Linear equations (solving by substitution) I'm having trouble solving this equation with fractions.
$$\frac 23 x+\frac 16 y=\frac 23 \\
-y=12-2x$$

Notice, we have $$\frac{2}{3}x+\frac{1}{6}y=\frac{2}{3}\tag 1$$ $$-y=12-2x\tag 2$$
$$\implies y=2x-12$$ setting this value of $y$ in the eq(1), we get $$\frac{2}{3}x+\frac{1}{6}(2x-12)=\frac{2}{3}$$
 $$2x+x-6=2$$ $$3x=2+6=8$$ $$x=\frac{8}{3}$$ Setting the value of $x$ in (2), we get $$-y=12-2\frac{8}{3}=\frac{36-16}{3}=\frac{20}{3}$$ $$y=-\frac{20}{3}$$
Hence, we get 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{8}{3}, \ y=-\frac{20}{3}}}$$
