In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer
I `14x^2-37x+24=0`
II `28y^2-53y+20=-4`

Options

A. if `xgey`
B. if `xgty`
C. if `xley`
D. if x=y or the relationship can’t be established.

Answer

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

### Step 1: Solve the first equation \( 14x^2 - 37x + 24 = 0 \)

We can use the quadratic formula to find the roots of the equation, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 14 \), \( b = -37 \), and \( c = 24 \).

#### Step 1.1: Calculate the discriminant

\[ b^2 - 4ac = (-37)^2 - 4 \cdot 14 \cdot 24 \]
\[ = 1369 - 1344 = 25 \]

#### Step 1.2: Substitute values into the quadratic formula

\[ x = \frac{37 \pm \sqrt{25}}{2 \cdot 14} \]
\[ = \frac{37 \pm 5}{28} \]

#### Step 1.3: Calculate the two possible values of \( x \)

1. \( x_1 = \frac{37 + 5}{28} = \frac{42}{28} = \frac{3}{2} = 1.5 \)
2. \( x_2 = \frac{37 - 5}{28} = \frac{32}{28} = \frac{8}{7} \approx 1.14 \)

### Step 2: Solve the second equation \( 28y^2 - 53y + 20 = -4 \)

First, we rearrange the equation:

\[ 28y^2 - 53y + 20 + 4 = 0 \]
\[ 28y^2 - 53y + 24 = 0 \]

#### Step 2.1: Calculate the discriminant for the second equation

Using the quadratic formula again, where \( a = 28 \), \( b = -53 \), and \( c = 24 \):

\[ b^2 - 4ac = (-53)^2 - 4 \cdot 28 \cdot 24 \]
\[ = 2809 - 2688 = 121 \]

#### Step 2.2: Substitute values into the quadratic formula

\[ y = \frac{53 \pm \sqrt{121}}{2 \cdot 28} \]
\[ = \frac{53 \pm 11}{56} \]

#### Step 2.3: Calculate the two possible values of \( y \)

1. \( y_1 = \frac{53 + 11}{56} = \frac{64}{56} = \frac{8}{7} \approx 1.14 \)
2. \( y_2 = \frac{53 - 11}{56} = \frac{42}{56} = \frac{3}{4} = 0.75 \)

### Step 3: Compare the values of \( x \) and \( y \)

We found:
- \( x_1 = 1.5 \) and \( x_2 \approx 1.14 \)
- \( y_1 \approx 1.14 \) and \( y_2 = 0.75 \)

#### Comparison:
1. For \( x = 1.5 \) and \( y = 0.75 \): \( x > y \)
2. For \( x \approx 1.14 \) and \( y \approx 1.14 \): \( x = y \)
3. For \( x \approx 1.14 \) and \( y = 0.75 \): \( x > y \)

### Conclusion

From the comparisons, we can conclude that \( x \) is either greater than or equal to \( y \).

### Final Answer
**Option 1: \( x \geq y \)**

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