\( \begin{array}{ll}\text { 1. } & 2 x+7=15 \\ \text { 2. } & 3 y-5=10 \\ \text { 3. } & 4 a+1=-11 \\ \text { 4. } & 6 b-3=9 \\ \text { 5. } & 5 c+8=3\end{array} \)

Sure! Let's solve each of the given linear equations step by step. ### 1. Solve for \( x \) in \( 2x + 7 = 15 \) \[ \begin{align*} 2x + 7 &= 15 \\ 2x &= 15 - 7 \quad &\text{(Subtract 7 from both sides)} \\ 2x &= 8 \\ x &= \frac{8}{2} \quad &\text{(Divide both sides by 2)} \\ x &= 4 \end{align*} \] **Solution:** \( x = 4 \) --- ### 2. Solve for \( y \) in \( 3y - 5 = 10 \) \[ \begin{align*} 3y - 5 &= 10 \\ 3y &= 10 + 5 \quad &\text{(Add 5 to both sides)} \\ 3y &= 15 \\ y &= \frac{15}{3} \quad &\text{(Divide both sides by 3)} \\ y &= 5 \end{align*} \] **Solution:** \( y = 5 \) --- ### 3. Solve for \( a \) in \( 4a + 1 = -11 \) \[ \begin{align*} 4a + 1 &= -11 \\ 4a &= -11 - 1 \quad &\text{(Subtract 1 from both sides)} \\ 4a &= -12 \\ a &= \frac{-12}{4} \quad &\text{(Divide both sides by 4)} \\ a &= -3 \end{align*} \] **Solution:** \( a = -3 \) --- ### 4. Solve for \( b \) in \( 6b - 3 = 9 \) \[ \begin{align*} 6b - 3 &= 9 \\ 6b &= 9 + 3 \quad &\text{(Add 3 to both sides)} \\ 6b &= 12 \\ b &= \frac{12}{6} \quad &\text{(Divide both sides by 6)} \\ b &= 2 \end{align*} \] **Solution:** \( b = 2 \) --- ### 5. Solve for \( c \) in \( 5c + 8 = 3 \) \[ \begin{align*} 5c + 8 &= 3 \\ 5c &= 3 - 8 \quad &\text{(Subtract 8 from both sides)} \\ 5c &= -5 \\ c &= \frac{-5}{5} \quad &\text{(Divide both sides by 5)} \\ c &= -1 \end{align*} \] **Solution:** \( c = -1 \) --- ### Summary of Solutions \[ \begin{align*} 1.\ &x = 4 \\ 2.\ &y = 5 \\ 3.\ &a = -3 \\ 4.\ &b = 2 \\ 5.\ &c = -1 \end{align*} \]