5.06 Solve rational equations

Determine whether x = 3 is a viable solution, an extraneous solution, or not a solution at all.

Substitute x=3 in the equation and evaluate the addition, subtraction, and division.

\frac{3 + 5}{3 - 3} = \frac{8}{0} = \text{undefined}

x = 3 is not a solution to this equation.

Graph the corresponding function, f\left(x\right)=\dfrac{x + 5}{x - 3}, and explain how any viable solutions, extraneous solutions or restricted domain values are displayed in the graph.

By solving the denominator equal to zero we can determine any domain restrictions of the function:

And by solving the equation \dfrac{x+5}{x-3}=0, we have:

Since x=-5 does not make the denominator of the function equal to zero, it is a viable solution.

Use technology to graph the function and analyze the function at x=3 and x=-5.

We can see that as the function approaches the undefined value at x=3 from the left, f\left(x\right) \to -\infty, and as the function approaches x=3 from the right, f\left(x\right) \to \infty. Hence, we see a vertical asymptote.

When x=-5, the value of the function is equal to zero and we can see an x-intercept at \left(-5,0\right) on the graph.

Recall for a rational function, if a value of x=a leads to a zero denominator but not a zero numerator or is a zero of lower multiplicity in the numerator, a vertical asymptote forms on the graph at x=a. We also have if a value of x=b leads to a zero numerator but not a zero denominator the graph of the function will have an x-intercept at x=b.

Determine whether x = 3 is a viable solution, an extraneous solution, or not a solution at all.

We can solve the equation for x as shown and get the solution x=3:

It looks like x=3 is a solution to the equation. However, we still need to examine the denominator.

We can see that x = 3 is a zero of both the numerator and denominator of the rational expression:\frac{\left( 3-3 \right) ^{2}}{3-3}= \frac{0}{0} = \text{undefined}

Although x=3 will be an algebraic solution to the equation, it is also a value which makes the equation undefined.

So x = 3 is an extraneous solution to this equation.

Unlike \dfrac{x+5}{x-3}=0, x=3 would be a solution of the equation \dfrac{\left(x - 3\right)^{2}}{x - 3} = 0 but it will lead to a zero denominator, which makes the type of solution different.

Graph the corresponding function, f\left(x\right)=\dfrac{\left(x - 3\right)^{2}}{x - 3}, and explain how any viable solutions, extraneous solutions or restricted domain values are displayed in the graph.

Use technology to graph the function and analyze the function at x=3.

We know that x=3 is an extraneous solution by analyzing any restrictions on the function. Without that, the graph of f\left(x\right)=\dfrac{\left(x - 3\right)^{2}}{x - 3} looks as though it includes the point \left(3, 0 \right), but when tracing the function, the point is excluded.

We consider the extraneous solution x=3 a removable point of discontinuity because it is a zero of both the numerator and denominator, and the zero in the numerator is not of lower multiplicity than the denominator. So, the graph should include a removable point of discontinuity or a hole, which we can manually enter to show this restriction on the domain:

Solve the equation.

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

We also want to factor the denominator, however, to check restrictions on the variable that could make each solution viable or extraneous.

When finding solutions to the equation, it's best not to simplify common factors in the numerator and denominator, as that may make us miss any possible extraneous solutions.

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 \frac{\left(-8\right)^{2} + 5\left(-8\right) - 24}{\left(-8\right)^{2} - 9\left(-8\right) + 18} = \frac{64 - 40 - 24}{64 + 72 + 18} = \frac{0}{154} = 0

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

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

Graph the corresponding function, f\left(x\right)=\dfrac{x^{2} + 5x - 24}{x^{2} - 9x + 18}, and explain how any viable solutions, extraneous solutions or restricted domain values are displayed in the graph.

Use technology to graph the function and determine the behavior of the function at the points of interest, x=-8, x=3, and x=6.

We know that x=-8 is a viable solution, and we see on the graph that the x-intercept is at \left( -8, 0 \right).

