Question

select the correct answer.
it takes leo and mary 6 minutes to wash a car. if leo washes the car alone, it takes 5 minutes longer than it takes mary working alone. when x represents the number of minutes it takes mary to wash the car working alone, the situation is represented by this equation:
\\(\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}\\)
which value is a solution of the rational equation that must be discarded because of the context?
a. \\(x = -10\\)
b. \\(x = -5\\)
c. \\(x = -3\\)
d. \\(x = 0\\)

Explanation:

Step1: Analyze the context

In the context of time taken to wash a car, \( x \) (time for Mary) and \( x + 5 \) (time for Leo) must be positive because time cannot be negative or zero. So we need to check which solution makes \( x \) or \( x + 5 \) non - positive.

Step2: Check each option

- Option A: If \( x=- 10\), then \( x+5=-10 + 5=-5\). Both \( x=-10\) (time for Mary) and \( x + 5=-5\) (time for Leo) are negative, which is not possible in the context of time.
- Option B: If \( x=-5\), then \( x + 5=-5 + 5 = 0\). Time for Leo is zero, which is not possible as it takes some time to wash a car. But let's also check the equation. The original equation is \(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\). If \( x=-5\), the term \(\frac{1}{x + 5}\) is undefined (division by zero), so \( x=-5\) is not a solution of the equation (it's an extraneous solution from the equation's domain perspective, but we are looking for a solution of the rational equation that must be discarded due to context. However, \( x=-5\) is not a solution of the equation because it makes the equation undefined.
- Option C: If \( x=-3\), then \( x+5=-3 + 5 = 2\). \( x=-3\) is negative (time for Mary is negative) which is not possible in the context of time, but let's check if it's a solution of the equation. Substitute \( x=-3\) into \(\frac{1}{x}+\frac{1}{x + 5}\): \(\frac{1}{-3}+\frac{1}{2}=\frac{-2 + 3}{-6}=\frac{1}{-6} eq\frac{1}{6}\). So \( x=-3\) is not a solution of the equation.
- Option D: If \( x = 0\), the term \(\frac{1}{x}\) is undefined (division by zero), so \( x = 0\) is not a solution of the equation.

Now, we first solve the rational equation \(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\).
Multiply through by \(6x(x + 5)\) (the least common denominator) to get rid of the fractions:
\(6(x + 5)+6x=x(x + 5)\)
Expand: \(6x+30 + 6x=x^{2}+5x\)
Simplify: \(12x+30=x^{2}+5x\)
Rearrange to standard quadratic form: \(x^{2}+5x-12x - 30=0\)
\(x^{2}-7x - 30=0\)
Factor: \(x^{2}-10x+3x - 30=0\)
\(x(x - 10)+3(x - 10)=0\)
\((x - 10)(x+3)=0\)
So the solutions of the equation are \(x = 10\) and \(x=-3\).

Among the options, \(x=-10\) is not a solution of the equation (since the solutions are \(x = 10\) and \(x=-3\)), but if we consider the context, \(x=-10\) gives negative times for both, and \(x=-3\) gives a negative time for Mary. But the question is asking for a solution of the rational equation that must be discarded because of the context. The solutions of the rational equation are \(x = 10\) and \(x=-3\). \(x=-3\) is a solution of the equation but in the context, time can't be negative, so \(x=-3\) is a solution of the equation that must be discarded? Wait, no, let's re - evaluate. Wait, when we solved the equation, we got \(x = 10\) and \(x=-3\). Now, let's check the options again.

