Solve the following quadratic equation and mark and answer as per instructions.
I. ` 9 x + 3y = 15`
II. ` 4 x + 5y = 14`

Options

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

Answer

To solve the given system of equations, we will use the method of elimination. The equations are:

I. \( 9x + 3y = 15 \)
II. \( 4x + 5y = 14 \)

### Step 1: Eliminate one variable
We will eliminate \( y \) by making the coefficients of \( y \) in both equations equal. The coefficients are 3 and 5. We can do this by multiplying the first equation by 5 and the second equation by 3.

- Multiply Equation I by 5:
\[ 5(9x + 3y) = 5(15) \implies 45x + 15y = 75 \]

- Multiply Equation II by 3:
\[ 3(4x + 5y) = 3(14) \implies 12x + 15y = 42 \]

### Step 2: Subtract the equations
Now we have:
1. \( 45x + 15y = 75 \) (Equation III)
2. \( 12x + 15y = 42 \) (Equation IV)

Next, we can subtract Equation IV from Equation III to eliminate \( y \):
\[ (45x + 15y) - (12x + 15y) = 75 - 42 \]
This simplifies to:
\[ 33x = 33 \]

### Step 3: Solve for \( x \)
Now, divide both sides by 33:
\[ x = 1 \]

### Step 4: Substitute \( x \) back to find \( y \)
Now that we have \( x \), we can substitute it back into either of the original equations to find \( y \). We will use Equation II:
\[ 4x + 5y = 14 \]
Substituting \( x = 1 \):
\[ 4(1) + 5y = 14 \implies 4 + 5y = 14 \]
Subtract 4 from both sides:
\[ 5y = 10 \]
Now divide by 5:
\[ y = 2 \]

### Step 5: Compare \( x \) and \( y \)
We found \( x = 1 \) and \( y = 2 \). Now we compare the two:
\[ x < y \quad \text{(since } 1 < 2\text{)} \]

### Final Answer
The relationship between \( x \) and \( y \) is:
\[ x < y \]