Question

1. \

$$\begin{cases}x - 2y = -4 \\\\ 3x - 2y = 12\\end{cases}$$

Explanation:

Step1: Subtract the two equations

We have the system of equations:

$$\begin{cases}x - 2y = -4 \\ 3x - 2y = 12\end{cases}$$

Subtract the first equation from the second equation:
$$(3x - 2y)-(x - 2y)=12-(-4)$$
Simplify the left - hand side: $3x - 2y - x + 2y=2x$, and the right - hand side: $12 + 4 = 16$. So we get $2x=16$.

Step2: Solve for x

Divide both sides of the equation $2x = 16$ by 2:
$$x=\frac{16}{2}=8$$

Step3: Substitute x into the first equation to solve for y

Substitute $x = 8$ into the equation $x-2y=-4$:
$$8-2y=-4$$
Subtract 8 from both sides: $-2y=-4 - 8=-12$
Divide both sides by - 2: $y=\frac{-12}{-2}=6$

Answer:

The solution of the system of equations is $x = 8,y = 6$