Question

solving a rational equation
consider the rational equation
\\(\frac{7}{10w} = \frac{1}{10} + \frac{1}{w^2}\\)

which step is correct when solving the rational equation?
a divide all terms in the equation by 10w
b multiply all terms in the equation by 10w
c multiply all terms in the equation by \\(10w^2\\)
d multiply only \\(\frac{7}{10w}\\) and \\(\frac{1}{w^2}\\) by \\(10w^2\\)

which choice, if any, is an extraneous solution for the equation?
a 5
b 2
c -2
d no extraneous solution exists

Explanation:

First Sub - Question (Which step is correct...)

Step 1: Recall how to solve rational equations

To solve a rational equation, we eliminate the denominators by multiplying each term by the least common denominator (LCD) of all the fractions. The denominators here are \(10w\), \(10\), and \(w^{2}\).
The prime factors of \(10w=2\times5\times w\), \(10 = 2\times5\), and \(w^{2}=w\times w\). The LCD is the product of the highest powers of all prime factors involved, so \(LCD = 10w^{2}\) (since for \(2\) and \(5\) the highest power is in \(10\), for \(w\) the highest power is \(w^{2}\)).

Step 2: Analyze each option

• Option A: Dividing by \(10w\) will not eliminate the denominators. For example, \(\frac{1}{w^{2}}\) when divided by \(10w\) will still have a denominator. So A is incorrect.
• Option B: Multiplying by \(10w\) will not eliminate the denominator \(w^{2}\) (because \(10w\times\frac{1}{w^{2}}=\frac{10w}{w^{2}}=\frac{10}{w}\), still a fraction with \(w\) in the denominator). So B is incorrect.
• Option C: Multiplying all terms by \(10w^{2}\) (the LCD) will eliminate all denominators. Let's check:

\(10w^{2}\times\frac{7}{10w}=7w\), \(10w^{2}\times\frac{1}{10} = w^{2}\), \(10w^{2}\times\frac{1}{w^{2}}=10\). So this step will convert the rational equation into a polynomial equation, which is the correct way to solve rational equations.

• Option D: We need to multiply all terms (including the middle term \(\frac{1}{10}\)) by the LCD, not just two of the fractions. So D is incorrect.

Step 1: First, solve the equation using the correct step from the first sub - question

From the first sub - question, we multiply each term by \(10w^{2}\):
\(10w^{2}\times\frac{7}{10w}=10w^{2}\times\frac{1}{10}+10w^{2}\times\frac{1}{w^{2}}\)
Simplify each term:
\(7w=w^{2} + 10\)
Rearrange to get a quadratic equation:
\(w^{2}-7w + 10=0\)

Step 2: Factor the quadratic equation

We factor \(w^{2}-7w + 10\). We need two numbers that multiply to \(10\) and add up to \(- 7\). The numbers are \(-2\) and \(-5\).
So \(w^{2}-7w + 10=(w - 2)(w - 5)=0\)
Set each factor equal to zero:
\(w-2 = 0\) or \(w - 5=0\)
So \(w = 2\) or \(w=5\)

Step 3: Check for extraneous solutions

We need to check if these solutions make the original denominators zero. The original denominators are \(10w\), \(10\), and \(w^{2}\).

• For \(w = 2\): \(10w=10\times2 = 20

eq0\), \(w^{2}=4
eq0\)

• For \(w = 5\): \(10w=10\times5=50

eq0\), \(w^{2}=25
eq0\)

Since both solutions do not make any of the original denominators zero, there is no extraneous solution.

Answer:

C. Multiply all terms in the equation by \(10w^{2}\)

Second Sub - Question (Which choice is an extraneous solution...)