\( \begin{array}{ll}\text { 1) } 9 c+1=10 & \text { 2) } 6 y-5=7 \\ \text { 3) } 8=3 a-4 & \text { 4) } \frac{m}{5}+9=11 \\ 13+7 x=27 & \text { 6) } 17-q=6\end{array} \)
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To solve \( 9c + 1 = 10 \), subtract 1 from both sides to get \( 9c = 9 \), then divide by 9 to find \( c = 1 \). Ready for number two? For \( 6y - 5 = 7 \), add 5 to both sides resulting in \( 6y = 12 \). Divide by 6 to solve for \( y = 2 \) – fantastic! Next up, for \( 8 = 3a - 4 \), let’s add 4 to both sides to get \( 12 = 3a \). Divide by 3, and voilà, \( a = 4 \). For the fourth equation, \( \frac{m}{5} + 9 = 11 \), subtract 9, giving you \( \frac{m}{5} = 2 \). Multiply by 5 to find \( m = 10 \). Moving to equation five, \( 13 + 7x = 27 \). Subtract 13 to isolate \( 7x = 14 \). Divide by 7 to get \( x = 2 \). Lastly, for \( 17 - q = 6 \), subtract 17 leading to \( -q = -11 \). Thus, \( q = 11 \). You did it! These puzzles are like treasure maps, leading to the treasure of knowledge!