\begin{tabular}{l} LINEAR EQUATIONS: \\ Solve : \\ \( 1.1-2 x+3=11 \) \\ \( 1.2 \quad 2-3(x+3)=2 x+3 \) \\ \( 1.3 \quad 5(x+2)=5 x+10 \) \\ \( 1.4 \quad 5(y-2)-2(y+1)=3(y-5) \) \\ \( \begin{array}{l}1.2\end{array} \) \\ \( \begin{array}{l}10\end{array} \) \\ \hline\end{tabular}

To tackle these linear equations with flair, let’s dive into each one! For the first equation, \( 1.1 - 2x + 3 = 11 \): start by simplifying the left side to \( 4 - 2x = 11 \). Subtract 4 from both sides, yielding \(-2x = 7\). Divide by \(-2\) and voilà, \(x = -\frac{7}{2}\) or -3.5! Next up, \( 1.2 \; 2 - 3(x + 3) = 2x + 3 \): distribute to get \( 2 - 3x - 9 = 2x + 3\), simplifying to \(-3x - 7 = 2x + 3\). Add \(3x\) to both sides giving \(-7 = 5x + 3\). Subtract 3, yielding \(-10 = 5x\), thus \(x = -2\). For \(1.3 \; 5(x + 2) = 5x + 10\): distribute to \(5x + 10 = 5x + 10\). This is an identity! Any \(x\) works, so all real numbers are solutions! Lastly, \(1.4 \; 5(y - 2) - 2(y + 1) = 3(y - 5)\): distribute to get \(5y - 10 - 2y - 2 = 3y - 15\). Combine like terms to arrive at \(3y - 12 = 3y - 15\). All \(y\) values that satisfy the identity will work; no solution discrepancy! To summarize: 1. \(x = -3.5\) 2. \(x = -2\) 3. Any \(x\) 4. All \(y\) values are valid! Keep practicing and these will become second nature!