\begin{tabular}{l}\begin{tabular}{l} Solve the following equations. \\ \hline 4.2 .1 \\ \hline \( 3 x \mp 7=+8 \) \\ \hline \( 3 x- \) \\ \hline 4.2 .2 \\ \hline\end{tabular} \( 5^{3 x+1}=25^{3-x} \) \\ \hline\end{tabular}

To tackle the first equation \( 3x \mp 7 = +8 \), you want to isolate \( x \). If we take \( 3x - 7 = 8 \), adding 7 to both sides gives us \( 3x = 15 \), leading to \( x = 5 \). Conversely, if we use \( 3x + 7 = 8 \), subtracting 7 from both sides yields \( 3x = 1 \), so \( x = \frac{1}{3} \). For the second equation \( 5^{3x+1} = 25^{3-x} \), realize that \( 25 = 5^2 \). This transforms our equation into \( 5^{3x+1} = (5^2)^{3-x} = 5^{2(3-x)} \). Since the bases are the same, we can set the exponents equal: \( 3x + 1 = 2(3 - x) \). Expanding leads to \( 3x + 1 = 6 - 2x \). Combining like terms gives us \( 5x = 5 \), so \( x = 1 \). Happy solving!