Solve 6x + 3y = 7xy and 3x + 9y = 11xy.

Options

A. `x = 1, y = 2/3`
B. `x =- 1, y =2/3`
C. `x = 2/3, y =1/2`
D. `x = 1, y= -3/2`

Answer

To solve the equations \(6x + 3y = 7xy\) and \(3x + 9y = 11xy\), we will follow these steps:

### Step 1: Rewrite the equations
We have the following two equations:
1. \(6x + 3y = 7xy\) (Equation 1)
2. \(3x + 9y = 11xy\) (Equation 2)

### Step 2: Divide both equations to simplify
To simplify these equations, we can divide both sides of each equation by \(xy\) (assuming \(x\) and \(y\) are not zero):

From Equation 1:
\[ \frac{6x}{xy} + \frac{3y}{xy} = \frac{7xy}{xy} \]
This simplifies to:
\[ \frac{6}{y} + \frac{3}{x} = 7 \quad \text{(Equation 3)} \]

From Equation 2:
\[ \frac{3x}{xy} + \frac{9y}{xy} = \frac{11xy}{xy} \]
This simplifies to:
\[ \frac{3}{y} + \frac{9}{x} = 11 \quad \text{(Equation 4)} \]

### Step 3: Multiply Equation 4 by 2
To eliminate the fractions, we can multiply Equation 4 by 2:
\[ 2\left(\frac{3}{y} + \frac{9}{x}\right) = 2 \cdot 11 \]
This gives us:
\[ \frac{6}{y} + \frac{18}{x} = 22 \quad \text{(Equation 5)} \]

### Step 4: Subtract Equation 3 from Equation 5
Now we will subtract Equation 3 from Equation 5:
\[ \left(\frac{6}{y} + \frac{18}{x}\right) - \left(\frac{6}{y} + \frac{3}{x}\right) = 22 - 7 \]
This simplifies to:
\[ \frac{18}{x} - \frac{3}{x} = 15 \]
Combining the terms gives:
\[ \frac{15}{x} = 15 \]

### Step 5: Solve for \(x\)
Now, we can solve for \(x\):
\[ 15 = 15x \implies x = 1 \]

### Step 6: Substitute \(x\) back to find \(y\)
Now that we have \(x = 1\), we can substitute this value back into Equation 3 to find \(y\):
\[ \frac{6}{y} + \frac{3}{1} = 7 \]
This simplifies to:
\[ \frac{6}{y} + 3 = 7 \]
Subtracting 3 from both sides gives:
\[ \frac{6}{y} = 4 \]
Now, solving for \(y\):
\[ 6 = 4y \implies y = \frac{6}{4} = \frac{3}{2} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 1, \quad y = \frac{3}{2} \]