In each of questions two eqautions numbered Iand II are given you have to solve both equations and give answer
I.`2x^2-13x+18=0`
II.`y^2-7y+12=0`

Options

A. If `xlty`
B. `if xley`
C. if `xgty`
D. if relation between x and y can not be established

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation \( 2x^2 - 13x + 18 = 0 \)

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = -13 \)
- \( c = 18 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-13)^2 - 4 \cdot 2 \cdot 18 = 169 - 144 = 25 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-(-13) \pm \sqrt{25}}{2 \cdot 2} = \frac{13 \pm 5}{4} \]

5. **Calculate the two possible values for \( x \)**:
- First value:
\[ x_1 = \frac{13 + 5}{4} = \frac{18}{4} = 4.5 \]
- Second value:
\[ x_2 = \frac{13 - 5}{4} = \frac{8}{4} = 2 \]

Thus, the solutions for the first equation are \( x = 4.5 \) and \( x = 2 \).

### Step 2: Solve the second equation \( y^2 - 7y + 12 = 0 \)

1. **Identify the coefficients**:
- \( a = 1 \)
- \( b = -7 \)
- \( c = 12 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-(-7) \pm \sqrt{1}}{2 \cdot 1} = \frac{7 \pm 1}{2} \]

5. **Calculate the two possible values for \( y \)**:
- First value:
\[ y_1 = \frac{7 + 1}{2} = \frac{8}{2} = 4 \]
- Second value:
\[ y_2 = \frac{7 - 1}{2} = \frac{6}{2} = 3 \]

Thus, the solutions for the second equation are \( y = 4 \) and \( y = 3 \).

### Summary of Solutions
- From the first equation: \( x = 4.5 \) and \( x = 2 \)
- From the second equation: \( y = 4 \) and \( y = 3 \)