Answer:
Consider the expression [tex]q=\frac{a}{b}[/tex]. q is called the quotient, a is is the dividend and b is the divisor.
Since q is a multiple of 6, then q has the form [tex]q=6k[/tex] for some integer k.
Since a is a multiple of 9, then a has the form [tex]a=9s[/tex] for some integer s.
Since b is a factor of 12, then if 12 can be expressed of the form [tex]12=c*b[/tex], for c an integer. Then b has the form [tex]b=\frac{12}{m}[/tex]
Replacing the preview expression in the initial expression we obtain:
[tex]6k=\frac{9s}{\frac{12}{m}}=\frac{9sm}{12}\\6*12k=9sm\\72k=9sm\\72k-9sm=0[/tex]
Then [tex]72k-9sm=0,\; for \; k,s,m\in \mathbb{Z}[/tex] is a equation to Isabel's problem.
