Problem 62
Question
Solve for the specified variable. $$A=B \sqrt[3]{\frac{C}{D}}-E \text { for } D$$
Step-by-Step Solution
Verified Answer
\( D = \frac{CB^3}{(A + E)^3} \)
1Step 1: Isolate the Term Involving D
Start by isolating the term involving \( D \) on one side of the equation. We have the equation \( A = B \sqrt[3]{\frac{C}{D}} - E \). Move \( E \) to the left side to get \( A + E = B \sqrt[3]{\frac{C}{D}} \).
2Step 2: Divide by B
To further isolate the cube root term, divide both sides of the equation by \( B \). This gives us \( \frac{A + E}{B} = \sqrt[3]{\frac{C}{D}} \).
3Step 3: Eliminate the Cube Root
To eliminate the cube root, raise both sides of the equation to the power of 3. This results in \( \left(\frac{A + E}{B}\right)^3 = \frac{C}{D} \).
4Step 4: Solve for D
Now, solve for \( D \) by taking the reciprocal of both sides. This gives \( D = \frac{C}{\left(\frac{A + E}{B}\right)^3} \).
5Step 5: Simplify the Expression
Simplify the expression: \( D = \frac{CB^3}{(A + E)^3} \). This is the simplified formula for \( D \).
Key Concepts
Isolating variablesCube rootRational expressionsSimplifying expressions
Isolating variables
To solve for a specific variable, such as in our case with "D," one of the key steps is isolating it on one side of the equation. Let’s simplify this idea using a simple strategy.
- Identify the terms involving the variable you want to solve for. Here, we're focusing on the term with "D", which is nested within a cube root.
- Rearrange the equation to get this term by itself on one side. This often involves adding or subtracting other terms from both sides of the equation. For instance, in the exercise, we move "E" to the left side of the equation to start the isolation process.
- Isolating a variable might also require dealing with coefficients or other operations (like division or multiplication) to further simplify the equation.
Cube root
The cube root is a mathematical operation that undoes cubing. In our exercise, we have a cube root operation represented by \(\sqrt[3]{x}\), where "x" is the expression involving "C" and "D." Understanding the cube root is crucial for manipulating our equation.
- To eliminate a cube root in an equation, raise both sides to the power of 3. This will "undo" the cube root.
- The cube root of any number \(y\), when cubed, will give back the number "y": \((\sqrt[3]{y})^3 = y\).
- By removing the cube root from one side, the equation becomes simpler and able to further operations.
Rational expressions
Rational expressions are fractions where the numerator and the denominator consist of polynomials. In the provided exercise, we encounter the rational expression \(\frac{C}{D}\). They are fundamental when dealing with equations.
- In algebra, manipulating rational expressions often involves finding a way to express the equation in a simplified form.
- When working with rational expressions, pay close attention to operations involving addition, subtraction, multiplication, and division as they follow specific algebraic rules.
- Take the reciprocal of both sides in an equation as a strategy to isolate or manipulate these expressions in your favor, as we did to solve for "D."
Simplifying expressions
Simplifying expressions is a crucial part of solving equations. It involves reducing an expression to its simplest form, often by combining like terms or reducing fractions.
- In our exercise, simplifying comes into play after finding the expression for "D." We started with \(D = \frac{C}{(\frac{A + E}{B})^3}\).
- By multiplying the numerator by \(B^3\) and expanding the denominator as \((A + E)^3\), we simplify the whole expression into a more workable form: \(D = \frac{CB^3}{(A + E)^3}\).
- Fully simplifying an expression makes it easier to substitute values or to understand the relationship between variables at a glance.
Other exercises in this chapter
Problem 62
Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt{\frac{1}{7}}$$
View solution Problem 62
The formula occurs in the indicated application. Solve for the specified variable. \(s=\frac{1}{2} g t^{2}+v_{0} t\) for \(t \quad\) (distance an object falls)
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Rationalize the denominator. $$\frac{16 x^{2}-y^{2}}{2 \sqrt{x}-\sqrt{y}}$$
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Verify the property. $$\overline{z^{2}}=(\bar{z})^{2}$$
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