I. ` 8x+y = 10`
II.` 4x+2y = 13`

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 system of equations:

I. \( 8x + y = 10 \)
II. \( 4x + 2y = 13 \)

we will follow these steps:

### Step 1: Simplify the second equation
First, we will manipulate the second equation to make it easier to work with. We can multiply the entire second equation by 2 to align the coefficients of \(y\) with the first equation.

\[ 4x + 2y = 13 \implies 2(4x + 2y) = 2(13) \implies 8x + 4y = 26 \]

### Step 2: Write down the modified equations
Now we have two equations:
1. \( 8x + y = 10 \) (Equation 1)
2. \( 8x + 4y = 26 \) (Modified Equation 2)

### Step 3: Eliminate \(x\)
Next, we will eliminate \(x\) by subtracting Equation 1 from Modified Equation 2.

\[ (8x + 4y) - (8x + y) = 26 - 10 \]

This simplifies to:

\[ 4y - y = 16 \implies 3y = 16 \]

### Step 4: Solve for \(y\)
Now we can solve for \(y\):

\[ y = \frac{16}{3} \]

### Step 5: Substitute \(y\) back to find \(x\)
Now that we have \(y\), we can substitute it back into Equation 1 to find \(x\):

\[ 8x + \frac{16}{3} = 10 \]

To isolate \(8x\), we subtract \(\frac{16}{3}\) from both sides:

\[ 8x = 10 - \frac{16}{3} \]

To perform the subtraction, we need a common denominator. The number 10 can be written as \(\frac{30}{3}\):

\[ 8x = \frac{30}{3} - \frac{16}{3} = \frac{30 - 16}{3} = \frac{14}{3} \]

### Step 6: Solve for \(x\)
Now we divide both sides by 8 to solve for \(x\):

\[ x = \frac{14}{3} \cdot \frac{1}{8} = \frac{14}{24} = \frac{7}{12} \]

### Summary of Solutions
Thus, we have:
\[ x = \frac{7}{12}, \quad y = \frac{16}{3} \]

### Step 7: Compare \(x\) and \(y\)
Now we can compare the values of \(x\) and \(y\):

- \(x = \frac{7}{12} \approx 0.583\)
- \(y = \frac{16}{3} \approx 5.333\)

Since \(y > x\), we can conclude that the correct option is that \(y\) is greater than \(x\).