The problem states that $y$ is the sum of two quantities. One quantity varies directly as $x^2$, and the other varies inversely as $x$. We are given that $y = 32$ when $x = 2$, and $y = 86$ when $x = 4$. We need to find: a. An equation for $y$ in terms of $x$. b. The value of $y$ when $x = 3$.

1. Problem Description

The problem states that $y$ is the sum of two quantities. One quantity varies directly as $x^2$, and the other varies inversely as $x$. We are given that $y = 32$ when $x = 2$, and $y = 86$ when $x = 4$. We need to find:

a. An equation for $y$ in terms of $x$.

b. The value of $y$ when $x = 3$.

2. Solution Steps

Let the quantity that varies directly as $x^2$ be $A x^2$, and the quantity that varies inversely as $x$ be $B/x$. Then we have:

$y = Ax^2 + \frac{B}{x}$

We are given two pairs of values for $x$ and $y$. We can use these to create a system of two equations with two unknowns, $A$ and $B$.

When $x = 2$, $y = 32$, so we have:

$32 = A(2^2) + \frac{B}{2}$

$32 = 4A + \frac{B}{2}$

Multiplying both sides by 2, we get:

$64 = 8A + B$ (Equation 1)

When $x = 4$, $y = 86$, so we have:

$86 = A(4^2) + \frac{B}{4}$

$86 = 16A + \frac{B}{4}$

Multiplying both sides by 4, we get:

$344 = 64A + B$ (Equation 2)

Now we can solve the system of equations:

$8A + B = 64$

$64A + B = 344$

Subtracting Equation 1 from Equation 2, we get:

$(64A + B) - (8A + B) = 344 - 64$

$56A = 280$

$A = \frac{280}{56}$

$A = 5$

Now, substitute $A = 5$ into Equation 1:

$8(5) + B = 64$

$40 + B = 64$

$B = 64 - 40$

$B = 24$

So the equation for $y$ in terms of $x$ is:

$y = 5x^2 + \frac{24}{x}$

Now, we need to find the value of $y$ when $x = 3$.

$y = 5(3^2) + \frac{24}{3}$

$y = 5(9) + 8$

$y = 45 + 8$

$y = 53$

3. Final Answer

a. The equation for $y$ in terms of $x$ is $y = 5x^2 + \frac{24}{x}$.

b. The value of $y$ when $x = 3$ is $53$.