The product of two rational numbers is $-\frac{40}{3}$. If one of them is $-\frac{5}{2}$, and the reciprocal of the other is x, then what is the value of 18 2 4x + ?

The correct answer is 3$\frac{3}{4}$.

Step 1: Find the second rational number.
Let the second rational number be $y$.
We are given that the product of two rational numbers is $-\frac{40}{3}$ and one of the numbers is $-\frac{5}{2}$.
Setting up the equation for their product:

$\left(-\frac{5}{2}\right)\times y=-\frac{40}{3}$

To find $y$, divide the product by the first number:

$y=\left(-\frac{40}{3}\right)÷\left(-\frac{5}{2}\right)$

Multiplying by the reciprocal of $-\frac{5}{2}$:

$y=\left(-\frac{40}{3}\right)\times \left(-\frac{2}{5}\right)$

$y=\frac{(-40)\times (-2)}{3\times 5}=\frac{80}{15}=\frac{16}{3}$

Step 2: Find the value of $x$.
We are given that $x$ is the reciprocal of the second rational number $y$.
Since $y=\frac{16}{3}$, its reciprocal $x$ is:

$x=\frac{1}{y}=\frac{3}{16}$

Step 3: Evaluate the required expression $\left(8x+2\frac{1}{4}\right)$.
First, convert the mixed fraction $2\frac{1}{4}$ to an improper fraction:

$2\frac{1}{4}=\frac{(2\times 4)+1}{4}=\frac{9}{4}$

Now substitute $x=\frac{3}{16}$ into the expression:

$8x+2\frac{1}{4}=8\left(\frac{3}{16}\right)+\frac{9}{4}$

$=\frac{3}{2}+\frac{9}{4}$

Find a common denominator (4) to add the fractions:

$=\frac{6}{4}+\frac{9}{4}=\frac{15}{4}$

Convert $\frac{15}{4}$ back into a mixed fraction:

$\frac{15}{4}=3\frac{3}{4}$

Thus, the final value is 3$\frac{3}{4}$.