Solve the given quadratic equations and mark the correct options based on your answer
I. ` 6 x + 7y = 15`
II. ` 3 x + 14 y = 19 . 5`

Options

A. ` x gt y `
B. x = y or no relation can be established between x and y
C. ` x le y `
D. ` x lt y `

Answer

To solve the given equations and find the values of \( x \) and \( y \), we will follow a systematic approach. The equations we need to solve are:

1. \( 6x + 7y = 15 \) (Equation I)
2. \( 3x + 14y = 19.5 \) (Equation II)

### Step 1: Multiply Equation I by 2
To eliminate \( y \), we can manipulate the equations. First, we will multiply Equation I by 2:

\[ 2(6x + 7y) = 2(15) \]

This gives us:

\[ 12x + 14y = 30 \quad \text{(Equation III)} \]

### Step 2: Write Equation II
Now, we can write Equation II as it is:

\[ 3x + 14y = 19.5 \quad \text{(Equation II)} \]

### Step 3: Subtract Equation II from Equation III
Next, we will subtract Equation II from Equation III to eliminate \( y \):

\[ (12x + 14y) - (3x + 14y) = 30 - 19.5 \]

This simplifies to:

\[ 12x - 3x = 30 - 19.5 \]

\[ 9x = 10.5 \]

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

\[ x = \frac{10.5}{9} = \frac{105}{90} = \frac{7}{6} \]

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

\[ 6\left(\frac{7}{6}\right) + 7y = 15 \]

This simplifies to:

\[ 7 + 7y = 15 \]

Subtracting 7 from both sides gives:

\[ 7y = 8 \]

### Step 6: Solve for \( y \)
Now, we can solve for \( y \):

\[ y = \frac{8}{7} \]

### Summary of Results
We have found:

\[ x = \frac{7}{6} \quad \text{and} \quad y = \frac{8}{7} \]

### Step 7: Compare \( x \) and \( y \)
To determine which is greater, we can convert both fractions to decimals:

- \( x = \frac{7}{6} \approx 1.1667 \)
- \( y = \frac{8}{7} \approx 1.1429 \)

Since \( 1.1667 > 1.1429 \), we conclude that:

\[ x > y \]

### Conclusion
Thus, the correct option based on our calculations is \( x > y \).