To classify the system of equations, let's work through them step-by-step.
We start with the following system of equations:
[tex]\[
\begin{aligned}
1) & \quad 2x = 1 - y \\
2) & \quad -3 + y = -2x + 1
\end{aligned}
\][/tex]
First, let's rewrite each equation in standard form (Ax + By = C).
### Rewriting the Equations
For the first equation:
[tex]\[
2x = 1 - y \implies 2x + y = 1
\][/tex]
For the second equation:
[tex]\[
-3 + y = -2x + 1 \implies y = -2x + 4 \implies 2x - y = -4 + 1 \implies 2x - y = 4
\][/tex]
So, our system of equations now looks like:
[tex]\[
\begin{aligned}
1) & \quad 2x + y = 1 \\
2) & \quad 2x - y = 4
\end{aligned}
\][/tex]
### Analyzing the System
We now have the system in a simpler, more recognizable form:
[tex]\[
\begin{aligned}
1) & \quad 2x + y = 1 \\
2) & \quad 2x - y = 4
\end{aligned}
\][/tex]
Let's solve this system of equations. We can use either substitution or elimination. Here, we will use elimination by adding the two equations together to eliminate [tex]\(y\)[/tex].
Adding the equations:
[tex]\[
(2x + y) + (2x - y) = 1 + 4 \implies 4x = 5 \implies x = \frac{5}{4}
\][/tex]
Now, let's substitute [tex]\(x = \frac{5}{4}\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]. We use the first equation:
[tex]\[
2\left(\frac{5}{4}\right) + y = 1 \implies \frac{10}{4} + y = 1 \implies \frac{5}{2} + y = 1 \implies y = 1 - \frac{5}{2} \implies y = 1 - 2.5 \implies y = -1.5
\][/tex]
So, the solution to the system is [tex]\(x = \frac{5}{4}\)[/tex] and [tex]\(y = -1.5\)[/tex].
### Conclusion
The system of equations has a unique solution [tex]\((x, y) = \left(\frac{5}{4}, -1.5\right)\)[/tex]. This means the lines represented by the equations intersect at a single point.
Therefore, the system of equations is:
[tex]\[ \boxed{\text{intersecting}} \][/tex]
