Solve using Cramer's rule:`(4)/(x+5)+(3)/(y+7)=-1&(6)/(x+5)-(6)/(y+7)=-5`

Answer

To solve the given system of equations using Cramer's Rule, we start with the equations:

1. \(\frac{4}{x+5} + \frac{3}{y+7} = -1\)
2. \(\frac{6}{x+5} - \frac{6}{y+7} = -5\)

### Step 1: Substitution
We substitute:
- Let \(u = \frac{1}{x+5}\)
- Let \(v = \frac{1}{y+7}\)

Now, we can rewrite the equations in terms of \(u\) and \(v\):

1. \(4u + 3v = -1\) (Equation 1)
2. \(6u - 6v = -5\) (Equation 2)

### Step 2: Write the system in matrix form
The system can be represented in matrix form as:

\[ \begin{bmatrix} 4 & 3 \\ 6 & -6 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} -1 \\ -5 \end{bmatrix} \]

### Step 3: Calculate the determinant \(D\)
To use Cramer's Rule, we first calculate the determinant \(D\) of the coefficient matrix:

\[ D = \begin{vmatrix} 4 & 3 \\ 6 & -6 \end{vmatrix} = (4)(-6) - (3)(6) = -24 - 18 = -42 \]

### Step 4: Calculate \(D_x\)
Next, we calculate \(D_x\) by replacing the first column of the coefficient matrix with the constants:

\[ D_x = \begin{vmatrix} -1 & 3 \\ -5 & -6 \end{vmatrix} = (-1)(-6) - (3)(-5) = 6 + 15 = 21 \]

### Step 5: Calculate \(D_y\)
Now, we calculate \(D_y\) by replacing the second column of the coefficient matrix with the constants:

\[ D_y = \begin{vmatrix} 4 & -1 \\ 6 & -5 \end{vmatrix} = (4)(-5) - (-1)(6) = -20 + 6 = -14 \]

### Step 6: Calculate \(u\) and \(v\)
Using Cramer's Rule, we find \(u\) and \(v\):

\[ u = \frac{D_x}{D} = \frac{21}{-42} = -\frac{1}{2} \]

\[ v = \frac{D_y}{D} = \frac{-14}{-42} = \frac{1}{3} \]

### Step 7: Back substitute to find \(x\) and \(y\)
Recall that:
- \(u = \frac{1}{x+5}\) implies \(x + 5 = \frac{1}{u}\)
- \(v = \frac{1}{y+7}\) implies \(y + 7 = \frac{1}{v}\)

Calculating \(x\):
\[ x + 5 = \frac{1}{-\frac{1}{2}} = -2 \implies x = -2 - 5 = -7 \]

Calculating \(y\):
\[ y + 7 = \frac{1}{\frac{1}{3}} = 3 \implies y = 3 - 7 = -4 \]

### Final Solution
Thus, the solution to the system is:
\[ x = -7, \quad y = -4 \]