In the following questions two equations numbered I and II are given. You have to solve both the equations and answer
`4x^2-21x+20=0`
`3y^2-19y+30=0`

Options

A. if `x gt y`
B. if `x ge y`
C. if `xlt y`
D. if x=y or the relationship between x and y cannot be established.

Answer

To solve the given equations step by step, we will first solve each equation separately and then analyze the relationship between the values of \( x \) and \( y \).

### Step 1: Solve the first equation \( 4x^2 - 21x + 20 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 4 \), \( b = -21 \), and \( c = 20 \).

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-21)^2 - 4 \cdot 4 \cdot 20 = 441 - 320 = 121 \]

4. **Calculate the roots**:
\[ x = \frac{21 \pm \sqrt{121}}{2 \cdot 4} = \frac{21 \pm 11}{8} \]
- First root:
\[ x_1 = \frac{21 + 11}{8} = \frac{32}{8} = 4 \]
- Second root:
\[ x_2 = \frac{21 - 11}{8} = \frac{10}{8} = \frac{5}{4} \]

### Step 2: Solve the second equation \( 3y^2 - 19y + 30 = 0 \)

1. **Identify the coefficients**:
- Here, \( a = 3 \), \( b = -19 \), and \( c = 30 \).

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-19)^2 - 4 \cdot 3 \cdot 30 = 361 - 360 = 1 \]

4. **Calculate the roots**:
\[ y = \frac{19 \pm \sqrt{1}}{2 \cdot 3} = \frac{19 \pm 1}{6} \]
- First root:
\[ y_1 = \frac{19 + 1}{6} = \frac{20}{6} = \frac{10}{3} \]
- Second root:
\[ y_2 = \frac{19 - 1}{6} = \frac{18}{6} = 3 \]

### Step 3: Analyze the relationship between \( x \) and \( y \)

- We have the values:
- From the first equation: \( x = 4 \) or \( x = \frac{5}{4} \) (which is \( 1.25 \))
- From the second equation: \( y = \frac{10}{3} \) (approximately \( 3.33 \)) or \( y = 3 \)

1. **Compare \( x = 4 \) with \( y \)**:
- \( 4 > 3 \)
- \( 4 > \frac{10}{3} \)

2. **Compare \( x = \frac{5}{4} \) with \( y \)**:
- \( \frac{5}{4} < 3 \)
- \( \frac{5}{4} < \frac{10}{3} \)

### Conclusion

- When \( x = 4 \), \( x > y \)
- When \( x = \frac{5}{4} \), \( x < y \)

Thus, the relationship between \( x \) and \( y \) varies based on the values of \( x \) derived from the first equation.