In the following two equations questions numbered (I)and (II) are given . You have to solve both equations and Give answer
I. ` 2 x^(2) + x - 28 = `0
II .` 2 y^(2) - 23 y + 56 = 0`

Options

A. A)If ` x gt y `
B. B)If `x ge y`
C. C)If ` y gt x`
D. D)If ` y ge x `

Answer

To solve the equations given in the question, we will follow these steps for each equation.

### Equation I: \( 2x^2 + x - 28 = 0 \)

**Step 1: Factor the equation.**
We need to express the quadratic equation in a factored form. We will look for two numbers that multiply to \( 2 \times -28 = -56 \) and add up to \( 1 \) (the coefficient of \( x \)).

The numbers \( 8 \) and \( -7 \) work since:
- \( 8 \times (-7) = -56 \)
- \( 8 + (-7) = 1 \)

Now we can rewrite the equation:
\[ 2x^2 + 8x - 7x - 28 = 0 \]

**Step 2: Group the terms.**
Group the first two terms and the last two terms:
\[ (2x^2 + 8x) + (-7x - 28) = 0 \]

**Step 3: Factor by grouping.**
Factor out the common terms:
\[ 2x(x + 4) - 7(x + 4) = 0 \]

Now we can factor out \( (x + 4) \):
\[ (x + 4)(2x - 7) = 0 \]

**Step 4: Solve for \( x \).**
Set each factor to zero:
1. \( x + 4 = 0 \) → \( x = -4 \)
2. \( 2x - 7 = 0 \) → \( 2x = 7 \) → \( x = \frac{7}{2} \)

So, the solutions for \( x \) are:
\[ x = -4 \quad \text{and} \quad x = \frac{7}{2} \]

### Equation II: \( 2y^2 - 23y + 56 = 0 \)

**Step 1: Factor the equation.**
We need to find two numbers that multiply to \( 2 \times 56 = 112 \) and add up to \( -23 \).

The numbers \( -16 \) and \( -7 \) work since:
- \( -16 \times -7 = 112 \)
- \( -16 + (-7) = -23 \)

Now we can rewrite the equation:
\[ 2y^2 - 16y - 7y + 56 = 0 \]

**Step 2: Group the terms.**
Group the first two terms and the last two terms:
\[ (2y^2 - 16y) + (-7y + 56) = 0 \]

**Step 3: Factor by grouping.**
Factor out the common terms:
\[ 2y(y - 8) - 7(y - 8) = 0 \]

Now we can factor out \( (y - 8) \):
\[ (y - 8)(2y - 7) = 0 \]

**Step 4: Solve for \( y \).**
Set each factor to zero:
1. \( y - 8 = 0 \) → \( y = 8 \)
2. \( 2y - 7 = 0 \) → \( 2y = 7 \) → \( y = \frac{7}{2} \)

So, the solutions for \( y \) are:
\[ y = 8 \quad \text{and} \quad y = \frac{7}{2} \]

### Summary of Solutions
- For \( x \): \( x = -4 \) and \( x = \frac{7}{2} \)
- For \( y \): \( y = 8 \) and \( y = \frac{7}{2} \)

### Comparing Values
Now we compare the values of \( x \) and \( y \):
1. \( -4 < \frac{7}{2} \)
2. \( \frac{7}{2} = \frac{7}{2} \)
3. \( \frac{7}{2} < 8 \)

Thus, we can conclude:
\[ x \leq y \]

### Final Answer
The relation between \( x \) and \( y \) is:
\[ x \leq y \]

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