Problems from Substitution method

Development:

We start by clearing $x$ from the first equation: $${\displaystyle \frac{x}{2}}-{\displaystyle \frac{y}{3}}=12\Rightarrow {\displaystyle \frac{x}{2}}=12+{\displaystyle \frac{y}{3}}\Rightarrow x=2(12+{\displaystyle \frac{y}{3}})\Rightarrow x=24+{\displaystyle \frac{2y}{3}}$$

Then we replace it into the second equation to obtain a linear equation with one unknown: $${\displaystyle \frac{-2x}{3}}+{\displaystyle \frac{3y}{2}}=6\Rightarrow {\displaystyle \frac{-2(24+{\displaystyle \frac{2y}{3}})}{3}}+{\displaystyle \frac{3y}{2}}=6\Rightarrow {\displaystyle \frac{-48-{\displaystyle \frac{4y}{3}}}{3}}+{\displaystyle \frac{3y}{2}}=6\Rightarrow$$ $$-144-{\displaystyle \frac{12y}{3}}+{\displaystyle \frac{3y}{2}}=6\Rightarrow -{\displaystyle \frac{12y}{3}}+{\displaystyle \frac{3y}{2}}=6+144\Rightarrow$$ $$-{\displaystyle \frac{24y}{6}}+{\displaystyle \frac{9y}{6}}=150\Rightarrow -{\displaystyle \frac{15y}{6}}=150\Rightarrow -15y=150\cdot 6\Rightarrow -15y=900\Rightarrow$$ $$y=-{\displaystyle \frac{900}{15}}=-60$$ We can now use the value we obtain for $y$ in the first equation to obtain the value of $x$: $$x=24+{\displaystyle \frac{2y}{3}}=24+{\displaystyle \frac{2(-60)}{3}}=24+{\displaystyle \frac{-120}{3}}=24-40=-16$$

Solution:

$x=-16;y=-60$

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Development:

First of all, we can see that the equations can be simplified. In the first case, we need to put all the variables on one side of the equation and all the constants on the other side: $$\begin{array}{c}2x+3y=21+4x\\ 8x-4y=6-2y\end{array}\}\Rightarrow \begin{array}{c}2x-4x+3y=21\\ 8x-4y+2y=6\end{array}\}\Rightarrow \begin{array}{c}-2x+3y=21\\ 8x-2y=6\end{array}\}$$

We should observe that all coefficients in the second equation can be divided by $2$, which facilitates obtaining an equation for $y$ in terms of $x$: $$\begin{array}{c}-2x+3y=21\\ 4x-y=3\end{array}\}$$

From the second equation we obtain $y$ in terms of $x$: $$-y=3-4x\Rightarrow y=4x-3$$

We then replace it in the first equation and we solve for $x$: $$-2x+3(4x-3)=21\Rightarrow -2x+12x-9=21\Rightarrow 10x=21+9\Rightarrow$$ $$\Rightarrow 10x=30\Rightarrow x={\displaystyle \frac{30}{10}}=3$$

We only have to use the solution we found for $x$ into $y$ in order to know the value of $y$: $$4\cdot 3-y=3\Rightarrow 12-y=3\Rightarrow -y=3-12\Rightarrow -y=-9\Rightarrow y=9$$

Solution:

$x=3;y=9$

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