` 2x^(2) - 13x+20 = 0, 2y^(2) - 7y + 6 = 0`

Options

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

Explanation

To solve the given equations \(2x^2 - 13x + 20 = 0\) and \(2y^2 - 7y + 6 = 0\), we will factor each quadratic equation and find the values of \(x\) and \(y\).

### Step 1: Solve the first equation \(2x^2 - 13x + 20 = 0\)

1. **Factor the quadratic**:
We need to rewrite the equation in a factored form. We look for two numbers that multiply to \(2 \times 20 = 40\) and add to \(-13\). The numbers are \(-8\) and \(-5\).

\[ 2x^2 - 8x - 5x + 20 = 0 \]
Grouping the terms:
\[ (2x^2 - 8x) + (-5x + 20) = 0 \]
Factoring by grouping:
\[ 2x(x - 4) - 5(x - 4) = 0 \]
This gives us:
\[ (2x - 5)(x - 4) = 0 \]

2. **Set each factor to zero**:
\[ 2x - 5 = 0 \quad \text{or} \quad x - 4 = 0 \]
Solving these gives:
\[ 2x = 5 \implies x = \frac{5}{2} \quad \text{and} \quad x = 4 \]

### Step 2: Solve the second equation \(2y^2 - 7y + 6 = 0\)

1. **Factor the quadratic**:
We need to find two numbers that multiply to \(2 \times 6 = 12\) and add to \(-7\). The numbers are \(-3\) and \(-4\).

\[ 2y^2 - 3y - 4y + 6 = 0 \]
Grouping the terms:
\[ (2y^2 - 3y) + (-4y + 6) = 0 \]
Factoring by grouping:
\[ y(2y - 3) - 2(2y - 3) = 0 \]
This gives us:
\[ (2y - 3)(y - 2) = 0 \]

2. **Set each factor to zero**:
\[ 2y - 3 = 0 \quad \text{or} \quad y - 2 = 0 \]
Solving these gives:
\[ 2y = 3 \implies y = \frac{3}{2} \quad \text{and} \quad y = 2 \]

### Step 3: Summary of solutions

The values obtained are:
- For \(x\): \(x = \frac{5}{2} = 2.5\) and \(x = 4\)
- For \(y\): \(y = \frac{3}{2} = 1.5\) and \(y = 2\)

### Step 4: Determine the relationship between \(x\) and \(y\)

Now we compare the values of \(x\) and \(y\):
- For \(x = 2.5\) and \(y = 1.5\): \(2.5 > 1.5\)
- For \(x = 4\) and \(y = 2\): \(4 > 2\)

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

### Final Result:
Thus, the relationship is \(x > y\).

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