Question

solving a rational equation
solve: \\(\frac{1}{3a^2} - \frac{1}{a} = \frac{1}{6a^2}\\)
the solution is \\(a = \square\\).

Explanation:

Step1: Find common denominator

The denominators are \(3a^2\), \(a\), and \(6a^2\). The least common denominator (LCD) is \(6a^2\) (since \(6a^2\) is a multiple of \(3a^2\) and \(a\) (as \(6a^2=6a\times a\))). Multiply each term by \(6a^2\) to eliminate denominators:

$$ 6a^2\times\frac{1}{3a^2}-6a^2\times\frac{1}{a}=6a^2\times\frac{1}{6a^2} $$

Step2: Simplify each term

Simplify each product:

• For \(6a^2\times\frac{1}{3a^2}\), the \(a^2\) cancels, and \(6\div3 = 2\), so this term becomes \(2\).
• For \(6a^2\times\frac{1}{a}\), one \(a\) cancels, leaving \(6a\), so this term becomes \(- 6a\) (note the subtraction sign).
• For \(6a^2\times\frac{1}{6a^2}\), both \(6\) and \(a^2\) cancel, so this term becomes \(1\).

So the equation simplifies to:

$$ 2-6a = 1 $$

Step3: Solve for \(a\)

Subtract \(2\) from both sides:

$$ -6a=1 - 2=-1 $$

Then divide both sides by \(-6\):

$$ a=\frac{-1}{-6}=\frac{1}{6} $$

We should check for extraneous solutions (since \(a = 0\) would make the original denominators zero, but \(\frac{1}{6}
eq0\), so it's valid).

Answer:

\(\frac{1}{6}\)