Answer: No, she is not correct.
Step-by-step explanation:
Since we have given that
[tex]6x+y=36-----------(1)\\\\5x-y=8---------------(2)[/tex]
First we check the consistency of system of equations:
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}[/tex]
here , [tex]a_1=6,b_1=1,c_1=36\\a_2=5,b_2=-1,c_2=8[/tex]
so, it becomes,
[tex]\frac{6}{5}\neq \frac{1}{-1}\neq \frac{36}{8}[/tex]
So, it is consistent and it is an intersecting lines.
So, it would have a unique solution.
From Eq(1), we have,
[tex]6x+y=36\\\\y=36-6x[/tex]
Put it in eq(2), we have
[tex]5x-y=8\\\\5x-(36-6x)=8\\\\5x-36+6x=8\\\\11x-36=8\\\\11x=36+8\\\\11x=44\\\\x=\frac{44}{11}=4[/tex]
So, x=4 and
[tex]y=36-6x=36-6\times 4=36-24=12[/tex]
so, the solution point of this line will be (4,12).
No, she is not correct.