Question

1. \\(\frac{1}{p} = \frac{6p + 3}{p^2} + \frac{1}{p^2}\\)

Explanation:

Step1: Combine like terms on the right

The right - hand side has two terms with denominator \(p^{2}\). We can combine \(\frac{6p + 3}{p^{2}}+\frac{1}{p^{2}}\) by adding the numerators: \(\frac{6p+3 + 1}{p^{2}}=\frac{6p + 4}{p^{2}}\)
So the equation becomes \(\frac{1}{p}=\frac{6p + 4}{p^{2}}\)

Step2: Eliminate denominators (note \(p

eq0\))
Multiply both sides of the equation by \(p^{2}\) (since \(p = 0\) would make the original equation undefined) to get rid of the denominators.
\(p^{2}\times\frac{1}{p}=p^{2}\times\frac{6p + 4}{p^{2}}\)
Simplify both sides: \(p=6p + 4\)

Step3: Solve for \(p\)

Subtract \(6p\) from both sides: \(p-6p=6p + 4-6p\)
\(- 5p=4\)
Divide both sides by \(-5\): \(p=-\frac{4}{5}\)

Answer:

\(p =-\frac{4}{5}\)