In each of these questions , two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 7 x^(2) - 19 x + 10 = 0 `
II ` 8 y^(2) + 2y - 3 = 0`

Answer

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

### Step 1: Solve Equation I
The first equation is:
\[ 7x^2 - 19x + 10 = 0 \]

To solve this quadratic equation, we will use the factorization method.

1. **Identify the coefficients**:
- \( a = 7 \)
- \( b = -19 \)
- \( c = 10 \)

2. **Calculate the product \( ac \)**:
\[ ac = 7 \times 10 = 70 \]

3. **Find two numbers that multiply to \( ac \) and add to \( b \)**:
We need two numbers that multiply to 70 and add to -19. The numbers are -14 and -5 because:
\[ -14 \times -5 = 70 \]
\[ -14 + (-5) = -19 \]

4. **Rewrite the equation**:
\[ 7x^2 - 14x - 5x + 10 = 0 \]

5. **Group the terms**:
\[ (7x^2 - 14x) + (-5x + 10) = 0 \]

6. **Factor by grouping**:
\[ 7x(x - 2) - 5(x - 2) = 0 \]
\[ (7x - 5)(x - 2) = 0 \]

7. **Set each factor to zero**:
\[ 7x - 5 = 0 \quad \text{or} \quad x - 2 = 0 \]

8. **Solve for \( x \)**:
- From \( 7x - 5 = 0 \):
\[ 7x = 5 \]
\[ x = \frac{5}{7} \]
- From \( x - 2 = 0 \):
\[ x = 2 \]

So, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = \frac{5}{7} \]

### Step 2: Solve Equation II
The second equation is:
\[ 8y^2 + 2y - 3 = 0 \]

We will also use the factorization method here.

1. **Identify the coefficients**:
- \( a = 8 \)
- \( b = 2 \)
- \( c = -3 \)

2. **Calculate the product \( ac \)**:
\[ ac = 8 \times -3 = -24 \]

3. **Find two numbers that multiply to \( ac \) and add to \( b \)**:
We need two numbers that multiply to -24 and add to 2. The numbers are 6 and -4 because:
\[ 6 \times -4 = -24 \]
\[ 6 + (-4) = 2 \]

4. **Rewrite the equation**:
\[ 8y^2 + 6y - 4y - 3 = 0 \]

5. **Group the terms**:
\[ (8y^2 + 6y) + (-4y - 3) = 0 \]

6. **Factor by grouping**:
\[ 2y(4y + 3) - 1(4y + 3) = 0 \]
\[ (2y - 1)(4y + 3) = 0 \]

7. **Set each factor to zero**:
\[ 2y - 1 = 0 \quad \text{or} \quad 4y + 3 = 0 \]

8. **Solve for \( y \)**:
- From \( 2y - 1 = 0 \):
\[ 2y = 1 \]
\[ y = \frac{1}{2} \]
- From \( 4y + 3 = 0 \):
\[ 4y = -3 \]
\[ y = -\frac{3}{4} \]

So, the solutions for \( y \) are:
\[ y = \frac{1}{2} \quad \text{and} \quad y = -\frac{3}{4} \]

### Summary of Solutions
- For \( x \): \( 2 \) and \( \frac{5}{7} \)
- For \( y \): \( \frac{1}{2} \) and \( -\frac{3}{4} \)