To classify the system of equations given by:
[tex]\[
\begin{aligned}
-\frac{1}{3}x &= -6 - y \\
-1 + y &= \frac{1}{3}x - 5
\end{aligned}
\][/tex]
we can analyze these equations step by step.
First, let's rewrite both equations in the standard form [tex]\(Ax + By = C\)[/tex].
Step 1: Rewrite the first equation in standard form.
Starting with the first equation:
[tex]\[
-\frac{1}{3}x = -6 - y
\][/tex]
We can rearrange it as:
[tex]\[
-\frac{1}{3}x + y = -6
\][/tex]
Multiplying through by 3 to clear the fraction:
[tex]\[
-x + 3y = -18
\][/tex]
So, the first equation in standard form is:
[tex]\[
-x + 3y = -18
\][/tex]
Step 2: Rewrite the second equation in standard form.
Starting with the second equation:
[tex]\[
-1 + y = \frac{1}{3}x - 5
\][/tex]
We can rearrange it as:
[tex]\[
y - \frac{1}{3}x = -4
\][/tex]
Multiplying through by 3 to clear the fraction:
[tex]\[
3y - x = -12
\][/tex]
So, the second equation in standard form is:
[tex]\[
-x + 3y = -12
\][/tex]
Step 3: Compare the two equations.
The system of equations in standard form is:
[tex]\[
\begin{aligned}
-x + 3y &= -18 \\
-x + 3y &= -12
\end{aligned}
\][/tex]
We can observe that both equations have the same left-hand side: [tex]\(-x + 3y\)[/tex]. However, their right-hand sides are different ([tex]\(-18\)[/tex] and [tex]\(-12\)[/tex]).
Step 4: Determine the type of system.
Since the two equations have the same coefficients for [tex]\(x\)[/tex] and [tex]\(y\)[/tex], but different constants, we conclude that they are parallel lines. Parallel lines do not intersect and therefore, the system of equations has no solutions.
Conclusion:
The system of equations is parallel.
