Answer:
Given the equation:
\(6q = 3s -9\)
Add 9 to both sides we have;
\(6q+9= 3s -9+9\)
Simplify:
\(6q+9= 3s\)
Divide by 3 to both sides we have;
\(\frac{1}{3}(6q+9)=s\)
Using the distributive property, \(a \cdot (b+c) = a\cdot b+ a\cdot c\)
\(2q+3 = s\)
or
\(s = 2q+3\) where, q is independent variable and s is the dependent variable.
Use notation: s = f(q)
Therefore, the equation in function notation:
\(f(q) = 2q+3\), where q is the independent variable
