Solve the simultaneous equations :
`(6x+3y)/(xy) = 6, (2x+4y)/(xy) = 5`

Answer

To solve the simultaneous equations:

1. \(\frac{6x + 3y}{xy} = 6\)
2. \(\frac{2x + 4y}{xy} = 5\)

we will follow these steps:

### Step 1: Simplify the first equation
Starting with the first equation:

\[ \frac{6x + 3y}{xy} = 6 \]

Multiply both sides by \(xy\):

\[ 6x + 3y = 6xy \]

Rearranging gives:

\[ 6xy - 6x - 3y = 0 \]

### Step 2: Factor the first equation
We can rearrange the equation:

\[ 6xy - 6x - 3y = 0 \]

Factoring out common terms, we can write:

\[ 3(2xy - 2x - y) = 0 \]

This simplifies to:

\[ 2xy - 2x - y = 0 \]

### Step 3: Simplify the second equation
Now, let's simplify the second equation:

\[ \frac{2x + 4y}{xy} = 5 \]

Multiply both sides by \(xy\):

\[ 2x + 4y = 5xy \]

Rearranging gives:

\[ 5xy - 2x - 4y = 0 \]

### Step 4: Factor the second equation
We can rearrange the second equation:

\[ 5xy - 2x - 4y = 0 \]

Factoring out common terms, we can write:

\[ y(5x - 4) - 2x = 0 \]

This simplifies to:

\[ 5xy - 4y - 2x = 0 \]

### Step 5: Substitute and solve
Now we have two equations:

1. \(2xy - 2x - y = 0\)
2. \(5xy - 2x - 4y = 0\)

Let's express \(y\) from the first equation:

From \(2xy - 2x - y = 0\):

\[ y(2x - 1) = 2x \]

Thus,

\[ y = \frac{2x}{2x - 1} \]

Now substitute \(y\) into the second equation:

\[ 5x\left(\frac{2x}{2x - 1}\right) - 2x - 4\left(\frac{2x}{2x - 1}\right) = 0 \]

### Step 6: Clear the fractions
Multiply through by \(2x - 1\) to eliminate the fraction:

\[ 10x^2 - 2x(2x - 1) - 8x = 0 \]

Expanding gives:

\[ 10x^2 - 4x^2 + 2x - 8x = 0 \]

Combining like terms:

\[ 6x^2 - 6x = 0 \]

Factoring out \(6x\):

\[ 6x(x - 1) = 0 \]

### Step 7: Solve for \(x\)
Setting each factor to zero gives:

1. \(6x = 0 \Rightarrow x = 0\) (not valid in this context)
2. \(x - 1 = 0 \Rightarrow x = 1\)

### Step 8: Find \(y\)
Substituting \(x = 1\) back into the equation for \(y\):

\[ y = \frac{2(1)}{2(1) - 1} = \frac{2}{1} = 2 \]

### Final Solution
Thus, the solution to the simultaneous equations is:

\[ x = 1, \quad y = 2 \]