The values of x and y satisfying `(x+3)(y-5)=xy+39` and `(x-2)(y+3)=xy-40` respectively are:

Options

A. 6 and 8
B. 6 and -8
C. `-6` and 8
D. `-6` and -8

Answer

To solve the equations `(x+3)(y-5)=xy+39` and `(x-2)(y+3)=xy-40`, we will follow these steps:

### Step 1: Expand the first equation
We start with the first equation:
\[ (x + 3)(y - 5) = xy + 39 \]
Expanding the left side:
\[ xy - 5x + 3y - 15 = xy + 39 \]

### Step 2: Simplify the first equation
Now, we can cancel \(xy\) from both sides:
\[ -5x + 3y - 15 = 39 \]
Adding 15 to both sides gives:
\[ -5x + 3y = 54 \quad \text{(Equation 1)} \]

### Step 3: Expand the second equation
Now, let's look at the second equation:
\[ (x - 2)(y + 3) = xy - 40 \]
Expanding the left side:
\[ xy + 3x - 2y - 6 = xy - 40 \]

### Step 4: Simplify the second equation
Cancelling \(xy\) from both sides:
\[ 3x - 2y - 6 = -40 \]
Adding 6 to both sides gives:
\[ 3x - 2y = -34 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
Now we have the two equations:
1. \(-5x + 3y = 54\)
2. \(3x - 2y = -34\)

We can solve these equations using the method of substitution or elimination. Here, we will use the elimination method.

### Step 6: Multiply equations to align coefficients
To eliminate \(y\), we can multiply Equation 1 by 2 and Equation 2 by 3:
\[ 2(-5x + 3y) = 2(54) \implies -10x + 6y = 108 \quad \text{(Equation 3)} \]
\[ 3(3x - 2y) = 3(-34) \implies 9x - 6y = -102 \quad \text{(Equation 4)} \]

### Step 7: Add the equations
Now, we add Equation 3 and Equation 4:
\[ (-10x + 6y) + (9x - 6y) = 108 - 102 \]
This simplifies to:
\[ -x = 6 \]
Thus,
\[ x = -6 \]

### Step 8: Substitute \(x\) back to find \(y\)
Now, substitute \(x = -6\) back into Equation 1:
\[ -5(-6) + 3y = 54 \]
This simplifies to:
\[ 30 + 3y = 54 \]
Subtracting 30 from both sides gives:
\[ 3y = 24 \]
Thus,
\[ y = 8 \]

### Final Answer
The values of \(x\) and \(y\) are:
\[ x = -6, \quad y = 8 \]