If `(5x^(2)+8x-2)/((3x-2)(x^(2)-x+3)) =(A)/(3x-2)+(Bx+C)/(x^(2)-x+3)`, then `A+B+C=`………..

Answer

To solve the equation

\[ \frac{5x^2 + 8x - 2}{(3x - 2)(x^2 - x + 3)} = \frac{A}{3x - 2} + \frac{Bx + C}{x^2 - x + 3}, \]

we will first combine the right-hand side into a single fraction.

### