Solve the equations given below and give answer
1) if `x lt y` 2) if `x gt y`
3) if `c ge y` 4) if `x le y`
5) if x=y or no relation can be established
I. `2x^2-15=7x` II. `17y=-7-6y^2`

Options

A. if `x lt y`
B. if `x gt y`
C. if `x ge y`
D. if x=y or no relation can be established

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation \(2x^2 - 15 = 7x\)

1. Rearrange the equation:
\[ 2x^2 - 7x - 15 = 0 \]

2. Factor the quadratic:
- We need two numbers that multiply to \(2 \times -15 = -30\) and add to \(-7\).
- The numbers are \(-10\) and \(3\).
- Rewrite the equation:
\[ 2x^2 - 10x + 3x - 15 = 0 \]

3. Group the terms:
\[ (2x^2 - 10x) + (3x - 15) = 0 \]
\[ 2x(x - 5) + 3(x - 5) = 0 \]

4. Factor out the common term:
\[ (2x + 3)(x - 5) = 0 \]

5. Set each factor to zero:
\[ 2x + 3 = 0 \quad \text{or} \quad x - 5 = 0 \]
\[ x = -\frac{3}{2} \quad \text{or} \quad x = 5 \]

### Step 2: Solve the second equation \(17y = -7 - 6y^2\)

1. Rearrange the equation:
\[ 6y^2 + 17y + 7 = 0 \]

2. Factor the quadratic:
- We need two numbers that multiply to \(6 \times 7 = 42\) and add to \(17\).
- The numbers are \(14\) and \(3\).
- Rewrite the equation:
\[ 6y^2 + 14y + 3y + 7 = 0 \]

3. Group the terms:
\[ (6y^2 + 14y) + (3y + 7) = 0 \]
\[ 2y(3y + 7) + 1(3y + 7) = 0 \]

4. Factor out the common term:
\[ (3y + 7)(2y + 1) = 0 \]

5. Set each factor to zero:
\[ 3y + 7 = 0 \quad \text{or} \quad 2y + 1 = 0 \]
\[ y = -\frac{7}{3} \quad \text{or} \quad y = -\frac{1}{2} \]

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

1. Values of \(x\):
- \(x_1 = 5\)
- \(x_2 = -\frac{3}{2} = -1.5\)

2. Values of \(y\):
- \(y_1 = -\frac{7}{3} \approx -2.33\)
- \(y_2 = -\frac{1}{2} = -0.5\)

### Step 4: Analyze the relationships

1. **If \(x < y\)**:
- For \(x = 5\) and \(y = -\frac{7}{3}\), \(5 > -\frac{7}{3}\).
- For \(x = 5\) and \(y = -\frac{1}{2}\), \(5 > -\frac{1}{2}\).
- For \(x = -1.5\) and \(y = -\frac{7}{3}\), \(-1.5 > -\frac{7}{3}\).
- For \(x = -1.5\) and \(y = -\frac{1}{2}\), \(-1.5 < -0.5\).

2. **If \(x > y\)**:
- For \(x = 5\), \(x\) is always greater than both \(y\) values.
- For \(x = -1.5\) and \(y = -\frac{7}{3}\), \(-1.5 > -2.33\).
- For \(x = -1.5\) and \(y = -\frac{1}{2}\), \(-1.5 < -0.5\).

3. **If \(c \geq y\)**:
- This depends on the value of \(c\) which is not provided.

4. **If \(x \leq y\)**:
- This is true for \(x = -1.5\) and \(y = -\frac{1}{2}\).

5. **If \(x = y\) or no relation can be established**:
- No values of \(x\) and \(y\) are equal.

### Conclusion

From the analysis, we can conclude:
- \(x\) can be greater than \(y\) in most cases.
- The only case where \(x\) is less than \(y\) is when \(x = -1.5\) and \(y = -\frac{1}{2}\).