Hello!
First, let's try to solve the following system:
[tex]\begin{cases}5x+6y=1\text{ equation A} \\ 10x+12y=2\text{ equation B}\end{cases}[/tex]
Let's solve it:
• In equation A, we will ,isolate the variable x,:
[tex]\begin{gathered} 5x+6y=1 \\ 5x=1-6y \\ \\ x=\frac{1-6y}{5} \end{gathered}[/tex]
• Now, let's ,replace where's x ,in equation B ,by the value above,:
[tex]\begin{gathered} 10x+12y=2 \\ 10\cdot(\frac{1-6y}{5})+12y=2 \\ \\ \frac{10}{5}-\frac{60y}{5}+12y=2 \\ \\ 2\cancel{-12y+12y}=2 \\ 2=2 \end{gathered}[/tex]
Notice that the variable y disappeared. So, we can say that this is a system with infinitely many solutions.
The solution set is:
[tex]\mleft\lbrace\frac{1-6y}{5},y\mright\rbrace[/tex]
Answer B.
