I.` 5x^(2) + 2x - 3 = 0" "II. 2y^(2) + 7y + 6 = 0`

Options

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

Explanation

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

### Step 1: Solve the first equation \(5x^2 + 2x - 3 = 0\)

We will factor the quadratic equation:
1. Multiply the coefficient of \(x^2\) (which is 5) by the constant term (which is -3):
\[ 5 \times -3 = -15 \]
2. We need two numbers that multiply to -15 and add up to 2 (the coefficient of \(x\)). These numbers are 5 and -3.
3. Rewrite the equation:
\[ 5x^2 + 5x - 3x - 3 = 0 \]
4. Group the terms:
\[ (5x^2 + 5x) + (-3x - 3) = 0 \]
5. Factor by grouping:
\[ 5x(x + 1) - 3(x + 1) = 0 \]
6. Factor out the common term \((x + 1)\):
\[ (5x - 3)(x + 1) = 0 \]
7. Set each factor to zero:
\[ 5x - 3 = 0 \quad \text{or} \quad x + 1 = 0 \]
Solving these gives:
\[ x = \frac{3}{5} \quad \text{or} \quad x = -1 \]

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

We will factor this quadratic equation:
1. Multiply the coefficient of \(y^2\) (which is 2) by the constant term (which is 6):
\[ 2 \times 6 = 12 \]
2. We need two numbers that multiply to 12 and add up to 7. These numbers are 4 and 3.
3. Rewrite the equation:
\[ 2y^2 + 4y + 3y + 6 = 0 \]
4. Group the terms:
\[ (2y^2 + 4y) + (3y + 6) = 0 \]
5. Factor by grouping:
\[ 2y(y + 2) + 3(y + 2) = 0 \]
6. Factor out the common term \((y + 2)\):
\[ (2y + 3)(y + 2) = 0 \]
7. Set each factor to zero:
\[ 2y + 3 = 0 \quad \text{or} \quad y + 2 = 0 \]
Solving these gives:
\[ y = -\frac{3}{2} \quad \text{or} \quad y = -2 \]

### Step 3: Establish the relationship between \(x\) and \(y\)

Now we have the values:
- For \(x\): \(x = \frac{3}{5}\) or \(x = -1\)
- For \(y\): \(y = -\frac{3}{2}\) or \(y = -2\)

**Comparing values:**
1. For \(x = \frac{3}{5}\) (which is positive), we compare it with both \(y\) values:
- \(\frac{3}{5} > -\frac{3}{2}\) (since \(-\frac{3}{2} = -1.5\))
- \(\frac{3}{5} > -2\)

2. For \(x = -1\):
- \(-1 > -\frac{3}{2}\) (since \(-\frac{3}{2} = -1.5\))
- \(-1 > -2\)

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

### Conclusion:
The relationship established is:
\[ x > y \]

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