$$y+1-(-8y\cdot3-4y+6) = 0$$
Answer
$$ y = \frac{5}{29} $$
Explanation
$$ \begin{aligned} y+1-(-8y\cdot3-4y+6) &= 0&& \text{simplify left side} \\[1 em]y+1-(-24y-4y+6) &= 0&& \\[1 em]y+1-(-28y+6) &= 0&& \\[1 em]y+1+28y-6 &= 0&& \\[1 em]29y-5 &= 0&& \text{ move the constants to the right } \\[1 em]29y &= 5&& \text{ divide both sides by $ 29 $ } \\[1 em]y &= \frac{5}{29}&& \\[1 em] \end{aligned} $$
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