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Chapter 8: Problem 72

Solve for the indicated variable. $$a=\frac{r t}{2 b} \text { for } b$$

Short Answer

Expert verified
The short answer is \(b = \frac{rt}{2a}\).

Step by step solution

01

Write the given equation

The equation given is \(a=\frac{rt}{2b}\).
02

Multiply both sides by \(2b\)

To get \(b\) out of the denominator, we'll multiply both sides of the equation by \(2b\). This gives us the equation: $$2ab = rt$$
03

Solve for \(b\)

We now want to isolate \(b\). To do this, we'll divide both sides by \(2a\): $$b = \frac{rt}{2a}$$ Now we have solved the given equation for the variable \(b\).

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Key Concepts

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

Solving for a Variable
In algebra, solving for a variable means finding the value of an unknown that makes an equation true. It involves manipulating the equation to get the unknown on one side and everything else on the other. For example, when given an equation like \(a=\frac{rt}{2b}\) and asked to solve for \(b\), we work to isolate \(b\). When isolating a variable, we perform opposite or inverse operations to both sides of the equation. This keeps the equation balanced, much like how a scale requires equal weight on both sides. The ultimate goal is to make \(b\) the subject of the equation.
Equations with Fractions
Equations with fractions can make algebra look complicated, but they are manageable with the right approach. The key is to eliminate fractions early on to simplify the equation. For example, in \(a=\frac{rt}{2b}\), \(b\) is in the denominator, making the equation a fraction. To make things simpler, multiply both sides by the denominator, which is \(2b\) in this case. Doing this, the equation becomes \(2ab = rt\), transforming the fraction equation into a simpler, fraction-free equation. This step is crucial because it prevents the fractions from bogging down our calculations later on.
Multiplication and Division in Algebra
Multiplication and division are foundational operations in algebra used to solve equations. They are particularly useful in equations with fractions. Consider the equation \(2ab = rt\) derived from multiplying both sides of an original fraction equation. To solve for \(b\), we need to isolate it. This requires dividing both sides by the coefficient of \(b\), which is \(2a\). This gives us \(b = \frac{rt}{2a}\). These steps highlight how multiplication and division serve as opposite operations: multiplication undoes division and vice versa. They are powerful tools that allow us to rearrange and solve equations effectively.

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Most popular questions from this chapter

Solve each equation. $$\frac{8}{3 g^{2}-7 g-6}+\frac{4}{g-3}=\frac{8}{3 g+2}$$

Identify the least common denominator of each group of rational expression, and rewrite each as an equivalent rational expression with the LCD as its denominator. \(\frac{8}{a+9}, \frac{a}{2 a+7}\)

Perform the indicated operations. $$\frac{n}{n^{2}+11 n+30}-\frac{3}{n^{2}+10 n+25}$$

Identify the least common denominator of each group of rational expression, and rewrite each as an equivalent rational expression with the LCD as its denominator. \(\frac{c}{c^{2}+11 c+28}, \frac{6}{c^{2}+14 c+49}\)

Determine the domain of each rational function. $$h(w)=\frac{w+7}{w^{2}-81}$$
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