/*! 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 64 Solve each proportion. $$ \f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 64

Solve each proportion. $$ \frac{-6}{r}=\frac{r}{-6} $$

Short Answer

Expert verified
The solutions are \(r = 6\) and \(r = -6\). Both satisfy the original proportion.

Step by step solution

01

Cross-Multiply

A proportion is an equation that shows two ratios are equivalent. To solve this proportion, cross-multiply the terms. This means multiplying the numerator of each fraction by the denominator of the other fraction.So, \(-6 \times -6 = r \times r\).This gives us the equation:\[36 = r^2\].
02

Solve the Quadratic Equation

Now that we have the equation \(r^2 = 36\), we need to solve for \(r\) by finding the square root of both sides.\(r = \pm \sqrt{36}\).This simplifies to either:\(r = 6\) or \(r = -6\).
03

Verify the Solutions

To verify our solutions, substitute each value of \(r\) back into the original proportion to see if they satisfy the equation.For \(r = 6\):\[\frac{-6}{6} = \frac{6}{-6} \rightarrow -1 = -1\]This holds true.For \(r = -6\):\[\frac{-6}{-6} = \frac{-6}{-6} \rightarrow 1 = 1\]This also holds true.

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.

Cross-Multiplication
Cross-multiplication is an essential tool when solving proportions. A proportion is an equation that states two ratios are equal. The method involves multiplying the numerator (the top number) of one fraction by the denominator (the bottom number) of the other fraction. This reveals a new equation without fractions, making it easier to solve.
Let's look at how this works with the example proportion:
  • Given proportion: \(\frac{-6}{r} = \frac{r}{-6}\).
  • By cross-multiplying, we calculate \(-6 \times -6 = r \times r\).
  • This simplification results in the equation \[36 = r^2\].
Using cross-multiplication helps us eliminate fractions quickly, laying the foundation for subsequent algebraic steps.
Quadratic Equations
Quadratic equations are those equations of the form \(ax^2 + bx + c = 0\), where "a", "b", and "c" are constants, and "x" represents the unknown variable. In our exercise, we derived a quadratic equation after cross-multiplication: \(r^2 = 36\).
To solve it, we need to find the value of "r" that satisfies this equation:
  • First, identify that \(r^2\) means the value squared equals 36.
  • The solution requires finding the square root of both sides.
  • The square roots of 36 are 6 and -6, thus \(r = \pm 6\).
Quadratic equations may seem complex, but solving them often involves straightforward techniques like square rooting, factoring, or using the quadratic formula.
Verification of Solutions
Verifying solutions is a verification step ensuring all mathematical workings are correct and applicable to the original problem. After solving the quadratic equation, we obtained solutions \(r = 6\) and \(r = -6\). We need to substitute these solutions back into the original proportion to ensure both satisfy the equation.
Verification process:
  • Substitute \(r = 6\) into \(\frac{-6}{r} = \frac{r}{-6}\), which results in \(\frac{-6}{6} = \frac{6}{-6}\), confirming \(-1 = -1\). This holds true.
  • Substitute \(r = -6\) into the same proportion, leading to \(\frac{-6}{-6} = \frac{-6}{-6}\), confirming \(1 = 1\). This is also true.
Verifying solutions helps ensure the accuracy and reliability of the derived results. Always remember to substitute back into the original equation as a confirmation step.

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 each equation. a. \(-\frac{2}{5}=\frac{3}{4 x}\) b. \(\frac{4}{x}-\frac{2}{5}=\frac{3}{4 x}\)

Solve each equation. a. \(\frac{1}{4}=\frac{2}{3 a}\) b. \(\frac{5}{6 a}+\frac{1}{4}=\frac{2}{3 a}\)

Perform the operations. Simplify, if possible. $$ \frac{x}{x-2}+\frac{4+2 x}{x^{2}-4} $$

Perform the operations. Simplify, if possible. $$ \frac{6}{s^{2}-9}-\frac{5}{s^{2}-s-6} $$

Perform the operations. Simplify, if possible. $$ \frac{x-7}{x^{2}+4 x-5}-\frac{x-9}{x^{2}+3 x-10} $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

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.