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11) \\(\frac{b + 2}{6b} - \frac{b - 2}{b} = \frac{1}{3}\\)

Question

  1. \\(\frac{b + 2}{6b} - \frac{b - 2}{b} = \frac{1}{3}\\)

Explanation:

Step1: Find common denominator

The denominators are \(6b\), \(b\), and \(3\). The least common denominator (LCD) is \(6b\). Multiply each term by \(6b\) to eliminate denominators:

$$ 6b\times\frac{b + 2}{6b}-6b\times\frac{b - 2}{b}=6b\times\frac{1}{3} $$

Simplify each term:

$$ (b + 2)-6(b - 2)=2b $$

Step2: Expand and simplify

Expand the left - hand side:

$$ b + 2-6b + 12=2b $$

Combine like terms:

$$ -5b+14 = 2b $$

Step3: Solve for \(b\)

Add \(5b\) to both sides:

$$ 14=2b + 5b $$
$$ 14 = 7b $$

Divide both sides by \(7\):

$$ b=\frac{14}{7}=2 $$

Step4: Check the solution

Substitute \(b = 2\) into the original equation:
Left - hand side: \(\frac{2 + 2}{6\times2}-\frac{2 - 2}{2}=\frac{4}{12}-0=\frac{1}{3}\)
Right - hand side: \(\frac{1}{3}\)
Since the left - hand side equals the right - hand side, \(b = 2\) is a valid solution.

Answer:

\(b = 2\)