Sure! Let's solve the given system of linear equations step-by-step:
Given the system:
[tex]\[
\left\{
\begin{array}{l}
2x + 3y = 1 \\
3x + 2y - y = 0
\end{array}
\right.
\][/tex]
First, let's simplify the second equation:
[tex]\[
3x + 2y - y = 0 \implies 3x + y = 0
\][/tex]
Now we have the simplified system of equations:
[tex]\[
\left\{
\begin{array}{l}
2x + 3y = 1 \\
3x + y = 0
\end{array}
\right.
\][/tex]
Next, we can solve the second equation for [tex]\( y \)[/tex]:
From [tex]\( 3x + y = 0 \)[/tex]:
[tex]\[
y = -3x
\][/tex]
Now substitute [tex]\( y = -3x \)[/tex] into the first equation:
[tex]\[
2x + 3(-3x) = 1 \\
2x - 9x = 1 \\
-7x = 1 \\
x = -\frac{1}{7}
\][/tex]
Now that we have the value of [tex]\( x \)[/tex], substitute [tex]\( x = -\frac{1}{7} \)[/tex] back into [tex]\( y = -3x \)[/tex] to find the value of [tex]\( y \)[/tex]:
[tex]\[
y = -3\left(-\frac{1}{7}\right) \\
y = \frac{3}{7}
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
x = -\frac{1}{7}, \quad y = \frac{3}{7}
\][/tex]
