/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Find an equation of the inverse ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 6

Find an equation of the inverse relation. $$2 x^{2}+5 y^{2}=4$$

Short Answer

Expert verified
The equation of the inverse relation is \( y = \pm \sqrt{\frac{4 - 5x^2}{2}} \).

Step by step solution

01

Interchange Variables

To find the inverse relation, start by interchanging the variables. Replace every instance of 'x' with 'y' and 'y' with 'x'. The original equation is \[2x^2 + 5y^2 = 4\]. This becomes \[2y^2 + 5x^2 = 4\].
02

Solve for the New 'y'

Isolate the new 'y' in terms of the new 'x'. Rewrite the equation as\[2y^2 = 4 - 5x^2\].
03

Simplify the Equation

Divide both sides by 2 to simplify: \[y^2 = \frac{4 - 5x^2}{2}\]. Next, take the square root of both sides, considering both the positive and negative roots:\[y = \pm \sqrt{\frac{4 - 5x^2}{2}}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Relation
An inverse relation swaps the roles of the dependent and independent variables in an equation.
That means if you start with an equation like \(y = f(x)\), the inverse relation will be \(x = f(y)\).
In the context of the given exercise, the original equation is \[2x^2 + 5y^2 = 4\].
When you swap \x\ and \y\, you essentially exchange their positions, creating a new equation representing the inverse relation.
This is useful because it helps to understand how \x\ and \y\ interact in different 'roles' within the equation.
Variable Interchange
Interchanging variables is a crucial step in finding the inverse relation of an equation.
In our example, the original equation is \[2x^2 + 5y^2 = 4\].
To find the inverse, we replace every instance of \x\ with \y\ and every instance of \y\ with \x\.
This transforms the equation into \[2y^2 + 5x^2 = 4\].
This step is vital because it essentially redefines the roles of the variables, helping us see the relationship from a different perspective.
Solving Equations
Once we have the inverse equation \[2y^2 + 5x^2 = 4\], the next step is to solve for the new \y\.
Our goal is to isolate \y\ in terms of \x\, which involves algebraic manipulation.
First, we rearrange the terms to get all the \y\ terms on one side:\[2y^2 = 4 - 5x^2\].
Next, we divide both sides by 2 to simplify:
\[y^2 = \frac{4 - 5x^2}{2}\].
Finally, we take the square root of both sides, remembering to consider both positive and negative roots:
\[y = \pm\ \sqrt{\frac{4 - 5x^2}{2}}\].
These steps ensure that we have correctly isolated \y\ in terms of \x\, giving us the inverse relation.
Quadratic Equations
In the given exercise, the original equation \[2x^2 + 5y^2 = 4\] is a quadratic equation.
Quadratic equations involve terms with variables raised to the second power.
Here, both \x\ and \y\ are squared, making the equation quadratic in nature.
When dealing with quadratic equations, it's essential to remember that solving them often leads to two potential solutions.
For example, when we isolated \y\ and took the square root, we considered both the positive and negative roots:
\[y = \pm\ \sqrt{\frac{4 - 5x^2}{2}}\].
This double-root nature is a fundamental property of quadratic equations and must always be taken into account to ensure all possible solutions are considered.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve using any method. Given that \(a=\left(\log _{125} 5\right)^{\log _{5} 125},\) find the value of \(\log _{3} a\)

Solve using any method. $$\ln x^{\ln x}=4$$

Use a graphing calculator to find the approximate solutions of the equation. $$4 \ln (x+3.4)=2.5$$

Solve. $$2 \ln x-\ln 5=\ln (x+10)$$

Use a graphing calculator to find the approximate solutions of the equation. $$0.082 e^{0.05 x}=0.034$$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.