Answer :

To solve the system of linear equations:
[tex]\[ \begin{cases} 2x + y = 1 \\ 3x + 2y = 0 \end{cases} \][/tex]
we will use the method of elimination.

Step 1: Multiply the first equation by 2 so that the coefficients of [tex]\( y \)[/tex] in both equations will be the same:
[tex]\[ 2(2x + y) = 2 \cdot 1 \][/tex]
[tex]\[ 4x + 2y = 2 \][/tex]

Now our system looks like this:
[tex]\[ \begin{cases} 4x + 2y = 2 \\ 3x + 2y = 0 \end{cases} \][/tex]

Step 2: Subtract the second equation from the first equation to eliminate [tex]\( y \)[/tex]:
[tex]\[ (4x + 2y) - (3x + 2y) = 2 - 0 \][/tex]
[tex]\[ 4x + 2y - 3x - 2y = 2 \][/tex]
[tex]\[ x = 2 \][/tex]

Step 3: Substitute [tex]\( x = 2 \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. Let's use the first equation:
[tex]\[ 2x + y = 1 \][/tex]
[tex]\[ 2(2) + y = 1 \][/tex]
[tex]\[ 4 + y = 1 \][/tex]
[tex]\[ y = 1 - 4 \][/tex]
[tex]\[ y = -3 \][/tex]

So, the solution to the system of equations is:
[tex]\[ x = 2 \][/tex]
[tex]\[ y = -3 \][/tex]

Therefore, the solution to the system is:
[tex]\[ \boxed{(2, -3)} \][/tex]