Let A = `{y :y in R, y^(2) = 25 " and " 2y=18}`, then

Options

A. `A = phi`
B. `A = {5,-5,9}`
C. `A = {5,9}`
D. {9}

Answer

To solve the problem, we need to find the set \( A = \{ y : y \in \mathbb{R}, y^2 = 25 \text{ and } 2y = 18 \} \).

### Step 1: Solve the first equation \( y^2 = 25 \)

To find the values of \( y \) that satisfy this equation, we take the square root of both sides:

\[ y = \pm \sqrt{25} \]

Calculating the square root gives us:

\[ y = 5 \quad \text{or} \quad y = -5 \]

### Step 2: Solve the second equation \( 2y = 18 \)

Next, we solve for \( y \) in the second equation:

\[ 2y = 18 \]

Dividing both sides by 2 gives:

\[ y = \frac{18}{2} = 9 \]

### Step 3: Find the intersection of the solutions

Now we have two potential solutions from the first equation: \( y = 5 \) and \( y = -5 \), and one solution from the second equation: \( y = 9 \).

We need to check which of these values satisfy both equations.

- For \( y = 5 \):
- Check \( 2y = 18 \):
\[ 2(5) = 10 \quad \text{(not equal to 18)} \]

- For \( y = -5 \):
- Check \( 2y = 18 \):
\[ 2(-5) = -10 \quad \text{(not equal to 18)} \]

- For \( y = 9 \):
- Check \( y^2 = 25 \):
\[ 9^2 = 81 \quad \text{(not equal to 25)} \]

### Conclusion

None of the values \( 5, -5, \) or \( 9 \) satisfy both equations simultaneously. Therefore, the set \( A \) is empty.

Thus, we conclude that:

\[ A = \emptyset \]

### Final Answer

The final answer is that the set \( A \) is empty.

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