Wait, maybe I made a mistake in solving the equation. Let's re - solve \(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\)

Multiply both sides by \(6x(x + 5)\):

\(6(x + 5)+6x=x(x + 5)\)

\(6x+30+6x=x^{2}+5x\)

\(12x + 30=x^{2}+5x\)

\(x^{2}-7x - 30=0\)

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 1\), \(b=-7\), \(c=-30\)

\(x=\frac{7\pm\sqrt{(-7)^{2}-4(1)(-30)}}{2(1)}=\frac{7\pm\sqrt{49 + 120}}{2}=\frac{7\pm\sqrt{169}}{2}=\frac{7\pm13}{2}\)

So \(x=\frac{7 + 13}{2}=\frac{20}{2}=10\) and \(x=\frac{7-13}{2}=\frac{-6}{2}=-3\)

Now, the solutions of the equation are \(x = 10\) and \(x=-3\). Now, in the context, \(x\) (time for Mary) must be positive, and \(x + 5\) (time for Leo) must be positive.

For \(x=-3\): \(x=-3\) (negative, invalid time for Mary), \(x + 5=2\) (positive, but Mary's time is negative). So \(x=-3\) is a solution of the equation but must be discarded due to context. But let's check the options. Option C is \(x=-3\), but wait, the options are A. \(x=-10\), B. \(x=-5\), C. \(x=-3\), D. \(x = 0\)

Wait, maybe I made a mistake in the initial analysis. Let's check the equation's domain. The domain of the equation \(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\) is \(x eq0\) and \(x eq - 5\) (because these values make the denominators zero). So \(x=-5\) and \(x = 0\) are not in the domain of the equation, so they are not solutions of the equation.

The solutions of the equation are \(x = 10\) and \(x=-3\). Now, \(x=-3\) is a solution of the equation, but in the context, \(x\) represents the number of minutes it takes Mary to wash the car alone, so \(x>0\) and \(x + 5>0\) (since Leo also takes some positive time). So \(x=-3\) is a solution of the equation but must be discarded because it gives a negative time for Mary. But let's check the options again. Wait, the option A is \(x=-10\). Let's see if \(x=-10\) is a solution of the equation.

Substitute \(x=-10\) into the left - hand side of the equation \(\frac{1}{x}+\frac{1}{x + 5}\):

\(\frac{1}{-10}+\frac{1}{-10 + 5}=\frac{1}{-10}+\frac{1}{-5}=\frac{-1-2}{10}=\frac{-3}{10} eq\frac{1}{6}\). So \(x=-10\) is not a solution of the equation.

Wait, there is a mistake in my earlier reasoning. Let's start over.

We have the equation \(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\)

1. First, find the domain of the equation: \(x eq0\) and \(x eq - 5\) (because denominators \(x\) and \(x + 5\) cannot be zero).

2. Solve the equation:

Multiply both sides by \(6x(x + 5)\):

\(6(x + 5)+6x=x(x + 5)\)

\(6x+30 + 6x=x^{2}+5x\)

\(12x+30=x^{2}+5x\)

\(x^{2}-7x - 30=0\)

Factor: \(x^{2}-10x+3x - 30=0\)

\(x(x - 10)+3(x - 10)=0\)

\((x - 10)(x + 3)=0\)

So \(x = 10\) or \(x=-3\)

3. Now, check which of these solutions are in the domain and which must be discarded due to context.

- \(x = 10\): \(x=10>0\), \(x + 5=15>0\), so this is valid.
- \(x=-3\): \(x=-3<0\), which is not valid in the context of time (time cannot be negative), so \(x=-3\) is a solution of the equation that must be discarded due to context. But wait, the option C is \(x=-3\). But let's check the options again. Wait, maybe I made a mistake in the problem interpretation.

Wait, the question is "Which value is a solution of the rational equation that must be discarded because of the context?".

A solution of the rational equation is a value that satisfies the equation (after considering the domain? No, a solution of the rational equation is a value that makes the equation true, and we have to check if it's valid in the context.

We found that \(x=-3\) is a solution of the equation (since when we substitute \(x=-3\) into \(\frac{1}{x}+\frac{1}{x + 5}\), we get \(\frac{1}{-3}+\frac{1}{2}=\frac{-2 + 3}{-6}=\frac{1}{-6}\), wait, no! Wait, \(\frac{1}{-3}+\frac{1}{-3 + 5}=\frac{-1}{3}+\frac{1}{2}=\frac{-2 + 3}{6}=\frac{1}{6}\). Oh! I made a mistake in the calculation earlier. \(\frac{1}{-3}+\frac{1}{2}=\frac{-2 + 3}{6}=\frac{1}{6}\). So \(x=-3\) is a solution of the equation.

