In each of these questions, two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 12 x + 3 y = 14`
II. ` 4 x + 2y = 16`

Options

A. A)If ` x gt y `
B. B)If ` x lt y `
C. C) If ` x ge y`
D. D)If ` x le y `

Answer

To solve the given equations step by step, we will follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( 12x + 3y = 14 \) (Equation I)
2. \( 4x + 2y = 16 \) (Equation II)

### Step 2: Simplify Equation II
We can simplify Equation II by dividing all terms by 2:
\[ 2x + y = 8 \]
Now we have:
1. \( 12x + 3y = 14 \) (Equation I)
2. \( 2x + y = 8 \) (Equation II simplified)

### Step 3: Express y in terms of x from Equation II
From the simplified Equation II, we can express \( y \):
\[ y = 8 - 2x \]

### Step 4: Substitute y in Equation I
Now, substitute \( y \) in Equation I:
\[ 12x + 3(8 - 2x) = 14 \]
Expanding this gives:
\[ 12x + 24 - 6x = 14 \]

### Step 5: Combine like terms
Combine the \( x \) terms:
\[ 6x + 24 = 14 \]

### Step 6: Isolate x
Subtract 24 from both sides:
\[ 6x = 14 - 24 \]
\[ 6x = -10 \]
Now, divide by 6:
\[ x = -\frac{10}{6} = -\frac{5}{3} \]

### Step 7: Substitute x back to find y
Now substitute \( x \) back into the equation for \( y \):
\[ y = 8 - 2\left(-\frac{5}{3}\right) \]
Calculating this gives:
\[ y = 8 + \frac{10}{3} \]
To add these, convert 8 into a fraction:
\[ y = \frac{24}{3} + \frac{10}{3} = \frac{34}{3} \]

### Step 8: Conclusion
We have found:
\[ x = -\frac{5}{3}, \quad y = \frac{34}{3} \]
Now we can compare \( x \) and \( y \):
Since \( -\frac{5}{3} < \frac{34}{3} \), we conclude that \( x < y \).

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

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