In the following questions two equations numbered I and II are given. You have to solve both the equations
`"I. "x^2-14x+48=0" II. "y^2-23y+132=0`

Options

A. `xgty`
B. `xgey`
C. `xlty`
D. `xley`

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation \( x^2 - 14x + 48 = 0 \)

We can factor the quadratic equation. We need to find two numbers that multiply to \( 48 \) (the constant term) and add up to \( -14 \) (the coefficient of \( x \)).

The numbers that satisfy these conditions are \( -6 \) and \( -8 \).

Thus, we can factor the equation as:

\[ (x - 6)(x - 8) = 0 \]

### Step 2: Find the values of \( x \)

Setting each factor to zero gives us:

1. \( x - 6 = 0 \) → \( x = 6 \)
2. \( x - 8 = 0 \) → \( x = 8 \)

So, the values of \( x \) are \( 6 \) and \( 8 \).

### Step 3: Solve the second equation \( y^2 - 23y + 132 = 0 \)

Similar to the first equation, we need to factor this quadratic equation. We need to find two numbers that multiply to \( 132 \) and add up to \( -23 \).

The numbers that satisfy these conditions are \( -11 \) and \( -12 \).

Thus, we can factor the equation as:

\[ (y - 11)(y - 12) = 0 \]

### Step 4: Find the values of \( y \)

Setting each factor to zero gives us:

1. \( y - 11 = 0 \) → \( y = 11 \)
2. \( y - 12 = 0 \) → \( y = 12 \)

So, the values of \( y \) are \( 11 \) and \( 12 \).

### Step 5: Compare the values of \( x \) and \( y \)

Now we have the values:

- \( x = 6 \) or \( 8 \)
- \( y = 11 \) or \( 12 \)

We can see that both values of \( y \) (11 and 12) are greater than both values of \( x \) (6 and 8).

### Conclusion

Thus, we conclude that \( y \) is always greater than \( x \).