What if If you want to start with a specific
y
y-value and solve for
x
x, you can reverse the process.

The equation we are working with is:

y
=
x
+
2
3
y=
3

x+2
​

To solve for
x
x in terms of
y
y, follow these steps:

Step-by-step process:

Cube both sides to eliminate the cube root:

y
3
=
x
+
2
y
3
=x+2
Solve for
x
x by subtracting 2 from both sides:

x
=
y
3
−
2
x=y
3
−2

To solve for \( x \) in terms of \( y \), based on the equation:

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

Here is the step-by-step process:

### Step 1: Cube both sides to eliminate the cube root
We start by cubing both sides of the equation to remove the cube root:

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

### Step 2: Solve for \( x \)
Now, solve for \( x \) by subtracting 2 from both sides of the equation:

\[
x = y^3 - 2
\]

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

\[
x = y^3 - 2
\]

---

Do you want further details or have any questions?

Here are 5 related questions that expand on the solution:

1. What happens if you have an equation like \( y = \sqrt[3]{x - 5} \)? How would you solve for \( x \)?
2. How do you handle higher-order roots, like \( y = \sqrt[5]{x + 3} \)?
3. If given \( y = \sqrt{x + 4} \), how would the process change?
4. What is the difference in solving for \( x \) if the cube root is on the right side of the equation, like \( x = \sqrt[3]{y + 2} \)?
5. How would the solution change if the original equation were \( y = 3\sqrt[3]{x + 2} \)?

**Tip**: When working with equations involving roots, always check your solution by substituting back into the original equation to avoid extraneous solutions.

What if you want to start with a specific y-value and solve for x? The equation we are working with is: y = √³(x + 2). To solve for x in terms of y, follow these steps: Cube both sides to eliminate the cube root: y³ = x + 2. Solve for x by subtracting 2 from both sides: x = y³ - 2.

Learn how to solve the equation y = √³(x + 2) for x with a clear, step-by-step solution. This algebra problem covers essential concepts such as cubing both sides and solving for a variable in terms of another. Suitable for students in Grades 9-12.

Math Problem Statement

y

y 3

x

Solution

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