In each of the following questions, two equations are given. You have to solve them and
I. `12 x^(2) - 2x - 4 = 0` II. `10y^(2) + 2 = 9y`

Options

A. if `x gt y`
B. if `x ge y`
C. if `x lt y`
D. if `x = y` or relationship can not be established

Answer

To solve the given equations step by step, let's break it down:

### Step 1: Solve the first equation
The first equation is:
\[ 12x^2 - 2x - 4 = 0 \]

To solve this quadratic equation, we can use the factorization method.

1. **Rearranging the equation:**
We can factor the equation. We start by rewriting it:
\[ 12x^2 - 8x + 6x - 4 = 0 \]

2. **Grouping terms:**
Now, we can group the terms:
\[ (12x^2 - 8x) + (6x - 4) = 0 \]

3. **Factoring out common terms:**
From the first group, we can factor out \(4x\):
\[ 4x(3x - 2) + 2(3x - 2) = 0 \]

4. **Factoring the entire expression:**
Now we can factor out \((3x - 2)\):
\[ (4x + 2)(3x - 2) = 0 \]

5. **Setting each factor to zero:**
Now, we set each factor equal to zero:
\[ 4x + 2 = 0 \quad \text{or} \quad 3x - 2 = 0 \]

Solving these gives:
\[ 4x + 2 = 0 \Rightarrow 4x = -2 \Rightarrow x = -\frac{1}{2} \]
\[ 3x - 2 = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3} \]

### Step 2: Solve the second equation
The second equation is:
\[ 10y^2 + 2 = 9y \]

1. **Rearranging the equation:**
We can rearrange it to:
\[ 10y^2 - 9y + 2 = 0 \]

2. **Grouping terms:**
We can factor this equation as well:
\[ 10y^2 - 5y - 4y + 2 = 0 \]

3. **Factoring out common terms:**
From the first group, we can factor out \(5y\):
\[ 5y(2y - 1) - 2(2y - 1) = 0 \]

4. **Factoring the entire expression:**
Now we can factor out \((2y - 1)\):
\[ (5y - 2)(2y - 1) = 0 \]

5. **Setting each factor to zero:**
Now, we set each factor equal to zero:
\[ 5y - 2 = 0 \quad \text{or} \quad 2y - 1 = 0 \]

Solving these gives:
\[ 5y - 2 = 0 \Rightarrow 5y = 2 \Rightarrow y = \frac{2}{5} \]
\[ 2y - 1 = 0 \Rightarrow 2y = 1 \Rightarrow y = \frac{1}{2} \]

### Step 3: Compare the values of x and y
Now we have the values:
- For \(x\): \(x_1 = -\frac{1}{2}\) and \(x_2 = \frac{2}{3}\)
- For \(y\): \(y_1 = \frac{2}{5}\) and \(y_2 = \frac{1}{2}\)

Now we will compare these values:

1. **Comparing \(x_1\) and \(y_1\):**
\(-\frac{1}{2} < \frac{2}{5}\) (since \(-0.5 < 0.4\))

2. **Comparing \(x_1\) and \(y_2\):**
\(-\frac{1}{2} < \frac{1}{2}\) (since \(-0.5 < 0.5\))

3. **Comparing \(x_2\) and \(y_1\):**
\(\frac{2}{3} > \frac{2}{5}\) (since \(0.666 > 0.4\))

4. **Comparing \(x_2\) and \(y_2\):**
\(\frac{2}{3} > \frac{1}{2}\) (since \(0.666 > 0.5\))

### Conclusion
From the comparisons, we see that:
- When \(x = -\frac{1}{2}\), \(y\) is greater.
- When \(x = \frac{2}{3}\), \(x\) is greater.

Since we have one case where \(x\) is greater and another where \(y\) is greater, we cannot establish a consistent relationship between \(x\) and \(y\).

### Final Answer
Thus, the relationship cannot be established, which corresponds to option 4.

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