In each of the following questions, two equations (I) and (II) are given . Solve the equations and mark the correct option :
I. ` 3 x^(2) - 23 x + 40 = 0 `
II. ` 5y^(2) - 17 y + 14 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If ` x le y`

Answer

To solve the given equations and compare the values of \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation \( 3x^2 - 23x + 40 = 0 \)

To factor the quadratic equation, we need to find two numbers that multiply to \( 3 \times 40 = 120 \) and add up to \( -23 \).

The numbers that satisfy this condition are \( -15 \) and \( -8 \).

We can rewrite the equation as:
\[ 3x^2 - 15x - 8x + 40 = 0 \]

Now, we can group the terms:
\[ (3x^2 - 15x) + (-8x + 40) = 0 \]

Factoring each group:
\[ 3x(x - 5) - 8(x - 5) = 0 \]

Now factor out the common term \( (x - 5) \):
\[ (3x - 8)(x - 5) = 0 \]

Setting each factor to zero gives us:
1. \( 3x - 8 = 0 \) → \( x = \frac{8}{3} \)
2. \( x - 5 = 0 \) → \( x = 5 \)

So, the solutions for \( x \) are \( x = 5 \) and \( x = \frac{8}{3} \).

### Step 2: Solve the second equation \( 5y^2 - 17y + 14 = 0 \)

Similarly, we need to factor this quadratic equation. We need two numbers that multiply to \( 5 \times 14 = 70 \) and add up to \( -17 \).

The numbers that satisfy this condition are \( -10 \) and \( -7 \).

We can rewrite the equation as:
\[ 5y^2 - 10y - 7y + 14 = 0 \]

Now, we can group the terms:
\[ (5y^2 - 10y) + (-7y + 14) = 0 \]

Factoring each group:
\[ 5y(y - 2) - 7(y - 2) = 0 \]

Now factor out the common term \( (y - 2) \):
\[ (5y - 7)(y - 2) = 0 \]

Setting each factor to zero gives us:
1. \( 5y - 7 = 0 \) → \( y = \frac{7}{5} \)
2. \( y - 2 = 0 \) → \( y = 2 \)

So, the solutions for \( y \) are \( y = 2 \) and \( y = \frac{7}{5} \).

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

We have the following values:
- For \( x \): \( 5 \) and \( \frac{8}{3} \)
- For \( y \): \( 2 \) and \( \frac{7}{5} \)

Now we compare:
1. \( 5 > 2 \)
2. \( 5 > \frac{7}{5} \)
3. \( \frac{8}{3} \approx 2.67 > 2 \)
4. \( \frac{8}{3} > \frac{7}{5} \approx 1.4 \)

In both cases, \( x \) is greater than \( y \).

### Conclusion

Thus, we conclude that \( x > y \).