Solve the following system of equations with the substitution method:
$$\left.\begin{array}{c} \dfrac{x}{2}-\dfrac{y}{3}=12 \\ \dfrac{-2x}{3}+\dfrac{3y}{2}=6 \end{array} \right\}$$
Development:
We start by clearing $$x$$ from the first equation: $$$\dfrac{x}{2}-\dfrac{y}{3}=12 \Rightarrow \dfrac{x}{2}=12+\dfrac{y}{3} \Rightarrow x=2\Big(12+\dfrac{y}{3}\Big) \Rightarrow x=24+\dfrac{2y}{3}$$$
Then we replace it into the second equation to obtain a linear equation with one unknown: $$$\dfrac{-2x}{3}+\dfrac{3y}{2}=6 \Rightarrow \dfrac{-2\Big(24+\dfrac{2y}{3}\Big)}{3}+\dfrac{3y}{2}=6 \Rightarrow \dfrac{-48-\dfrac{4y}{3}}{3}+\dfrac{3y}{2}=6 \Rightarrow$$$ $$$-144-\dfrac{12y}{3}+\dfrac{3y}{2}=6 \Rightarrow -\dfrac{12y}{3}+\dfrac{3y}{2}=6+144 \Rightarrow$$$ $$$-\dfrac{24y}{6}+\dfrac{9y}{6}=150 \Rightarrow -\dfrac{15y}{6}=150 \Rightarrow -15y=150\cdot6 \Rightarrow -15y=900 \Rightarrow$$$ $$$y=-\dfrac{900}{15}=-60$$$ We can now use the value we obtain for $$y$$ in the first equation to obtain the value of $$x$$: $$$x=24+\dfrac{2y}{3}=24+\dfrac{2(-60)}{3}=24+\dfrac{-120}{3}=24-40=-16$$$
Solution:
$$x=-16; y=-60$$