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 finding the least common denominator (LCD) of all the fractions and then multiplying each term in the equation by the LCD. The denominators here are \(10w\), \(10\), and \(w^{2}\). The prime factors of \(10\) are \(2\times5\), for \(w\) (from \(10w\)) and \(w^{2}\) (from \(w^{2}\)), the highest power of \(w\) is \(w^{2}\). So the LCD of \(10w\), \(10\), and \(w^{2}\) is \(10w^{2}\)? Wait, no, wait. Wait, the denominators are \(10w\), \(10\), and \(w^{2}\). Let's factor each denominator:

• \(10w=2\times5\times w\)
• \(10 = 2\times5\)
• \(w^{2}=w\times w\)

The LCD is the product of the highest powers of all prime factors and variables. So for \(2\) and \(5\), the highest power is \(2\times5 = 10\), for \(w\) the highest power is \(w^{2}\). Wait, no, wait the first term is \(\frac{7}{10w}\), the second is \(\frac{1}{10}\), the third is \(\frac{1}{w^{2}}\). Wait, maybe I made a mistake. Wait, let's list the denominators: \(10w\), \(10\), \(w^{2}\). The LCD is the least common multiple of these denominators. Let's find LCM of \(10w\), \(10\), \(w^{2}\).

LCM of \(10\) and \(10\) is \(10\), LCM of \(w\) and \(w^{2}\) is \(w^{2}\), so LCM of \(10w\), \(10\), \(w^{2}\) is \(10w^{2}\)? Wait, no, wait \(10w\) is \(10\times w\), \(10\) is \(10\times1\), \(w^{2}\) is \(w\times w\). So to get a common denominator, we need to multiply each term by a value that will cancel out each denominator. Let's check each option:

• Option A: Dividing by \(10w\) will not eliminate the denominators. So A is wrong.
• Option B: Multiplying by \(10w\). Let's see, for the term \(\frac{1}{w^{2}}\), multiplying by \(10w\) gives \(\frac{10w}{w^{2}}=\frac{10}{w}\), which still has a denominator. So B is wrong.
• Option C: Multiplying all terms by \(10w^{2}\). Let's check each term:
• For \(\frac{7}{10w}\times10w^{2}=7w\)
• For \(\frac{1}{10}\times10w^{2}=w^{2}\)
• For \(\frac{1}{w^{2}}\times10w^{2}=10\)

So this eliminates all denominators.

• Option D: Multiplying only two terms by \(10w^{2}\) is incorrect because we need to multiply all terms in the equation to maintain equality. So D is wrong.

Step 2: Conclusion on the first sub - question

So the correct step is to multiply all terms by \(10w^{2}\), which is option C.

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

First, we solve the equation \(\frac{7}{10w}=\frac{1}{10}+\frac{1}{w^{2}}\) by multiplying each term by \(10w^{2}\) (from the correct step above):

Step 1: Multiply 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:

• Left - hand side: \(10w^{2}\times\frac{7}{10w}=7w\)
• Middle term: \(10w^{2}\times\frac{1}{10}=w^{2}\)
• Right - hand term: \(10w^{2}\times\frac{1}{w^{2}} = 10\)

So the equation becomes \(7w=w^{2}+10\)

Step 2: Rearrange into standard quadratic form

\(w^{2}-7w + 10=0\)

Step 3: Factor the quadratic equation

We need two numbers that multiply to \(10\) and add up to \(-7\). The numbers are \(-2\) and \(-5\). So \((w - 2)(w - 5)=0\)

Step 4: Solve for \(w\)

Setting each factor equal to zero:

• \(w - 2=0\Rightarrow w = 2\)
• \(w - 5=0\Rightarrow w = 5\)

Now we check for extraneous solutions. An extraneous solution is a solution that makes the original denominator equal to zero. The original denominators are \(10w\), \(10\), and \(w^{2}\).

• For \(w = 2\): \(10w=20

eq0\), \(w^{2}=4
eq0\), so \(w = 2\) is valid.

• For \(w = 5\): \(10w = 50

eq0\), \(w^{2}=2…

Answer:

First Sub - Question: C. Multiply all terms in the equation by \(10w^{2}\)
Second Sub - Question: D. No extraneous solution exists.