You have the equation \(50z + 60x + 30y = 5290\) with the conditions \(z > y > x\) and \(4x > z > 2x\). We need to find integer values for \(x\), \(y\), and \(z\) that satisfy both the equation and the inequalities.

### Step-by-Step Solution

1. **Simplify the equation**:  
   Divide the entire equation by 10:
   \[
   5z + 6x + 3y = 529
   \]

2. **Analyze the inequalities**:
   - \(4x > z > 2x\): This tells us that \(z\) must be more than \(2x\) but less than \(4x\).
   - \(z > y > x\): \(z\) is greater than \(y\) and \(y\) is greater than \(x\).

3. **Express one variable in terms of the others**:  
   Let's express \(z\) in terms of \(x\) and \(y\) using the equation:
   \[
   z = \frac{529 - 6x - 3y}{5}
   \]
   For \(z\) to be an integer, \(529 - 6x - 3y\) must be divisible by 5.  
   Let \(a = 529 - 6x - 3y\) and \(a \equiv 0 \ (\text{mod} \ 5)\).

4. **Test the inequalities with possible integer values**:
   - From the inequality \(2x < z < 4x\), substitute \(z = \frac{529 - 6x - 3y}{5}\) and solve.
   - \(2x < \frac{529 - 6x - 3y}{5} < 4x\).

5. **Solve the inequalities**:

   Multiply the inequality by 5:
   \[
   10x < 529 - 6x - 3y < 20x
   \]

   Split this into two inequalities:
   \[
   529 - 6x - 3y > 10x \quad \text{(I)}
   \]
   \[
   529 - 6x - 3y < 20x \quad \text{(II)}
   \]

   - Inequality I: \(529 - 6x - 3y > 10x \Rightarrow 529 > 16x + 3y\).
   - Inequality II: \(529 - 6x - 3y < 20x \Rightarrow 529 < 26x + 3y\).

6. **Check possible integer solutions**:
   We can test integer values for \(x\) and \(y\) that satisfy the inequalities \(16x + 3y < 529 < 26x + 3y\) and the conditions \(z > y > x\) and \(2x < z < 4x\).

   - Let \(x = 10\) (a reasonable choice, considering the constraints):
     \[
     16(10) + 3y < 529 < 26(10) + 3y
     \]
     \[
     160 + 3y < 529 < 260 + 3y
     \]
     Subtract 3y from all parts:
     \[
     160 < 529 - 3y < 260
     \]
     Solve for \(y\):
     \[
     -260 < -3y < -160
     \]
     Divide by -3 (reverse inequality):
     \[
     \frac{160}{3} < y < \frac{260}{3} \quad \Rightarrow \quad 53.33 < y < 86.67
     \]

     Since \(y\) must be an integer and \(z > y > x\), let's test \(y = 60\):
     - Substituting \(x = 10\) and \(y = 60\) back:
       \[
       5z + 6(10) + 3(60) = 529
       \]
       \[
       5z + 60 + 180 = 529
       \]
       \[
       5z = 289 \quad \Rightarrow \quad z = 57.8 \quad (\text{not an integer, so no solution here})
       \]

     Let's try \(y = 53\):
     - Substituting \(x = 10\) and \(y = 53\) back:
       \[
       5z + 60 + 159 = 529
       \]
       \[
       5z = 310 \quad \Rightarrow \quad z = 62 \quad (\text{valid integer solution})
       \]

Thus, \(z = 62\), \(y = 53\), and \(x = 10\) satisfy all the conditions.

### Final Answer
The values satisfying the equation and all the given conditions are:
\[
z = 62, \ y = 53, \ x = 10
\]

Do you want more details or have any questions about this solution?

---

1. How can we solve such equations with more complicated inequalities?
2. What if the inequalities involve other mathematical operations?
3. What are some ways to check solutions efficiently?
4. How can we use programming to solve such problems?
5. What other methods can be used to solve systems of equations?

**Tip:** When dealing with multiple inequalities, break them down into simpler inequalities and solve step by step.

Explore how to solve the linear equation 50z + 60x + 30y = 5290 with given inequalities z > y > x and 4x > z > 2x. This problem involves finding integer solutions that satisfy both the equation and the inequalities, suitable for high school students.

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