Answer: Option 'B' is correct.
Step-by-step explanation:
Since we have given that
[tex]x+5y=6\\\\and\\\\3x+30y=36[/tex]
We need to find that " How many solutions are there ?"
First we compare the coefficients of x and y :
[tex]a_1x+b_1y=c_1\\\\and,\\\\a_2x+b_2y=c_2[/tex]
Now,
[tex]\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}[/tex]
Here,
[tex]a_1=1,a_2=5,c_1=-6\\\\and\\\\a_2=3,b_2=30,c_2=36[/tex]
so, it becomes,
[tex]\frac{1}{3}=\frac{5}{30}=\frac{-6}{-36}\\\\\frac{1}{3}\neq \frac{1}{6}=\frac{1}{6}[/tex]
So, it becomes intersecting lines . So, they have unique solution.
Hence, Option 'B' is correct.
