Question

assignment 4.5
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solve each equation. remember to check fo

1. \\(\frac{1}{p^2} - \frac{1}{p} = \frac{2}{p}\\)

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

1. \\(\frac{1}{x} - \frac{4}{3x^2} = \frac{1}{3x^2}\\)

Explanation:

Problem 1: Solve \(\boldsymbol{\frac{1}{p^2} - \frac{1}{p} = \frac{2}{p}}\)

Step 1: Eliminate denominators ( \(p

eq 0\) )
Multiply all terms by \(p^2\) (the least common denominator) to get rid of the fractions:
\(p^2 \cdot \frac{1}{p^2} - p^2 \cdot \frac{1}{p} = p^2 \cdot \frac{2}{p}\)
Simplify each term:
\(1 - p = 2p\)

Step 2: Solve for \(p\)

Add \(p\) to both sides:
\(1 = 3p\)
Divide both sides by \(3\):
\(p = \frac{1}{3}\)

Step 3: Check the solution

Substitute \(p = \frac{1}{3}\) into the original equation:
Left - hand side (LHS): \(\frac{1}{(\frac{1}{3})^2} - \frac{1}{\frac{1}{3}} = 9 - 3 = 6\)
Right - hand side (RHS): \(\frac{2}{\frac{1}{3}} = 6\)
Since \(LHS = RHS\), \(p=\frac{1}{3}\) is valid (and \(p
eq0\) is satisfied).

Problem 3: Solve \(\boldsymbol{\frac{1}{p} = \frac{6p + 3}{p^2} + \frac{1}{p^2}}\)

Step 1: Combine fractions on the right ( \(p

eq 0\) )
First, combine the two fractions on the right - hand side:
\(\frac{6p + 3}{p^2}+\frac{1}{p^2}=\frac{6p + 3+1}{p^2}=\frac{6p + 4}{p^2}\)
So the equation becomes \(\frac{1}{p}=\frac{6p + 4}{p^2}\)

Step 2: Eliminate denominators

Multiply both sides by \(p^2\) (since \(p
eq0\)):
\(p^2\cdot\frac{1}{p}=p^2\cdot\frac{6p + 4}{p^2}\)
Simplify: \(p = 6p+4\)

Step 3: Solve for \(p\)

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

Step 4: Check the solution

Substitute \(p = -\frac{4}{5}\) into the original equation:
LHS: \(\frac{1}{-\frac{4}{5}}=-\frac{5}{4}\)
RHS: \(\frac{6(-\frac{4}{5})+3}{(-\frac{4}{5})^2}+\frac{1}{(-\frac{4}{5})^2}=\frac{-\frac{24}{5}+3}{\frac{16}{25}}+\frac{1}{\frac{16}{25}}=\frac{-\frac{24}{5}+\frac{15}{5}}{\frac{16}{25}}+\frac{25}{16}=\frac{-\frac{9}{5}}{\frac{16}{25}}+\frac{25}{16}=-\frac{45}{16}+\frac{25}{16}=-\frac{20}{16}=-\frac{5}{4}\)
Since \(LHS = RHS\), \(p = -\frac{4}{5}\) is valid (and \(p
eq0\) is satisfied).

Problem 5: Solve \(\boldsymbol{\frac{1}{x}-\frac{4}{3x^2}=\frac{1}{3x^2}}\)

Step 1: Eliminate denominators ( \(x

eq 0\) )
Multiply all terms by \(3x^2\) (the least common denominator):
\(3x^2\cdot\frac{1}{x}-3x^2\cdot\frac{4}{3x^2}=3x^2\cdot\frac{1}{3x^2}\)
Simplify each term:
\(3x-4 = 1\)

Step 2: Solve for \(x\)

Add \(4\) to both sides:
\(3x=1 + 4=5\)
Divide both sides by \(3\):
\(x=\frac{5}{3}\)

Step 3: Check the solution

Substitute \(x = \frac{5}{3}\) into the original equation:
LHS: \(\frac{1}{\frac{5}{3}}-\frac{4}{3(\frac{5}{3})^2}=\frac{3}{5}-\frac{4}{3\times\frac{25}{9}}=\frac{3}{5}-\frac{4}{\frac{25}{3}}=\frac{3}{5}-\frac{12}{25}=\frac{15 - 12}{25}=\frac{3}{25}\)
RHS: \(\frac{1}{3(\frac{5}{3})^2}=\frac{1}{3\times\frac{25}{9}}=\frac{3}{25}\)
Since \(LHS = RHS\), \(x=\frac{5}{3}\) is valid (and \(x
eq0\) is satisfied).

Answer:

s:

1. \(p=\boldsymbol{\frac{1}{3}}\)
2. \(p=\boldsymbol{-\frac{4}{5}}\)
3. \(x=\boldsymbol{\frac{5}{3}}\)