Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between x and y and give answer
I. ` 2 x^(2) - 13 x + 21 = 0 `
II. `5 y ^(2) - 22 y + 21 = 0 `

Options

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

Answer

To solve the given equations and find the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve the first equation for \( x \)
The first equation is:
\[ 2x^2 - 13x + 21 = 0 \]

We will factor this equation. We need two numbers that multiply to \( 2 \times 21 = 42 \) and add to \( -13 \). The numbers are \( -7 \) and \( -6 \).

Rewriting the equation:
\[ 2x^2 - 7x - 6x + 21 = 0 \]

Grouping the terms:
\[ (2x^2 - 7x) + (-6x + 21) = 0 \]

Factoring by grouping:
\[ x(2x - 7) - 3(2x - 7) = 0 \]

Factoring out the common term:
\[ (2x - 7)(x - 3) = 0 \]

Setting each factor to zero gives:
1. \( 2x - 7 = 0 \) → \( x = \frac{7}{2} \)
2. \( x - 3 = 0 \) → \( x = 3 \)

Thus, the values of \( x \) are:
\[ x = 3 \quad \text{and} \quad x = \frac{7}{2} \]

### Step 2: Solve the second equation for \( y \)
The second equation is:
\[ 5y^2 - 22y + 21 = 0 \]

We need two numbers that multiply to \( 5 \times 21 = 105 \) and add to \( -22 \). The numbers are \( -15 \) and \( -7 \).

Rewriting the equation:
\[ 5y^2 - 15y - 7y + 21 = 0 \]

Grouping the terms:
\[ (5y^2 - 15y) + (-7y + 21) = 0 \]

Factoring by grouping:
\[ 5y(y - 3) - 7(y - 3) = 0 \]

Factoring out the common term:
\[ (y - 3)(5y - 7) = 0 \]

Setting each factor to zero gives:
1. \( y - 3 = 0 \) → \( y = 3 \)
2. \( 5y - 7 = 0 \) → \( y = \frac{7}{5} \)

Thus, the values of \( y \) are:
\[ y = 3 \quad \text{and} \quad y = \frac{7}{5} \]

### Step 3: Determine the relationship between \( x \) and \( y \)
We have the following values:
- For \( x \): \( 3 \) and \( \frac{7}{2} \)
- For \( y \): \( 3 \) and \( \frac{7}{5} \)

Now, we will compare these values:
1. When \( x = 3 \) and \( y = 3 \): \( x = y \)
2. When \( x = \frac{7}{2} = 3.5 \) and \( y = \frac{7}{5} = 1.4 \): \( x > y \)

### Conclusion
From the above comparisons, we find that:
- \( x \) can be equal to \( y \) (when both are \( 3 \)).
- \( x \) can also be greater than \( y \) (when \( x = \frac{7}{2} \) and \( y = \frac{7}{5} \)).

Thus, the final relationship we can conclude is:
\[ x \geq y \]

### Final Answer
The relationship between \( x \) and \( y \) is:
\[ x \geq y \]

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