Pregunta
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\( \begin{array}{ll}\text { 1) } 9 c+1=10 & \text { 2) } 6 y-5=7 \\ \text { 3) } 8=3 a-4 & \text { 4) } \frac{m}{5}+9=11 \\ 13+7 x=27 & \text { 6) } 17-q=6\end{array} \)

Ask by Griffin Wright. in the United States
Jan 22,2025

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Here are the solutions to the equations: 1. \( c = 1 \) 2. \( y = 2 \) 3. \( a = 4 \) 4. \( m = 10 \) 5. \( x = 2 \) 6. \( q = 11 \)

Solución

Sure, let's solve each of the equations step by step to find the values of the variables \( c \), \( y \), \( a \), \( m \), \( x \), and \( q \). --- ### 1) \( 9c + 1 = 10 \) **Steps to Solve:** 1. Subtract 1 from both sides: \[ 9c = 10 - 1 \] \[ 9c = 9 \] 2. Divide both sides by 9: \[ c = \frac{9}{9} \] \[ c = 1 \] **Solution:** \( c = 1 \) --- ### 2) \( 6y - 5 = 7 \) **Steps to Solve:** 1. Add 5 to both sides: \[ 6y = 7 + 5 \] \[ 6y = 12 \] 2. Divide both sides by 6: \[ y = \frac{12}{6} \] \[ y = 2 \] **Solution:** \( y = 2 \) --- ### 3) \( 8 = 3a - 4 \) **Steps to Solve:** 1. Add 4 to both sides: \[ 8 + 4 = 3a \] \[ 12 = 3a \] 2. Divide both sides by 3: \[ a = \frac{12}{3} \] \[ a = 4 \] **Solution:** \( a = 4 \) --- ### 4) \( \frac{m}{5} + 9 = 11 \) **Steps to Solve:** 1. Subtract 9 from both sides: \[ \frac{m}{5} = 11 - 9 \] \[ \frac{m}{5} = 2 \] 2. Multiply both sides by 5: \[ m = 2 \times 5 \] \[ m = 10 \] **Solution:** \( m = 10 \) --- ### 5) \( 13 + 7x = 27 \) **Steps to Solve:** 1. Subtract 13 from both sides: \[ 7x = 27 - 13 \] \[ 7x = 14 \] 2. Divide both sides by 7: \[ x = \frac{14}{7} \] \[ x = 2 \] **Solution:** \( x = 2 \) --- ### 6) \( 17 - q = 6 \) **Steps to Solve:** 1. Subtract 17 from both sides: \[ -q = 6 - 17 \] \[ -q = -11 \] 2. Multiply both sides by -1: \[ q = 11 \] **Solution:** \( q = 11 \) --- ### **Summary of Solutions:** 1. \( c = 1 \) 2. \( y = 2 \) 3. \( a = 4 \) 4. \( m = 10 \) 5. \( x = 2 \) 6. \( q = 11 \) If you have any further questions or need additional explanations, feel free to ask!

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Extra Insights

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!

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