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Solve this using the y formula and move the x to the right so solve for x and their can’t be decimal when listing 5 value for x or y

The given equation is:

\[
y = \sqrt[3]{x} + 2
\]

To solve for \(x\), we need to isolate \(x\) on one side of the equation. Here's how to proceed step by step:

1. **Subtract 2 from both sides** to move the constant term:
   \[
   y - 2 = \sqrt[3]{x}
   \]

2. **Cube both sides** to eliminate the cube root:
   \[
   (y - 2)^3 = x
   \]

Thus, the equation for \(x\) in terms of \(y\) is:

\[
x = (y - 2)^3
\]

### Now, let's calculate some values of \(x\) for integer values of \(y\) (without decimals):

For \(y = 0, 1, 2, 3, 4\):
- When \(y = 0\):  
  \[
  x = (0 - 2)^3 = (-2)^3 = -8
  \]
- When \(y = 1\):  
  \[
  x = (1 - 2)^3 = (-1)^3 = -1
  \]
- When \(y = 2\):  
  \[
  x = (2 - 2)^3 = 0^3 = 0
  \]
- When \(y = 3\):  
  \[
  x = (3 - 2)^3 = 1^3 = 1
  \]
- When \(y = 4\):  
  \[
  x = (4 - 2)^3 = 2^3 = 8
  \]

Therefore, the pairs of \(x\) and \(y\) values are:
- \(y = 0 \Rightarrow x = -8\)
- \(y = 1 \Rightarrow x = -1\)
- \(y = 2 \Rightarrow x = 0\)
- \(y = 3 \Rightarrow x = 1\)
- \(y = 4 \Rightarrow x = 8\)

Would you like more details or have any other questions?

### Relative questions:
1. What would the solution for \(x\) be if \(y = -1\)?
2. How does the behavior of the equation change for negative values of \(y\)?
3. Can you explain why cubing eliminates the cube root in this context?
4. What happens to the value of \(x\) as \(y\) increases significantly, say for \(y = 10\)?
5. How would you solve this equation if there were an additional term in the cube root expression?

### Tip:
When solving equations involving roots, always consider inversing the root operation to simplify the equation (e.g., cube both sides for cube roots).

Solve this using the y formula and move the x to the right so solve for x and there can’t be decimal when listing 5 values for x or y.

Learn how to solve the equation y = ∛x + 2 by isolating x and calculating values of x and y without decimals. This step-by-step solution covers cube roots and their inverse operations, making it suitable for students in Grades 9-11.

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