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The problem you provided involves solving the following system of linear equations:

\[
\begin{cases}
x\sqrt{2} - 2y = \sqrt{2} \\
x - y\sqrt{2} = 2
\end{cases}
\]

Let's solve this system step by step:

### Step 1: Express \( x \) from the second equation
From the second equation \( x - y\sqrt{2} = 2 \), we can isolate \( x \):
\[
x = y\sqrt{2} + 2
\]

### Step 2: Substitute \( x \) into the first equation
Now substitute \( x = y\sqrt{2} + 2 \) into the first equation:
\[
(y\sqrt{2} + 2)\sqrt{2} - 2y = \sqrt{2}
\]
Distribute \( \sqrt{2} \) on the left side:
\[
y(\sqrt{2})^2 + 2\sqrt{2} - 2y = \sqrt{2}
\]
Since \( (\sqrt{2})^2 = 2 \), this simplifies to:
\[
2y + 2\sqrt{2} - 2y = \sqrt{2}
\]

### Step 3: Simplify and solve for \( y \)
The \( 2y \) terms cancel out, leaving:
\[
2\sqrt{2} = \sqrt{2}
\]
This is a contradiction because \( 2\sqrt{2} \neq \sqrt{2} \).

### Conclusion:
The system has no solution because it leads to a contradiction.

Would you like more details or have any questions about this process? Here are five related questions you can explore:

1. How do we solve a system of equations using substitution in general?
2. What does it mean if solving a system leads to a contradiction?
3. How can we use matrices to solve systems of linear equations?
4. What are some common types of systems of equations (e.g., consistent, inconsistent)?
5. How does the value of square roots affect the solution of algebraic equations?

**Tip**: Always check if a system of equations leads to a contradiction or has infinite solutions before concluding the result.

Solve the system of equations: x√2 - 2y = √2 and x - y√2 = 2

Learn how to solve the system of equations involving square roots: x√2 - 2y = √2 and x - y√2 = 2. This problem involves the substitution method and square root properties, suitable for high school students in Grades 9-12.

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