In each of these questions, two equations I. and II. are given . you have to solve both the equations and answer the following questions
I. `10 x^(2) - 29 x + 21 = 0 `
II. ` 2 y ^(2) - 19 y + 45 = 0 `

Options

A. x = y or no relation
B. ` x lt y`
C. ` x le y `
D. ` x gt y `

Answer

To solve the given equations step by step, we will first tackle each quadratic equation separately.

### Step 1: Solve Equation I: \(10x^2 - 29x + 21 = 0\)

1. **Identify the coefficients**:
- \(a = 10\)
- \(b = -29\)
- \(c = 21\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-29)^2 - 4(10)(21) = 841 - 840 = 1 \]

4. **Substitute the values into the quadratic formula**:
\[ x = \frac{-(-29) \pm \sqrt{1}}{2(10)} = \frac{29 \pm 1}{20} \]

5. **Calculate the two possible values for \(x\)**:
- First value:
\[ x_1 = \frac{29 + 1}{20} = \frac{30}{20} = 1.5 \]
- Second value:
\[ x_2 = \frac{29 - 1}{20} = \frac{28}{20} = 1.4 \]

### Step 2: Solve Equation II: \(2y^2 - 19y + 45 = 0\)

1. **Identify the coefficients**:
- \(a = 2\)
- \(b = -19\)
- \(c = 45\)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-19)^2 - 4(2)(45) = 361 - 360 = 1 \]

4. **Substitute the values into the quadratic formula**:
\[ y = \frac{-(-19) \pm \sqrt{1}}{2(2)} = \frac{19 \pm 1}{4} \]

5. **Calculate the two possible values for \(y\)**:
- First value:
\[ y_1 = \frac{19 + 1}{4} = \frac{20}{4} = 5 \]
- Second value:
\[ y_2 = \frac{19 - 1}{4} = \frac{18}{4} = 4.5 \]

### Summary of Solutions:
- The solutions for \(x\) are \(1.5\) and \(1.4\).
- The solutions for \(y\) are \(5\) and \(4.5\).

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

- Compare \(x_1 = 1.5\) with \(y_1 = 5\):
\[ 1.5 < 5 \]
- Compare \(x_1 = 1.5\) with \(y_2 = 4.5\):
\[ 1.5 < 4.5 \]
- Compare \(x_2 = 1.4\) with \(y_1 = 5\):
\[ 1.4 < 5 \]
- Compare \(x_2 = 1.4\) with \(y_2 = 4.5\):
\[ 1.4 < 4.5 \]

### Conclusion:
In all comparisons, \(x\) is less than \(y\). Therefore, the relationship can be expressed as:
\[ x < y \]