Since x=6 is undefined, but not extraneous, there is a vertical asymptote at x=6.

We know that x=3 is an extraneous solution, so there is a removable point of discontinuity at x=3. This is not obvious with a visible inspection of the computer-generated graph. When we draw a sketch of the graph, we show the visible hole in the graph at x=3 in order to draw attention to the fact that the function is undefined at this point.

Notice that if we had simplified the equation in part (a) to \dfrac{\left( x + 8 \right)}{\left( x - 6 \right)}, we may have missed the fact that x=3 is an extraneous solution, and may not have recognized there was a point of discontinuity in the graph.

Solve the equation for x.

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 initial 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.

Graph the corresponding function, f\left(x\right)=\dfrac{1}{x + 6} - \dfrac{2}{2x - 5}, and explain how any viable solutions, extraneous solutions or restricted domain values are displayed in the graph.

Before graphing with technology, we can algebraically determine key features of the function.

We know there are no removable points of discontinuity because no values of x will make the numerator of the function equal to zero.

When x+6=0 or 2x-5=0, the graph will have vertical asymptotes and the function will be undefined. So, when x=-6 and x=\dfrac{5}{2}, the function is undefined and there will be two vertical asymptotes.

We determined that the equation \dfrac{1}{x+6}=\dfrac{2}{2x-5} has no solutions, so f\left(x\right) will have no x-intercepts.

Examining the graph, we see there is a horizontal asymptote at f \left(x \right)=0, which means there are no solutions to f \left(x \right)=0. Two vertical asymptotes are located at x=-6 and x=\dfrac{5}{2}.

Write the average cost function, A\left(x\right), of producing x car batteries in a month.

The average cost function is A \left(x \right)= \dfrac{ \text{total cost of producing batteries for the month}}{\text{total number of batteries produced}}.

Since the production costs \$10\,000 and each car battery produced costs \$10, the expression that represents the total cost for the month of producing batteries is 10\,000+10x, for x car batteries produced that month.

The average cost per battery would be calculated by dividing the total by the number of batteries, x, so the average cost function isA\left(x\right)=\frac{10\,000+10x}{x}

State the domain and range of the function.

Consider the context of the problem and what the variables represent. A \left(x \right) is the average cost per battery and x is the number of batteries.

Then, consider any values of x that lead to non-viable solutions, extraneous solutions, or that make the function undefined.

Based on the context, we can initially state that the cost must be greater than or equal to \$0 and the number of batteries produced must also be greater than or equal to 0. Also in the real-life context of producing batteries, we would only have whole number values for x, since we can't make a negative number of batteries or make a fraction of a battery.

When x=0, the function is undefined since the denominator is equal to zero, so x=0 is excluded from the domain of the function.

When 10 \,000 + 10x =0, the solution for x will be a negative number of car batteries. This occurs when x=-1 000. Since we've already stated that the domain are positive values of x, -1 000 is already excluded from the domain and considered a non-viable solution.

The domain of the function can be defined as the integers on the interval \left[1, \infty \right). No maximum number of batteries was indicated in the problem.

Since this is an inverse relationship, as more batteries are produced, the average cost will decrease. Thus, when x=1 the cost of production will be its most expensive, at A \left(1 \right) = \dfrac{10\,000+ 10 \left(1 \right)}{1}= \dfrac{10 \,010}{1}= \$10\,010. As the cost can never be less than \$0, the range of the function can be defined as \left( 0, 10\,010 \right ], in interval notation. Although, in practice this would be a set of discrete values due to the domain being whole numbers.

How many car batteries does the Apoleisk Company need to produce each month so that the average cost of production per car battery is \$60?

Since we are attempting to find A \left(x\right)=60, solve the equation 60 = \dfrac{10 \,000 + 10}{x}.

The Apoleisk Company needs to produce 200 car batteries each month so that the average cost of production per battery is \$60.