In each of these questions two equations numbered I and n are given. You have to solve both the equations and give answer
I. `5x^2-13x+6=0`
II. `2y^2-7y+6=0`

Options

A. `xley`
B. `xgty`
C. `xgey`
D. the relationship can’t be established

Answer

To solve the given equations step by step, we will start with each equation separately.

### Step 1: Solve the first equation \(5x^2 - 13x + 6 = 0\)

1. **Identify the coefficients**:
- \(a = 5\)
- \(b = -13\)
- \(c = 6\)

2. **Factor the quadratic equation**:
We need to find two numbers that multiply to \(a \cdot c = 5 \cdot 6 = 30\) and add up to \(b = -13\).
- The numbers are \(-10\) and \(-3\) because \(-10 \cdot -3 = 30\) and \(-10 + -3 = -13\).

3. **Rewrite the equation**:
\[ 5x^2 - 10x - 3x + 6 = 0 \]

4. **Group the terms**:
\[ (5x^2 - 10x) + (-3x + 6) = 0 \]

5. **Factor by grouping**:
\[ 5x(x - 2) - 3(x - 2) = 0 \]
\[ (5x - 3)(x - 2) = 0 \]

6. **Set each factor to zero**:
- \(5x - 3 = 0 \Rightarrow x = \frac{3}{5}\)
- \(x - 2 = 0 \Rightarrow x = 2\)

### Step 2: Solve the second equation \(2y^2 - 7y + 6 = 0\)

1. **Identify the coefficients**:
- \(a = 2\)
- \(b = -7\)
- \(c = 6\)

2. **Factor the quadratic equation**:
We need two numbers that multiply to \(a \cdot c = 2 \cdot 6 = 12\) and add up to \(b = -7\).
- The numbers are \(-3\) and \(-4\) because \(-3 \cdot -4 = 12\) and \(-3 + -4 = -7\).

3. **Rewrite the equation**:
\[ 2y^2 - 3y - 4y + 6 = 0 \]

4. **Group the terms**:
\[ (2y^2 - 3y) + (-4y + 6) = 0 \]

5. **Factor by grouping**:
\[ y(2y - 3) - 2(2y - 3) = 0 \]
\[ (2y - 3)(y - 2) = 0 \]

6. **Set each factor to zero**:
- \(2y - 3 = 0 \Rightarrow y = \frac{3}{2}\)
- \(y - 2 = 0 \Rightarrow y = 2\)

### Summary of Solutions:
- For the first equation, \(x = \frac{3}{5}\) and \(x = 2\).
- For the second equation, \(y = \frac{3}{2}\) and \(y = 2\).