$$x\cdot4+x\cdot3-43x\cdot2-x+42 = 0$$
Answer
$$ x = \frac{21}{40} $$
Explanation
$$ \begin{aligned} x\cdot4+x\cdot3-43x\cdot2-x+42 &= 0&& \text{simplify left side} \\[1 em]4x+3x-86x-x+42 &= 0&& \\[1 em]-80x+42 &= 0&& \text{ move the constants to the right } \\[1 em]-80x &= -42&& \text{ divide both sides by $ -80 $ } \\[1 em]x &= \frac{-42}{-80}&& \\[1 em]x &= \frac{21}{40}&& \\[1 em] \end{aligned} $$
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