To determine which equations have the same solution set as the equation \(\frac{2}{3} - x + \frac{1}{6} = 6x\), we can manipulate the original equation by applying algebraic properties without solving explicitly.
First, let's simplify the original equation:
- Combine the terms on the left side: \[ \frac{2}{3} + \frac{1}{6} - x = 6x \] To combine \(\frac{2}{3}\) and \(\frac{1}{6}\), find a common denominator (which is 6): \[ \frac{4}{6} + \frac{1}{6} - x = 6x \quad \Rightarrow \quad \frac{5}{6} - x = 6x \]
Now we can analyze the options given and see if they can be transformed into an equivalent equation with the same solution set as \(\frac{5}{6} - x = 6x\).
Examining each equation:
-
\(4 - 6x + 1 = 36x\)
- Combine like terms: \(5 - 6x = 36x\)
- Rearranging gives: \(5 = 42x\) (Same as option 7, which is equivalent).
-
\(\frac{5}{6} - x = 6x - x\)
- Simplifies to: \(\frac{5}{6} - x = 5x\)
- Rearranging gives: \(\frac{5}{6} = 6x\) (not equivalent to the original).
-
\(4 - x + 1 = 6x\)
- Combine like terms: \(5 - x = 6x\)
- Rearranging gives: \(5 = 7x\) (not equivalent to the original).
-
\(\frac{5}{6} + x = 6x + x\)
- Simplifies to: \(\frac{5}{6} + x = 7x\)
- Rearranging gives: \(\frac{5}{6} = 6x\) (not equivalent to the original).
-
\(5 = 30x\)
- Rearranging gives: \(x = \frac{1}{6}\) (not equivalent to the original).
-
\(5 = 42x\)
- Rearranging gives: \(x = \frac{5}{42}\) (not equivalent to the original).
Conclusion: The equations with the same solution set are:
- Option 1: \(4 - 6x + 1 = 36x\) (which simplifies to \(5 = 42x\))
- None of the other options have the same solution set as \(\frac{5}{6} - x = 6x\).