In each of the following questions, two equations are given. You have to solve them
`"I. "10x^2-7x+1=0" II. "35y^2-12y+1=0`

Options

A. `xgty`
B. `xgey`
C. `xlty`
D. `xley`

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation \(10x^2 - 7x + 1 = 0\)

1. **Rewrite the equation**: We want to factor the quadratic equation. We can look for two numbers that multiply to \(10 \times 1 = 10\) and add to \(-7\).
2. **Factor the equation**: We can rewrite \(-7x\) as \(-5x - 2x\):
\[ 10x^2 - 5x - 2x + 1 = 0 \]
Now, group the terms:
\[ (10x^2 - 5x) + (-2x + 1) = 0 \]
Factor out the common terms:
\[ 5x(2x - 1) - 1(2x - 1) = 0 \]
Now, factor by grouping:
\[ (5x - 1)(2x - 1) = 0 \]
3. **Find the roots**: Set each factor to zero:
\[ 5x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{5} \]
\[ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2} \]
So, the solutions for \(x\) are \(x = \frac{1}{5}\) and \(x = \frac{1}{2}\).

### Step 2: Solve the second equation \(35y^2 - 12y + 1 = 0\)

1. **Rewrite the equation**: We will look for two numbers that multiply to \(35 \times 1 = 35\) and add to \(-12\).
2. **Factor the equation**: We can rewrite \(-12y\) as \(-7y - 5y\):
\[ 35y^2 - 7y - 5y + 1 = 0 \]
Now, group the terms:
\[ (35y^2 - 7y) + (-5y + 1) = 0 \]
Factor out the common terms:
\[ 7y(5y - 1) - 1(5y - 1) = 0 \]
Now, factor by grouping:
\[ (7y - 1)(5y - 1) = 0 \]
3. **Find the roots**: Set each factor to zero:
\[ 7y - 1 = 0 \quad \Rightarrow \quad y = \frac{1}{7} \]
\[ 5y - 1 = 0 \quad \Rightarrow \quad y = \frac{1}{5} \]
So, the solutions for \(y\) are \(y = \frac{1}{5}\) and \(y = \frac{1}{7}\).

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

Now we have:
- \(x = \frac{1}{5}, \frac{1}{2}\)
- \(y = \frac{1}{5}, \frac{1}{7}\)

1. **Compare the values**:
- For \(x = \frac{1}{5}\) and \(y = \frac{1}{5}\), we have \(x = y\).
- For \(x = \frac{1}{2}\) and \(y = \frac{1}{5}\), we have \(x > y\).

Since \(x = \frac{1}{5}\) is equal to \(y = \frac{1}{5}\) and \(x = \frac{1}{2}\) is greater than \(y = \frac{1}{5}\), we conclude that \(x\) is greater than or equal to \(y\).

### Final Answer:
Thus, the answer is \(x \geq y\).

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