Solve the equation \(\frac{2y-1}{3} - \frac{3y-1}{4} = 1\)
\(\frac{4(2y-1)-3(3y-1)}{12}=12 \Rightarrow 8y - 4 - 9y + 3 = 12 \\ \Rightarrow -y-1=12\Rightarrow -y=13\Rightarrow y=-13\)
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