Final Answers:

1. $x = -8$
2. $x = 8,\ x = -2$
3. $x = 7$
4. $r = 4,\ r = -1$
5. $p = 1,\ p = 4$

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