To solve the equation \(\frac{2}{3x} + \frac{1}{4} = \frac{25}{6x} - \frac{1}{3}\), we will perform the following steps:
Step a: Identify restrictions on the variable
The equations have denominators that can become zero if the variable \(x\) takes certain values. We look for any term in the denominator:
- For \(\frac{2}{3x}\), the denominator is \(3x\). This term is undefined when \(3x = 0\) which gives \(x = 0\).
- For \(\frac{25}{6x}\), the denominator is \(6x\). This term is also undefined when \(6x = 0\), which also gives \(x = 0\).
Thus, the restriction on the variable is: \[ x \neq 0 \]
Step b: Solve the equation
We now proceed to solve the equation while keeping the restriction in mind.
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Start with the original equation: \[ \frac{2}{3x} + \frac{1}{4} = \frac{25}{6x} - \frac{1}{3} \]
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To eliminate the fractions, find a common denominator for all terms. The least common multiple (LCM) of the denominators \(3x\), \(4\), and \(6x\) is \(12x\). So we will multiply each term by \(12x\): \[ 12x \cdot \frac{2}{3x} + 12x \cdot \frac{1}{4} = 12x \cdot \frac{25}{6x} - 12x \cdot \frac{1}{3} \]
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Simplifying each term:
- For the first term: \(12x \cdot \frac{2}{3x} = 4 \cdot 2 = 8\)
- For the second term: \(12x \cdot \frac{1}{4} = 3x\)
- For the third term: \(12x \cdot \frac{25}{6x} = 2 \cdot 25 = 50\)
- For the fourth term: \(12x \cdot \frac{1}{3} = 4x\)
Putting it all together: \[ 8 + 3x = 50 - 4x \]
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Combine like terms. Start by adding \(4x\) to both sides: \[ 8 + 3x + 4x = 50 \] This simplifies to: \[ 8 + 7x = 50 \]
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Subtract \(8\) from both sides: \[ 7x = 50 - 8 \] Simplifying gives: \[ 7x = 42 \]
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Finally, divide both sides by \(7\): \[ x = \frac{42}{7} = 6 \]
Final Answer
a. The restriction on the variable is \(x \neq 0\).
b. The solution to the equation is \(x = 6\).