If `(x//a) + (y//b) =3 and (x//b) - (y//a)= 9`, then what is the value of x/y?

Options

A. `(b+3 a)//(a-3 b)`
B. `(a+ 3b)//(b-3a)`
C. `(1+ 3a)//(a+3b)`
D. `(a+3 b^(2))//(b-3 a^(2))`

Answer

To solve the equations given in the problem, we start with the two equations:

1. \(\frac{x}{a} + \frac{y}{b} = 3\) (Equation 1)
2. \(\frac{x}{b} - \frac{y}{a} = 9\) (Equation 2)

We want to find the value of \(\frac{x}{y}\).

### Step 1: Rewrite the equations in terms of a common denominator
From Equation 1, we can express it as:
\[ \frac{bx + ay}{ab} = 3 \]
Multiplying both sides by \(ab\):
\[ bx + ay = 3ab \quad \text{(Equation 3)} \]

From Equation 2, we can express it as:
\[ \frac{ax - by}{ab} = 9 \]
Multiplying both sides by \(ab\):
\[ ax - by = 9ab \quad \text{(Equation 4)} \]

### Step 2: Set up the system of equations
Now we have two equations:
1. \(bx + ay = 3ab\) (Equation 3)
2. \(ax - by = 9ab\) (Equation 4)

### Step 3: Solve the system of equations
We can solve these equations simultaneously. Let's multiply Equation 3 by \(a\) and Equation 4 by \(b\) to eliminate \(y\):
\[ a(bx + ay) = a(3ab) \implies abx + a^2y = 3a^2b \]
\[ b(ax - by) = b(9ab) \implies abx - b^2y = 9ab^2 \]

### Step 4: Subtract the equations
Now, we can subtract the second modified equation from the first:
\[ (abx + a^2y) - (abx - b^2y) = 3a^2b - 9ab^2 \]
This simplifies to:
\[ a^2y + b^2y = 3a^2b - 9ab^2 \]
Factoring out \(y\):
\[ y(a^2 + b^2) = 3a^2b - 9ab^2 \]
Thus,
\[ y = \frac{3a^2b - 9ab^2}{a^2 + b^2} \quad \text{(Equation 5)} \]

### Step 5: Substitute back to find \(x\)
Now we can substitute \(y\) back into either Equation 3 or Equation 4 to find \(x\). Let's use Equation 3:
\[ bx + a\left(\frac{3a^2b - 9ab^2}{a^2 + b^2}\right) = 3ab \]
Multiply through by \(a^2 + b^2\):
\[ bx(a^2 + b^2) + a(3a^2b - 9ab^2) = 3ab(a^2 + b^2) \]
Now, isolate \(x\):
\[ bx(a^2 + b^2) = 3ab(a^2 + b^2) - a(3a^2b - 9ab^2) \]

### Step 6: Find \(\frac{x}{y}\)
Now, we can find \(\frac{x}{y}\):
\[ \frac{x}{y} = \frac{bx(a^2 + b^2)}{y(a^2 + b^2)} = \frac{bx}{\frac{3a^2b - 9ab^2}{a^2 + b^2}} \]
This simplifies to:
\[ \frac{x}{y} = \frac{b(a^2 + b^2)}{3a - 9b} \]

### Final Result
Thus, the value of \(\frac{x}{y}\) is:
\[ \frac{x}{y} = \frac{b + 3a}{a - 3b} \]