Now, in the context, \(x\) is the number of minutes it takes Mary to wash the car alone. So \(x>0\) (because time cannot be negative). Also, \(x + 5\) (time for Leo) must be \(>0\), so \(x>-5\).

For \(x=-3\): \(x=-3\) is negative, so it's not valid in the context (even though it's a solution of the equation).

For \(x=-10\): Let's substitute \(x=-10\) into the equation: \(\frac{1}{-10}+\frac{1}{-10 + 5}=\frac{-1}{10}+\frac{-1}{5}=\frac{-1-2}{10}=\frac{-3}{10} eq\frac{1}{6}\), so \(x=-10\) is not a solution of the equation.

For \(x=-5\): The term \(\frac{1}{x + 5}\) is undefined, so \(x=-5\) is not a solution of the equation.

For \(x = 0\): The term \(\frac{1}{x}\) is undefined, so \(x = 0\) is not a solution of the equation.

So the solution of the rational equation that must be discarded because of the context is \(x=-3\)? But wait, the option A is \(x=-10\). Wait, maybe I made a mistake in the sign when solving the equation.

Wait, let's re - solve the equation:

\(\frac{1}{x}+\frac{1}{x + 5}=\frac{1}{6}\)

\(\frac{x + 5+x}{x(x + 5)}=\frac{1}{6}\)

\(\frac{2x + 5}{x(x + 5)}=\frac{1}{6}\)

Cross - multiply: \(6(2x + 5)=x(x + 5)\)

\(12x+30=x^{2}+5x\)

\(x^{2}-7x - 30=0\)

Which is the same as before. So the solutions are \(x = 10\) and \(x=-3\).

Now, \(x=-3\) is a solution of the equation. In the context, \(x\) (Mary's time) must be positive. \(x=-3\) is negative, so it must be discarded. But the option C is \(x=-3\). But let's check the options again.

Wait, the problem says "Which value is a solution of the rational equation that must be discarded because of the context?".

A solution of the rational equation: \(x=-3\) is a solution (as we saw from the correct substitution: \(\frac{1}{-3}+\frac{1}{-3 + 5}=\frac{-1}{3}+\frac{1}{2}=\frac{-2 + 3}{6}=\frac{1}{6}\)). And in the context, \(x=-3\) is negative, so it must be discarded. But let's check the options. Option C is \(x=-3\), option A is \(x=-10\).

Wait, maybe there is a mistake in my context analysis. Let's think about the time for Leo: \(x + 5\). If \(x=-10\), then \(x + 5=-5\), which is negative. If \(x=-3\), \(x + 5 = 2\) (positive). So \(x=-10\) makes both times negative, \(x=-3\) makes Mary's time negative and Leo's time positive. But the equation's solution \(x=-3\) is a solution, but in the context, Mary's time can't be negative. So \(x=-3\) is a solution of the equation that must be discarded. But let's check the answer options again.

Wait, the options are:

A. \(x=-10\)

B. \(x=-5\)

C. \(x=-3\)

D. \(x = 0\)

We know that \(x=-3\) is a solution of the equation (since it satisfies the equation) and must be discarded because \(x=-3\) (Mary's time) is negative. So the answer should be C. But wait, let's check the initial calculation when I thought \(x=-3\) didn't satisfy the equation. I made a mistake earlier. \(\frac{1}{-3}+\frac{1}{-3 + 5}=\frac{-1}{3}+\frac{1}{2}=\frac{-2 + 3}{6}=\frac{1}{6}\), which is equal to the right - hand side. So \(x=-3\) is a solution. And since time can't be negative, we discard it.

Answer:

C. \(x=-3\)