$$4a-a\cdot3-3+2a\cdot3-5a\cdot2+8 = 0$$
Answer
$$ a = \frac{5}{3} $$
Explanation
$$ \begin{aligned} 4a-a\cdot3-3+2a\cdot3-5a\cdot2+8 &= 0&& \text{simplify left side} \\[1 em]a-3+6a-10a+8 &= 0&& \\[1 em]a-3-4a+8 &= 0&& \\[1 em]-3a+5 &= 0&& \text{ move the constants to the right } \\[1 em]-3a &= -5&& \text{ divide both sides by $ -3 $ } \\[1 em]a &= \frac{-5}{-3}&& \\[1 em]a &= \frac{5}{3}&& \\[1 em] \end{aligned} $$
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