Answer :
To classify the system of equations:
[tex]\[ \begin{aligned} 2x &= -2 - y \quad \text{(Equation 1)} \\ 5 + y &= -2x + 6 \quad \text{(Equation 2)} \end{aligned} \][/tex]
we can follow these steps:
1. Rewrite the equations in standard form:
- Equation 1: [tex]\( 2x + y = -2 \)[/tex]
- Equation 2: [tex]\( 2x + y = 1 \)[/tex]
2. Compare the equations:
- Notice that both equations have the same left-hand side ([tex]\( 2x + y \)[/tex]) but different right-hand sides ([tex]\( -2 \)[/tex] and [tex]\( 1 \)[/tex]).
3. Since the left-hand sides are identical but the right-hand sides are different, the two lines represented by these equations are parallel. Parallel lines have the same slope but different intercepts and hence, they never intersect.
Therefore, the system of equations does not have a common solution and the system is classified as:
[tex]\[ \boxed{parallel} \][/tex]
[tex]\[ \begin{aligned} 2x &= -2 - y \quad \text{(Equation 1)} \\ 5 + y &= -2x + 6 \quad \text{(Equation 2)} \end{aligned} \][/tex]
we can follow these steps:
1. Rewrite the equations in standard form:
- Equation 1: [tex]\( 2x + y = -2 \)[/tex]
- Equation 2: [tex]\( 2x + y = 1 \)[/tex]
2. Compare the equations:
- Notice that both equations have the same left-hand side ([tex]\( 2x + y \)[/tex]) but different right-hand sides ([tex]\( -2 \)[/tex] and [tex]\( 1 \)[/tex]).
3. Since the left-hand sides are identical but the right-hand sides are different, the two lines represented by these equations are parallel. Parallel lines have the same slope but different intercepts and hence, they never intersect.
Therefore, the system of equations does not have a common solution and the system is classified as:
[tex]\[ \boxed{parallel} \][/tex]