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Chapter 4: Problem 3

If \(x=y=z\) and \(x y z=216, x\) is: A. 72 B. 9 C. 12 D 6 E. 4

Short Answer

Expert verified
The value of \(x\) is \textbf{6}\).

Step by step solution

01

Analyze the given information

It is given that \(x = y = z\) which means all three variables are equal. Also, \(x \cdot y \cdot z = 216\). We need to find the value of \(x\).
02

Substitute the equal variables

Since \(x = y = z\), substitute \(x\) for \(y\) and \(z\) in the equation. Therefore, \(x \cdot x \cdot x = 216\), which simplifies to \(x^3 = 216\).
03

Solve for x

To find \(x\), take the cube root of both sides of the equation: \(\root{3}{x^3} = \root{3}{216}\). Therefore, \(x = 6\).
04

Verify the solution

Verify by cubing \(x = 6\). \( 6^3 = 6 \cdot 6 \cdot 6 = 216\), which matches the given condition. Hence, the solution is correct.

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Key Concepts

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

Cube Roots
Cube roots are the opposite of cubing a number, which means they undo the operation of raising a number to the power of three. If you have an expression like \(x^3 = 216\), you can find \(x\) by figuring out the cube root of 216. The cube root of a number \(N\), written as \( \root{3}{N} \), is the value that, when multiplied by itself three times, gives \(N\). In our case, \( \root{3}{216} = 6\) because \(6 \times 6 \times 6 = 216 \).

Generally, cube roots can be found using a calculator or by recognizing common cubes:
  • \( \root{3}{1} = 1 \)
  • \( \root{3}{8} = 2 \)
  • \( \root{3}{27} = 3 \)
and so on. These basics will help you identify cube roots quickly.
Equations
An equation is a statement that asserts the equality of two expressions. In the given problem, we had the equation \(x \times y \times z = 216\). Since we were told that \(x = y = z\), we substituted all variables with \(x\), turning the equation into \(x \times x \times x = 216\) or \(x^3 = 216\).

The goal when solving an equation is to isolate the variable you are solving for. Here, we isolated \(x\) by taking the cube root of both sides of the equation, thus solving \(x^3 = 216\) by finding \( \root{3}{216}\). This technique uses the properties of equality and operations to maintain balance on both sides of the equation.
Algebra
Algebra involves the use of symbols and letters to represent numbers and quantities in formulas and equations. In our problem, we used algebra to represent three unknowns with the same variable \(x\), simplifying our equation to \(x^3 = 216\).

One fundamental concept in algebra is substitution, which allows us to replace variables with actual values or other variables. This makes the equations simpler and more manageable. In the given problem, substitution and recognizing equal variables allowed for an easier solution path.

Mastery of algebraic manipulation is essential. This includes
  • Solving for unknowns
  • Using properties of operations, like taking roots
  • Understanding variable relationships and their balances.
Grasping these concepts will help you solve more complex algebra problems with confidence.

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Most popular questions from this chapter

A sequence of numbers \(a_1, a_2, a_3\) etc. is generated using the following algorithm: \(a_{n+1}=\left(a_n\right)^2+2\). Does the number 35 appear in the sequence if: (1) 0 appears in the sequence (2) \(a_3>a_2\) A. 1 alone, not 2 alone B. 2 alone, not 1 alone C. 1 and 2 together (need both) Answer D. 1 alone or 2 alone E. 1 and 2 together are not sufficient

He was fond to swim but preferred to swim in the sea. A. fond to swim but preferred to swim B. fond of swimming but preferred to swim C. fond of swimming but preferred of swim D. fond to swim but preferred of swimming E. fond to swimming but preferred to swimming

He left his speech on the plane and so was unable of giving his presentation; all the same he insisted to go to the conference. A. unable of giving his presentation; all the same he insisted to go B. unable to giving his presentation; all the same he insisted on go C. unable of giving his presentation; all the same he insisted on going D. unable to give his presentation; all the same he insisted on going E. unable to give his presentation; all the same he insisted to going

How many women graduates will NOT set up their own business if half the total of 800 graduates are women and 15% of all graduates start their own business? A. 120 B. 90 C. 60 Answer D. 340 E. 360

Is \(x^2>y^2 ?\) (1) \(x-y>0\) (2) \(x\) is negative A. 1 alone, not 2 alone B. 2 alone, not 1 alone C. 1 and 2 together (need both) D. 1 alone or 2 alone E. 1 and 2 together are not sufficient
See all solutions

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