Question

1. \frac{1}{3x^2} = \frac{x + 3}{2x^2} - \frac{1}{6x^2}

Explanation:

Step1: Find a common denominator

The denominators are \(3x^2\), \(2x^2\), and \(6x^2\). The least common denominator (LCD) of these is \(6x^2\).

Step2: Rewrite each fraction with the LCD

• For \(\frac{1}{3x^2}\), multiply numerator and denominator by \(2\): \(\frac{1\times2}{3x^2\times2}=\frac{2}{6x^2}\)
• For \(\frac{x + 3}{2x^2}\), multiply numerator and denominator by \(3\): \(\frac{(x + 3)\times3}{2x^2\times3}=\frac{3x+9}{6x^2}\)
• The third fraction \(\frac{1}{6x^2}\) already has the LCD.

Step3: Substitute back into the equation

The equation \(\frac{1}{3x^2}=\frac{x + 3}{2x^2}-\frac{1}{6x^2}\) becomes \(\frac{2}{6x^2}=\frac{3x + 9}{6x^2}-\frac{1}{6x^2}\)

Step4: Simplify the right - hand side

\(\frac{3x + 9-1}{6x^2}=\frac{3x+8}{6x^2}\)

Step5: Solve for \(x\)

Since the denominators are the same (\(x
eq0\)), we can set the numerators equal:
\(2=3x + 8\)
Subtract \(8\) from both sides: \(2-8 = 3x\), so \(-6=3x\)
Divide both sides by \(3\): \(x=- 2\)
We need to check if \(x = - 2\) makes the original denominators non - zero. For \(3x^2\), \(3\times(-2)^2=12
eq0\); for \(2x^2\), \(2\times(-2)^2 = 8
eq0\); for \(6x^2\), \(6\times(-2)^2=24
eq0\). So \(x=-2\) is a valid solution.

Answer:

\(x=-2\)