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Chapter 2: Problem 63

Solve for the specified variable or expression. $$ w=\frac{s}{f} \text { for } f $$

Short Answer

Expert verified
\(f = \frac{s}{w}\)

Step by step solution

01

Understand the Problem

The formula given is \(w = \frac{s}{f}\). We want to solve this equation for \(f\). This means we should get \(f\) on one side of the equation and all other terms on the opposite side.
02

Isolate the Division Term

Notice that \(f\) is in the denominator on the right side of the equation. To eliminate the fraction, we multiply both sides of the equation by \(f\) to get: \[ f imes w = s \]
03

Solve for the Variable

Now, to solve for \(f\), divide both sides by \(w\):\[ f = \frac{s}{w} \] This isolates \(f\) on one side of the equation.

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Key Concepts

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

Solving Equations
Solving equations is like solving a puzzle. The goal is to find the value of the unknown variable that makes the equation true. In algebra, this involves performing operations such as addition, subtraction, multiplication, or division to both sides of the equation to isolate the variable. The key is to keep the equation balanced, just like a scale.
  • Ensure you know what you are solving for, which is often referred to as the 'specified variable'.
  • Carefully apply operations to both sides of the equation to maintain equality.
  • Remember that whatever operation you perform on one side, you must perform on the other.
Once you've isolated the variable on one side of the equation, you've essentially 'solved' it, finding the value of that variable that satisfies the original equation.
Isolating Variables
Isolating a variable means re-arranging the equation so that the variable we are interested in is by itself on one side of the equation. This process often involves moving other terms to the opposite side by using inverse operations.
  • If the variable is being multiplied, divide both sides by that number to cancel it out.
  • If it's being divided, multiply both sides to eliminate the fraction, as was done in our original exercise with the formula.
  • Use addition or subtraction to move terms across the equation.
Consider the example where we isolated the variable "\(f\)" from the equation \(w = \frac{s}{f}\). By multiplying both sides by \(f\), we canceled out the fraction, which made it easier to move the whole term so we could solve for \(f\). Understanding this fundamental skill allows you to seek a solution in any algebraic formula.
Fractions in Algebra
Fractions can sometimes make algebra seem tricky, but they are just numbers expressed differently. In algebra, fractions often come into play when a variable is either in the numerator or the denominator.
  • When a variable is a denominator, consider multiplying both sides by the numerator to eliminate the fraction.
  • Conversely, if it's in a numerator, think about dividing to keep terms balanced and maintain equality.
  • Always simplify your equations by reducing fractions to their simplest form when possible.
In our specific problem, the fraction \(\frac{s}{f}\) needed simplification to solve for \(f\). By remembering that the inverse of division is multiplication, we managed to eliminate the fraction by multiplying with \(f\), an important technique when dealing with algebraic fractions. Once comfortable handling these expressions, you'll find that fractions become a useful tool to solve more complex equations.

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Most popular questions from this chapter

Solve each equation and inequality. Write the solution set of each inequality in interval notation and graph it. a. \(\frac{3}{8}+\frac{b}{3}>\frac{5}{12}\) b. \(\frac{3}{8}+\frac{b}{3}=\frac{5}{12}\)

Liquid mixture problems. A printer has ink that is \(8 \%\) cobalt blue color and ink that is \(22 \%\) cobalt blue color. How many ounces of each ink are needed to make one-half gallon ( 64 ounces) of ink that is \(15 \%\) cobalt blue color?

Solve each inequality or compound inequality. Write the solution set in interval notation and graph it. $$ -1 \leq-\frac{1}{2} n $$

Thermodynamics. The Gibbs free-energy formula is given by \(G=U-T S+p V .\) Solve the formula for the pressure \(p\).

Dry mixture problems. Lemon drops worth 3.80 dollar per pound are to be mixed with jelly beans that cost 2.40 dollar per pound to make 300 pounds of a mixture worth 2.96 dollar per pound. How many pounds of each candy should be used?
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