Question

select the correct answer.
it would take charlene and gina 18 minutes to organize charlene’s comic book collection if they work together. if gina works alone, it would take her 15 minutes longer than it would take charlene working alone.
when x represents the number of minutes it takes charlene to organize the comic book collection working alone, the situation is represented by this equation:
\\(\frac{1}{x} + \frac{1}{x + 15} = \frac{1}{18}\\).
which value is a solution of the rational equation that must be discarded because of the context?
a. \\(x = 0\\)
b. \\(x = -9\\)
c. \\(x = -15\\)
d. \\(x = -30\\)

Explanation:

Step1: Solve the rational equation

First, find a common denominator for the left - hand side of the equation \(\frac{1}{x}+\frac{1}{x + 15}=\frac{1}{18}\). The common denominator is \(x(x + 15)\).
Multiply each term by \(x(x + 15)\) to clear the fractions:
\(x(x + 15)\times\frac{1}{x}+x(x + 15)\times\frac{1}{x + 15}=x(x + 15)\times\frac{1}{18}\)
Simplify each term:
\((x + 15)+x=\frac{x(x + 15)}{18}\)
Combine like terms on the left - hand side:
\(2x+15=\frac{x^{2}+15x}{18}\)
Multiply both sides by 18 to eliminate the fraction:
\(18(2x + 15)=x^{2}+15x\)
Expand the left - hand side:
\(36x+270=x^{2}+15x\)
Rearrange the equation to form a quadratic equation:
\(x^{2}+15x-36x - 270 = 0\)
\(x^{2}-21x - 270=0\)
Factor the quadratic equation:
We need to find two numbers that multiply to \(- 270\) and add up to \(-21\). The numbers are \(-30\) and \(9\).
So, \(x^{2}-21x - 270=(x - 30)(x+9)=0\)
Set each factor equal to zero:
\(x - 30=0\) or \(x + 9=0\)
Solving for \(x\), we get \(x = 30\) or \(x=-9\)

Step2: Consider the context

In the context of the problem, \(x\) represents the number of minutes it takes Charlene to organize the comic book collection working alone. Also, we need to consider the domain of the original rational equation \(\frac{1}{x}+\frac{1}{x + 15}=\frac{1}{18}\). The denominators \(x\) and \(x + 15\) cannot be zero, so \(x
eq0\) and \(x
eq - 15\).
Now, we also know that Gina takes 15 minutes longer than Charlene to complete the work alone. If \(x=-9\), then the time it takes Gina to work alone is \(x + 15=-9 + 15 = 6\) minutes. But we know that working together they take 18 minutes, and if Gina takes 6 minutes alone and Charlene takes - 9 minutes (which is not possible in the context of time, since time cannot be negative), so we discard \(x=-9\)? Wait, no. Wait, let's re - examine the context. The time \(x\) (Charlene's time alone) and \(x + 15\) (Gina's time alone) must be positive because time cannot be negative.
If \(x=-9\), then Charlene's time is \(-9\) minutes (invalid, since time can't be negative). If \(x = 30\), then Gina's time is \(30 + 15=45\) minutes. Working together, the rate of Charlene is \(\frac{1}{30}\) (per minute) and the rate of Gina is \(\frac{1}{45}\) (per minute). The combined rate is \(\frac{1}{30}+\frac{1}{45}=\frac{3 + 2}{90}=\frac{5}{90}=\frac{1}{18}\), which matches the given equation.
Now let's check the options:

• Option A: If \(x = 0\), the denominator of \(\frac{1}{x}\) is zero, so \(x = 0\) is not in the domain of the original equation. But we are looking for a solution that must be discarded due to context.
• Option B: \(x=-9\) gives a negative time for Charlene, which is not possible in the context of the problem (time cannot be negative), so this solution must be discarded.
• Option C: If \(x=-15\), the denominator of \(\frac{1}{x + 15}\) is zero, so \(x=-15\) is not in the domain of the original equation.
• Option D: \(x=-30\), if we substitute \(x=-30\) into the original equation, \(\frac{1}{-30}+\frac{1}{-30 + 15}=\frac{-1}{30}+\frac{1}{-15}=\frac{-1-2}{30}=\frac{-3}{30}=-\frac{1}{10}

eq\frac{1}{18}\), so \(x=-30\) is not a solution of the equation.

Answer:

B. \(x=-9\)