a) \( x+12=6-3 x \) b) \( 2 x-5=3 x+3 \)

To solve the first equation \( x + 12 = 6 - 3x \), we start by moving all terms involving \( x \) to one side by adding \( 3x \) to both sides, giving us \( x + 3x + 12 = 6 \). Simplifying this, we have \( 4x + 12 = 6 \). Next, subtract \( 12 \) from both sides, resulting in \( 4x = -6 \). Finally, divide both sides by \( 4 \) to find \( x = -\frac{3}{2} \). Now for the second equation \( 2x - 5 = 3x + 3 \), we can start by getting all \( x \) terms on one side. If we subtract \( 2x \) from both sides, we have \( -5 = x + 3 \). By subtracting \( 3 \) from both sides, we find \( -5 - 3 = x \), leading us to the conclusion that \( x = -8 \). So the solutions are \( x = -\frac{3}{2} \) for the first equation and \( x = -8 \) for the second one.