In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `5x^(2)-18x+9=0`
(ii) `20y^(2)-13y+2=0`

Options

A. `xlty`
B. x=y
C. `xgty`
D. none of these

Answer

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

### Step 1: Solve the first equation \(5x^2 - 18x + 9 = 0\)

1. **Identify coefficients**: Here, \(a = 5\), \(b = -18\), and \(c = 9\).
2. **Calculate the product \(ac\)**:
\[ ac = 5 \times 9 = 45 \]
3. **Find two numbers that multiply to \(ac\) (45) and add to \(b\) (-18)**:
The numbers are \(-15\) and \(-3\) because:
\[ -15 \times -3 = 45 \quad \text{and} \quad -15 + (-3) = -18 \]
4. **Rewrite the equation**:
\[ 5x^2 - 15x - 3x + 9 = 0 \]
5. **Factor by grouping**:
\[ (5x^2 - 15x) + (-3x + 9) = 0 \]
Factor out common terms:
\[ 5x(x - 3) - 3(x - 3) = 0 \]
This gives:
\[ (5x - 3)(x - 3) = 0 \]
6. **Set each factor to zero**:
\[ 5x - 3 = 0 \quad \text{or} \quad x - 3 = 0 \]
7. **Solve for \(x\)**:
\[ x = \frac{3}{5} \quad \text{or} \quad x = 3 \]

### Step 2: Solve the second equation \(20y^2 - 13y + 2 = 0\)

1. **Identify coefficients**: Here, \(a = 20\), \(b = -13\), and \(c = 2\).
2. **Calculate the product \(ac\)**:
\[ ac = 20 \times 2 = 40 \]
3. **Find two numbers that multiply to \(ac\) (40) and add to \(b\) (-13)**:
The numbers are \(-8\) and \(-5\) because:
\[ -8 \times -5 = 40 \quad \text{and} \quad -8 + (-5) = -13 \]
4. **Rewrite the equation**:
\[ 20y^2 - 8y - 5y + 2 = 0 \]
5. **Factor by grouping**:
\[ (20y^2 - 8y) + (-5y + 2) = 0 \]
Factor out common terms:
\[ 4y(5y - 2) - 1(5y - 2) = 0 \]
This gives:
\[ (5y - 2)(4y - 1) = 0 \]
6. **Set each factor to zero**:
\[ 5y - 2 = 0 \quad \text{or} \quad 4y - 1 = 0 \]
7. **Solve for \(y\)**:
\[ y = \frac{2}{5} \quad \text{or} \quad y = \frac{1}{4} \]

### Final Solutions:
- For the first equation, the solutions are \(x = 3\) and \(x = \frac{3}{5}\).
- For the second equation, the solutions are \(y = \frac{2}{5}\) and \(y = \frac{1}{4}\).

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