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I. (x^(7//5) div 9) = 169 div x^(3//5) ...

I. ` (x^(7//5) div 9) = 169 div x^(3//5)`
II. ` y^(1//4) xx y^(1//4) xx 7 = 273 div y^(1//2) `

A

if`x gt y`

B

if `x ge y`

C

if `x lt y`

D

if `x le y`

Text Solution

AI Generated Solution

The correct Answer is:
Let's solve the equations step by step. ### Part I: Solve for x Given the equation: \[ \frac{x^{7/5}}{9} = \frac{169}{x^{3/5}} \] **Step 1: Cross-multiply to eliminate the fractions** \[ x^{7/5} \cdot x^{3/5} = 169 \cdot 9 \] **Hint:** When you have a fraction on both sides, you can cross-multiply to simplify the equation. **Step 2: Combine the powers of x on the left side** \[ x^{(7/5 + 3/5)} = 169 \cdot 9 \] \[ x^{10/5} = 169 \cdot 9 \] \[ x^2 = 169 \cdot 9 \] **Hint:** When multiplying powers with the same base, you can add the exponents. **Step 3: Calculate the right side** \[ 169 \cdot 9 = 1521 \] So, we have: \[ x^2 = 1521 \] **Hint:** Perform the multiplication carefully to ensure accuracy. **Step 4: Take the square root of both sides** \[ x = \pm \sqrt{1521} \] **Hint:** Remember that taking the square root gives both positive and negative solutions. **Step 5: Calculate the square root** \[ \sqrt{1521} = 39 \] Thus, \[ x = \pm 39 \] ### Part II: Solve for y Given the equation: \[ y^{1/4} \cdot y^{1/4} \cdot 7 = \frac{273}{y^{1/2}} \] **Step 1: Combine the powers of y on the left side** \[ y^{1/2} \cdot 7 = \frac{273}{y^{1/2}} \] **Hint:** When multiplying powers with the same base, you can add the exponents. **Step 2: Cross-multiply to eliminate the fraction** \[ 7y^{1/2} \cdot y^{1/2} = 273 \] \[ 7y = 273 \] **Hint:** Again, cross-multiplication helps to simplify the equation. **Step 3: Solve for y** \[ y = \frac{273}{7} \] \[ y = 39 \] ### Conclusion We found: - \( x = \pm 39 \) - \( y = 39 \) ### Final Relation Since \( x \) can be both positive and negative, but \( y \) is positive, we can state: \[ x \leq y \]

Let's solve the equations step by step. ### Part I: Solve for x Given the equation: \[ \frac{x^{7/5}}{9} = \frac{169}{x^{3/5}} \] ...
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