Question

In the following exercises, evaluate the rational expression for the given values.

$$\begin{gathered} \frac{x^2+3xy+2y^2}{2x^{3}y} \\ \left(a\right)x=1,y=-1 \\ \left(b\right)x=2,y=1 \\ \left(c\right)x=-1,y=-2 \end{gathered}$$

Short answer

The required answers are

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)0 \\ \mathrm{Part}\left(\mathrm{b}\right)\frac{3}{4} \\ \mathrm{Part}\left(\mathrm{c}\right)\frac{15}{4} \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information 

The given expression is$\frac{x^2+3xy+2y^2}{2x^{3}y}$

Part (a) Step 2. Calculation 

Substitute $x=1,y=-1$in the given expression,

$$\begin{gathered} \frac{x^2+3xy+2y^2}{2x^{3}y}=\frac{{\left(1\right)}^2+3\left(1\right)(-1)+2{(-1)}^2}{2{\left(1\right)}^3(-1)} \\ =\frac{1-3+2}{-2} \\ =\frac{0}{-2} \\ =0 \end{gathered}$$

Part (b) Step 1. Calculation  

Substitute $x=2,y=1$in the given expression,

$$\begin{gathered} \frac{x^2+3xy+2y^2}{2x^{3}y}=\frac{{\left(2\right)}^2+3\left(2\right)\left(1\right)+2{\left(1\right)}^2}{2{\left(2\right)}^3\left(1\right)} \\ =\frac{4+6+2}{16} \\ =\frac{12}{16} \\ =\frac{3}{4} \end{gathered}$$

Part (c) Step 1. Calculation   

Substitute $x=-1,y=-2$in the given expression,

$$\begin{gathered} \frac{x^2+3xy+2y^2}{2x^{3}y}=\frac{{(-1)}^2+3(-1)(-2)+2{(-2)}^2}{2{(-1)}^3(-2)} \\ =\frac{1+6+8}{4} \\ =\frac{15}{4} \end{gathered}$$