Step 1

Given,

The system of equations are

\(2x+3y=-4\)

\(3x+2y=-1\)

Step 2

Now,

The system of equations are

\(2x+3y=-4...(1)\)

\(3x+2y=-1...(2)\)

From equation (1), we have

\(\displaystyle\Rightarrow{2}{x}+{3}{y}=-{4}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{4}-{3}{y}}}{{{2}}}}\)...(3)

Putting equation (3) and (2), we have

\(\displaystyle\Rightarrow{3}{\left({\frac{{-{4}-{3}{y}}}{{{2}}}}\right)}+{2}{y}=-{1}\)

\(\displaystyle\Rightarrow{\frac{{-{12}-{9}{y}+{4}{y}}}{{{2}}}}=-{1}\)

\(\displaystyle\Rightarrow-{12}-{5}{y}=-{2}\)

\(\displaystyle\Rightarrow-{5}{y}={10}\)

\(\displaystyle\Rightarrow{y}=-{2}\)

Step 3

Putting the value of y in equation (3), we have

\(\displaystyle\Rightarrow{x}={\frac{{-{4}-{3}{\left(-{2}\right)}}}{{{2}}}}\)

\(\displaystyle\Rightarrow{x}={\frac{{-{4}+{6}}}{{{2}}}}\)

\(\displaystyle\Rightarrow{x}={1}\)

\(\displaystyle\therefore\) The values of the system of equations are \(x=1\) and \(y=−2.\)