/*! 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 13 Solve each equation. $$\text {... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 13

Solve each equation. $$\text { Solve } P=L+\frac{s}{f} i \text { for } s$$

Short Answer

Expert verified
The solution to the equation \( P=L+\frac{s}{f} i \) for \( s \) is \( s = \frac{f (P - L)}{i} \)

Step by step solution

01

Isolate the term with \( s \)

Subtract \( L \) from both sides of the equation to isolate the term that includes \( s \). This leads to: \( P - L = \frac{s}{f} i \)
02

Eliminate the fraction

To remove the fraction, multiply both sides by \( f \). Since \( f \) is in the denominator on the right side, multiplying by \( f \) will eliminate it. However, remember to also multiply the left side by \( f \) to keep both sides balanced. This results in: \( f (P - L) = s i \)
03

Solve for \( s \)

Finally, to get \( s \) on its own, divide both sides by \( i \). This gives: \( s = \frac{f (P - L)}{i} \)

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.

Isolation of Variables
When solving equations, one key technique is the isolation of variables. This means you want to get the variable of interest all by itself on one side of the equation. This makes it much easier to solve for that variable. In our example, we are solving for \( s \) in the equation \( P = L + \frac{s}{f}i \). The first step is to separate \( s \) from any other terms in the equation.

To isolate \( s \), we need to move other terms to the opposite side. Begin by subtracting \( L \) from both sides to move it away from \( \frac{s}{f}i \). This gives us:
  • \( P - L = \frac{s}{f}i \)
This isolated the term involving our variable \( s \) on one side, simplifying the next phases of solving the equation.
Eliminating Fractions
Fractions can be tricky when solving algebraic equations, so it's often helpful to eliminate them. In our example, \( \frac{s}{f}i \) includes a fraction, \( \frac{s}{f} \), that we need to handle.

To "clear" the fraction, multiply both sides of the equation by the denominator of the fraction. This step ensures that the equation remains balanced. For our scenario, multiplying both sides by \( f \) removes the fraction:
  • Original: \( P - L = \frac{s}{f}i \)
  • Eliminate Fraction: \( f(P - L) = si \)
By doing this, we achieve an equation without fractions, making it simpler to solve for \( s \) in the next step.
Algebraic Manipulation
Algebraic manipulation involves changing the form of an equation to make it easier to solve or understand. After eliminating fractions and isolating terms, further manipulation helps reach the solution for \( s \).

In the simple equation \( f(P - L) = si \), our goal is to find \( s \). Divide both sides by \( i \) to isolate \( s \):
  • Start with: \( f(P - L) = si \)
  • Solve for \( s \): \( s = \frac{f(P - L)}{i} \)
This final manipulation produces a clean expression for \( s \) in terms of \( P \), \( L \), \( f \), and \( i \). Algebraic manipulation is crucial because it allows you to express variables in a more useful, clear, or simple way.

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

Simplify or solve as appropriate. $$3 y(y+2)=3(y+1)(y-1)$$

Perform the operations. $$3 r s t+4 r s t+7 r s t$$

The radius of one millstone in the illustration is 3 meters greater than the radius of the other, and their areas differ by \(15 \pi\) square meters. Find the radius of the larger millstone.

In major league baseball, the distance between bases is 30 feet greater than it is in softball. The bases in major league baseball mark the corners of a square that has an area 4,500 square feet greater than for softball. Find the distance between the bases in baseball.

Give an example of a polynomial that is \(\ldots\) a monomial
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

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.