$$3(x-3)\cdot2-3(x-3)-7 = 0$$
Answer
$$ x = \frac{16}{3} $$
Explanation
$$ \begin{aligned} 3(x-3)\cdot2-3(x-3)-7 &= 0&& \text{simplify left side} \\[1 em](3x-9)\cdot2-(3x-9)-7 &= 0&& \\[1 em]6x-18-(3x-9)-7 &= 0&& \\[1 em]6x-18-3x+9-7 &= 0&& \\[1 em]3x-16 &= 0&& \text{ move the constants to the right } \\[1 em]3x &= 16&& \text{ divide both sides by $ 3 $ } \\[1 em]x &= \frac{16}{3}&& \\[1 em] \end{aligned} $$
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