Answer: Option 'b' is correct.
Step-by-step explanation:
Since we have given that
2y-x=2
⇒2y=2+x
⇒2y-2=x--------------(1)
x-2y=8 --------------------(2)
Put the eq(1) in eq(2), we get that,
[tex]x-2y=8\\\\2y-2-2y=8\\\\-2\neq 8[/tex]
Here,
[tex]a_1=-1,b_1=2,c_1=2\\\\a_2=1,b_2=-2,c_2=8[/tex]
Now,
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\\\frac{-1}{1}=\frac{2}{-2}=\frac{2}{8}\\\\-1= -1\neq \frac{1}{4}[/tex]
So, it is the parallel line.
So, it is inconsistent.
Hence, Option 'b' is correct.
