Solve the equation and check the solution.

StartRoot 3 y plus 3 EndRoot3y+3minus−11equals=5

We are given the equation:

\[
\sqrt{3y + 3} - 11 = 5
\]

### Step 1: Isolate the square root term

To start solving the equation, add 11 to both sides to isolate the square root term:

\[
\sqrt{3y + 3} = 5 + 11
\]

\[
\sqrt{3y + 3} = 16
\]

### Step 2: Eliminate the square root

To eliminate the square root, square both sides of the equation:

\[
(\sqrt{3y + 3})^2 = 16^2
\]

\[
3y + 3 = 256
\]

### Step 3: Solve for \( y \)

Now, subtract 3 from both sides:

\[
3y = 256 - 3
\]

\[
3y = 253
\]

Finally, divide both sides by 3:

\[
y = \frac{253}{3}
\]

\[
y = 84.33 \quad (\text{approximately})
\]

### Step 4: Check the solution

To verify, substitute \( y = 84.33 \) back into the original equation:

\[
\sqrt{3(84.33) + 3} - 11 = 5
\]

\[
\sqrt{253 + 3} - 11 = 5
\]

\[
\sqrt{256} - 11 = 5
\]

\[
16 - 11 = 5
\]

Since the left-hand side equals the right-hand side, the solution is correct.

Thus, the solution is:

\[
y = \frac{253}{3} \quad \text{or} \quad y \approx 84.33
\]

---

Would you like further details or have any questions?

### Related questions:
1. How do you isolate square root terms in equations?
2. What is the process to check solutions of square root equations?
3. Can squaring both sides of an equation introduce extraneous solutions?
4. How do you solve equations with fractional coefficients?
5. What are the general steps to solving radical equations?

### Tip:
Always check for extraneous solutions when solving radical equations by substituting the solution back into the original equation.

Solve the equation and check the solution. \( \sqrt{3y + 3} - 11 = 5 \)

This page covers the step-by-step solution for solving the radical equation \( \sqrt{3y + 3} - 11 = 5 \). Learn how to isolate square roots, solve for variables, and check solutions. This solution is suitable for Grades 8-10 and introduces key algebraic concepts and formulas.

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