The solution of the system of equations `(2x+ 5y)/(xy) = 6 and (4x- 5y)/(xy) +3= 0` (where `x ne 0, y ne 0`) is

Options

A. 1,2
B. 0,0
C. `-1, 2`
D. `1, -2`

Answer

To solve the system of equations given by

\[ \frac{2x + 5y}{xy} = 6 \]

and

\[ \frac{4x - 5y}{xy} + 3 = 0, \]

we will follow these steps:

### Step 1: Rewrite the equations

First, we can rewrite the first equation by multiplying both sides by \(xy\):

\[ 2x + 5y = 6xy \tag{1} \]

For the second equation, we can isolate the fraction:

\[ \frac{4x - 5y}{xy} = -3 \]

Now, multiply both sides by \(xy\):

\[ 4x - 5y = -3xy \tag{2} \]

### Step 2: Rearranging the equations

Now we have two equations:

1. \(2x + 5y = 6xy\)
2. \(4x - 5y = -3xy\)

We can rearrange both equations to express them in standard form:

From equation (1):

\[ 6xy - 2x - 5y = 0 \tag{3} \]

From equation (2):

\[ 4x + 3xy - 5y = 0 \tag{4} \]

### Step 3: Solve the equations

Now we will solve equations (3) and (4) simultaneously.

From equation (3):

\[ 6xy - 2x - 5y = 0 \]

We can express \(y\) in terms of \(x\):

\[ 6xy = 2x + 5y \]

Rearranging gives:

\[ 6xy - 5y = 2x \]

Factoring out \(y\):

\[ y(6x - 5) = 2x \]

Thus,

\[ y = \frac{2x}{6x - 5} \tag{5} \]

Now substitute equation (5) into equation (4):

\[ 4x + 3x\left(\frac{2x}{6x - 5}\right) - 5\left(\frac{2x}{6x - 5}\right) = 0 \]

### Step 4: Simplifying the equation

Multiply through by \(6x - 5\) to eliminate the denominator:

\[ 4x(6x - 5) + 3(2x)(x) - 5(2x) = 0 \]

Expanding this gives:

\[ 24x^2 - 20x + 6x^2 - 10x = 0 \]

Combining like terms:

\[ 30x^2 - 30x = 0 \]

Factoring out \(30x\):

\[ 30x(x - 1) = 0 \]

Thus, \(x = 0\) or \(x = 1\). Since \(x \neq 0\), we have:

\[ x = 1 \]

### Step 5: Finding \(y\)

Now substitute \(x = 1\) back into equation (5):

\[ y = \frac{2(1)}{6(1) - 5} = \frac{2}{1} = 2 \]

### Conclusion

The solution to the system of equations is:

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