Question

1. \\(\frac{5}{6x} = \frac{2}{3x} + \frac{x + 4}{3x^2}\\)

Explanation:

Step1: Find common denominator

The denominators are \(6x\), \(3x\), \(3x^2\). The least common denominator (LCD) is \(6x^2\). Multiply each term by \(6x^2\) to eliminate denominators:
\(6x^2\times\frac{5}{6x}=6x^2\times\frac{2}{3x}+6x^2\times\frac{x + 4}{3x^2}\)
Simplify each term: \(5x = 4x+2(x + 4)\)

Step2: Simplify and solve for \(x\)

Expand the right - hand side: \(5x=4x + 2x+8\)
Combine like terms: \(5x=6x + 8\)
Subtract \(6x\) from both sides: \(5x-6x=8\)
\(-x = 8\)
Multiply both sides by \(- 1\): \(x=-8\)

Step3: Check for extraneous solutions

We need to check if \(x = - 8\) makes the original denominators non - zero.
For \(6x\), when \(x=-8\), \(6x=6\times(-8)=-48
eq0\)
For \(3x\), when \(x = - 8\), \(3x=3\times(-8)=-24
eq0\)
For \(3x^2\), when \(x=-8\), \(3x^2=3\times(-8)^2=3\times64 = 192
eq0\)
So \(x=-8\) is a valid solution.

Answer:

\(x=-8\)