Question

(\frac{3p + 12}{p^{2}}=\frac{p - 1}{2p^{2}}+\frac{1}{p})

Explanation:

Step1: Identify the common denominator

The denominators are \(p^2\), \(2p^2\), and \(p\). The least common denominator (LCD) is \(2p^2\).

Step2: Multiply each term by the LCD

Multiply each term in the equation \(\frac{3p + 12}{p^2}=\frac{p - 1}{2p^2}+\frac{1}{p}\) by \(2p^2\) to eliminate the denominators:

$$ 2p^2\times\frac{3p + 12}{p^2}=2p^2\times\frac{p - 1}{2p^2}+2p^2\times\frac{1}{p} $$

Step3: Simplify each term

Simplify each term:

• For the first term: \(2p^2\times\frac{3p + 12}{p^2}=2(3p + 12)=6p + 24\)
• For the second term: \(2p^2\times\frac{p - 1}{2p^2}=p - 1\)
• For the third term: \(2p^2\times\frac{1}{p}=2p\)

So the equation becomes:

$$ 6p + 24=p - 1 + 2p $$

Step4: Combine like terms on the right side

Combine the \(p\) terms on the right side: \(p + 2p = 3p\), so the equation is:

$$ 6p + 24=3p - 1 $$

Step5: Subtract \(3p\) from both sides

Subtract \(3p\) from both sides:

$$ 6p - 3p + 24=3p - 3p - 1 $$

$$ 3p + 24=-1 $$

Step6: Subtract 24 from both sides

Subtract 24 from both sides:

$$ 3p + 24 - 24=-1 - 24 $$

$$ 3p=-25 $$

Step7: Divide by 3

Divide both sides by 3:

$$ p=\frac{-25}{3} $$

Step8: Check for extraneous solutions

We need to check if \(p = \frac{-25}{3}\) makes any original denominator zero. The denominators are \(p^2\), \(2p^2\), and \(p\). Since \(p=\frac{-25}{3}
eq0\), it is a valid solution.

Answer:

\(p = -\frac{25}{3}\)