Chapter 8: Q. 13 (page 917)

In the following exercises, evaluate the rational expression for the given values.
$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y} (a) x = 1, y = - 1 (b) x = 2, y = 1 (c) x = - 1, y = - 2$

Short Answer

Expert verified

The required answers are

$Part (a) 0 Part (b) \frac{3}{4} Part (c) \frac{15}{4}$

Step by step solution

Part(a) Step 1. Given Information

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

Part (a) Step 2. Calculation

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

$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y} = \frac{{(1)}^{2} + 3 (1) (- 1) + 2 {(- 1)}^{2}}{2 {(1)}^{3} (- 1)} = \frac{1 - 3 + 2}{- 2} = \frac{0}{- 2} = 0$

Part (b) Step 1. Calculation

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

$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y} = \frac{{(2)}^{2} + 3 (2) (1) + 2 {(1)}^{2}}{2 {(2)}^{3} (1)} = \frac{4 + 6 + 2}{16} = \frac{12}{16} = \frac{3}{4}$

Part (c) Step 1. Calculation

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

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

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In the following exercises, evaluate the rational expression for the given values.
$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y} (a) x = 1, y = - 1 (b) x = 2, y = 1 (c) x = - 1, y = - 2$

Short Answer

Step by step solution

Part(a) Step 1. Given Information

Part (a) Step 2. Calculation

Part (b) Step 1. Calculation

Part (c) Step 1. Calculation

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In the following exercises, evaluate the rational expression for the given values.x2+3xy+2y22x3y(a)x=1,y=-1(b)x=2,y=1(c)x=-1,y=-2

Short Answer

Step by step solution

Part(a) Step 1. Given Information

Part (a) Step 2. Calculation

Part (b) Step 1. Calculation

Part (c) Step 1. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

In the following exercises, evaluate the rational expression for the given values.
$\frac{x^{2} + 3 x y + 2 y^{2}}{2 x^{3} y} (a) x = 1, y = - 1 (b) x = 2, y = 1 (c) x = - 1, y = - 2$