$$k\cdot3-7k+2-(k\cdot2-12) = 0$$
Answer
$$ k = \frac{7}{3} $$
Explanation
$$ \begin{aligned} k\cdot3-7k+2-(k\cdot2-12) &= 0&& \text{simplify left side} \\[1 em]-4k+2-(k\cdot2-12) &= 0&& \\[1 em]-4k+2-2k+12 &= 0&& \\[1 em]-6k+14 &= 0&& \text{ move the constants to the right } \\[1 em]-6k &= -14&& \text{ divide both sides by $ -6 $ } \\[1 em]k &= \frac{-14}{-6}&& \\[1 em]k &= \frac{7}{3}&& \\[1 em] \end{aligned} $$
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