Question

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$$\begin{cases} 2x - 3y = 12 \\\\ 2x + y = 0 \\end{cases}$$

Explanation:

Step1: Subtract the two equations to eliminate \(x\)

We have the system:

$$
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$$

Subtract the second equation from the first equation: \((2x - 3y)-(2x + y)=12 - 0\)
Simplify the left - hand side: \(2x-3y - 2x - y=12\), which gives \(-4y = 12\)

Step2: Solve for \(y\)

From \(-4y=12\), divide both sides by \(- 4\): \(y=\frac{12}{-4}=-3\)

Step3: Substitute \(y = - 3\) into one of the original equations to solve for \(x\)

We substitute \(y=-3\) into the second equation \(2x + y=0\), we get \(2x+(-3)=0\)
Add 3 to both sides: \(2x=3\)
Divide both sides by 2: \(x=\frac{3}{2}=1.5\)

Answer:

The solution of the system of equations is \(x = \frac{3}{2}\), \(y=-3\) (or \((\frac{3}{2},-3)\))