In order for a rational expression to be equal to 0, its numerator must be equal to 0. So we will want to factor the numerator to solve this equation.

We also want to factor the denominator, however, to check whether each solution is viable or extraneous.

The expression in the numerator is x^2 + 5x - 24, which can be factored to \left(x + 8\right)\left(x - 3\right).

The expression in the denominator is x^2 - 9x + 18, which can be factored to \left(x - 6\right)\left(x - 3\right).

So we can rewrite the equation as\frac{\left(x + 8\right)\left(x - 3\right)}{\left(x - 6\right)\left(x - 3\right)} = 0

From the denominator, we can see that x cannot be equal to 3 or 6.

From the numerator, we can see that the possible solutions are x = -8 and x = 3.

Putting this together, we have that x = -8 is a viable solution, while x = 3 is an extraneous solution.

If we were to solve the equation by first multiplying both sides by x^2 - 9x + 18, and then factoring the remaining quadratic, we would still get the two possible solutions of x = -8 and x = 3.

In order to determine whether they were valid or extraneous, we could substitute into the initial expression \dfrac{x^2 + 5x - 24}{x^2 - 9x + 18}:

For x = -8:\quad\dfrac{\left(-8\right)^2 + 5\left(-8\right) - 24}{\left(-8\right)^2 - 9\left(-8\right) + 18} = \dfrac{64 - 40 - 24}{64 + 72 + 18} = \dfrac{0}{154} = 0

For x = 3:\quad\dfrac{3^2 + 5\left(3\right) - 24}{3^2 - 9\left(3\right) + 18} = \dfrac{9 + 15 - 24}{9 - 27 + 18} = \dfrac{0}{0}

So only the solution of x = -8 is valid. The solution of x = 3 produces an undefined expression and is therefore extraneous.

Since there are two rational expressions in this equation, we will start by rearranging the equation to have both rational expressions on the same side, equal to zero, and then combine them by making a common denominator.

At this point, we can see that the equation has no solutions (viable or otherwise), since there is no value of x that can make the numerator of the rational expression be 0.

It is also possible to draw this conclusion by rewriting the intial equation\frac{1}{x + 6} = \frac{2}{2x - 5}as\frac{2}{2x + 12} = \frac{2}{2x - 5}At this point the numerators are equal, while the functions in the denominators are linear expressions with the same slope (i.e. parallel lines). So the denominators will never be equal for any value of x, and therefore the equation has no solutions.

The information that we know about each chicken coop is:

1. • Area = 60\text{ ft}^2
   • Width = x\text{ ft}
2. • Area =96\text{ ft}^2
   • Width = \left(x + 3\right)\text{ ft}

and we know that they have the same length. We can use this information to draw a sketch:

So we can form an equation by getting an expression for the length of each coop in terms of x and setting them equal to each other, then solve that equation for x. We can then use the value of x to find the widths and length of the coops.

For a rectangle, we know that \text{Area} = \text{Length} \cdot \text{Width}, which we can rearrange to get \text{Length} = \text{Area} \div \text{Width}.

So the length of the first coop will be \dfrac{60}{x}, while the length of the second coop will be \dfrac{96}{x + 3}.

We can now set these lengths to be equal and solve for x as follows:

We know that the widths of the coops are given by x and x + 3. So the width of the smaller coop will be 5 feet, while the width of the larger coop will be x = 8 feet.

The length of each coop is the same, and we know that the length is equal to \dfrac{60}{x} from when we set up the equation. So the length of the two coops will be \dfrac{60}{5} = 12 feet.

In summary, one coop will be 12 \times 5 feet and the other will be 12 \times 8 feet.

Since the value of x that we solved for is not 0 or -3, we know that it is not an extraneous solution.

Also, since x represents a width, it must be a positive value, which it is. If the solution had been negative, it would have been a non-viable solution.