Final answer:
The equations provided are linear, and solving for x is straightforward by substitution. In a quadratic example, we solve for x using the quadratic formula or factorization. The context dictates which solution for x is feasible.
Step-by-step explanation:
The student is asking for the equation that models a situation and to solve it for the variable x. The given functional forms, f(x) = -3x + 45 and f(x) = 3x + 45, are linear equations. When x = 15 items, these expressions can be evaluated directly by substitution to find the value of f(x).
Example:
Using the first equation:
f(x) = -3x + 45
f(15) = (-3 * 15) + 45
f(15) = -45 + 45
f(15) = 0.
For the quadratic equation example, the solution for x is given as x = 0.00139, with the other solution x = -0.0024 being discarded as it does not make sense in a real-world context, since a negative amount of items is not possible.
To solve for x using the second example, which resembles a quadratic equation, the equilibrium equation is rewritten, and the values for x are found, presumably through factorization or application of the quadratic formula.
The least-squares regression line and equilibrium equation examples show how solving for x can depend on the context and the form of the equation, whether it is linear or quadratic.
Solving Models with Algebra
When solving models with algebra, we first identify the unknown, which in most of these cases is x. The known values are provided or assumed based on the context. Then, an appropriate equation is chosen, the known values are plugged in, and the equation is solved for the unknown.