\begin{tabular}{c|c}\( 7 x+2-2 \) & solve for \( x \) \\ \hline 0 & 4 \\ \hline 3 & -1 \\ \hline\end{tabular}
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To solve the equation \( 7x + 2 - 2 = 0 \), we first simplify it to \( 7x = 0 \). Therefore, \( x = 0 \). Now, plugging \( x = 0 \) into the equation for checking the second condition: \( 7(3) + 2 - 2 \), we simplify it to \( 21 + 2 - 2 = 21 \neq -1 \). So it looks like there's a misunderstanding on the context given because 3 doesn't satisfy the equation derived. Finally, for the row that gave -1, we substituted \( x = -1 \): \( 7(-1) + 2 - 2 = -7 + 2 - 2 = -7 \neq 3 \). Hence, only \( x = 0 \) satisfies the equation.
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