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Let's solve the two equations one by one:

### Equation 1:
\[
\sqrt{x - 8} = 2 - \sqrt{x}
\]

### Step 1: Isolate one of the square root terms
Move \(\sqrt{x}\) to the left side:
\[
\sqrt{x - 8} + \sqrt{x} = 2
\]

### Step 2: Square both sides to remove square roots
Squaring both sides:
\[
(\sqrt{x - 8} + \sqrt{x})^2 = 2^2
\]
This expands as:
\[
(x - 8) + 2\sqrt{(x - 8)(x)} + x = 4
\]
Simplify:
\[
x - 8 + 2\sqrt{x(x - 8)} + x = 4
\]
\[
2x - 8 + 2\sqrt{x(x - 8)} = 4
\]

### Step 3: Isolate the square root term
Move all terms except the square root to the right-hand side:
\[
2\sqrt{x(x - 8)} = 4 - 2x + 8
\]
\[
2\sqrt{x(x - 8)} = 12 - 2x
\]

### Step 4: Square both sides again
Squaring both sides to eliminate the remaining square root:
\[
4x(x - 8) = (12 - 2x)^2
\]
Expand both sides:
\[
4x^2 - 32x = 144 - 48x + 4x^2
\]
Simplify:
\[
-32x = 144 - 48x
\]
Move all terms involving \(x\) to one side:
\[
16x = 144
\]
\[
x = 9
\]

### Step 5: Check the solution in the original equation
Substitute \(x = 9\) into the original equation:
\[
\sqrt{9 - 8} = 2 - \sqrt{9}
\]
\[
\sqrt{1} = 2 - 3
\]
\[
1 = -1
\]
This is **false**, so \(x = 9\) is not a solution.

---

### Equation 2:
\[
x - 8 = 4 - 4\sqrt{x} + x
\]

### Step 1: Simplify the equation
Cancel out \(x\) from both sides:
\[
-8 = 4 - 4\sqrt{x}
\]

### Step 2: Isolate the square root term
\[
-12 = -4\sqrt{x}
\]
\[
\sqrt{x} = 3
\]

### Step 3: Square both sides
\[
x = 9
\]

### Step 4: Check the solution in the original equation
Substitute \(x = 9\) into the original equation:
\[
9 - 8 = 4 - 4\sqrt{9} + 9
\]
\[
1 = 4 - 12 + 9
\]
\[
1 = 1
\]
This is **true**, so the solution is \(x = 9\).

Thus, the second equation has a valid solution of \(x = 9\), but the first equation does not accept this as a solution.

Would you like more details or have any questions?

---

Here are 5 related questions to expand your understanding:
1. How would the solution change if the terms inside the square roots were different?
2. Can you solve a system of equations that involves both square roots and quadratic terms?
3. How does squaring both sides of an equation affect the validity of the solution?
4. What are some other methods to isolate square roots in complex algebraic equations?
5. How can you check for extraneous solutions when solving square root equations?

**Tip:** When solving equations involving square roots, always check for extraneous solutions, as squaring both sides can introduce false solutions.

Solve the system of equations: 
1. \(\sqrt{x - 8} = 2 - \sqrt{x}\) 
2. \(x - 8 = 4 - 4\sqrt{x} + x\)

Learn how to solve a system of equations involving square roots with detailed step-by-step solutions. This problem covers key algebraic concepts, including isolating square root terms and squaring both sides to remove square roots. Suitable for students in Grades 9-12.

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