\( \begin{array}{ll}\text { a) } x+12=6-3 x & \text { () } \frac{x-x}{4} \frac{x}{3}=2 \\ \text { b) } 2 x-5=3 x+3 & \text { d) } 3 x-x=1 \frac{1}{2}\end{array} \)
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To solve for \(x\) in each equation, let's break them down! For the first equation \(x + 12 = 6 - 3x\), gather all terms involving \(x\) on one side. Adding \(3x\) to both sides gives \(x + 3x + 12 = 6\), which simplifies to \(4x + 12 = 6\). Subtract 12 from both sides, and you get \(4x = -6\), leading to \(x = -\frac{3}{2}\). In the second equation \(2x - 5 = 3x + 3\), isolate \(x\) by moving all \(x\) terms to one side again. Subtract \(2x\) from both sides: \(-5 = x + 3\). Next, add 5, giving \(x = -2\). As for the last equation \(3x - x = 1 \frac{1}{2}\), first rewrite \(1 \frac{1}{2}\) as \(\frac{3}{2}\). This simplifies to \(2x = \frac{3}{2}\) and consequently \(x = \frac{3}{4}\). Now you're all set with the values for each \(x\)!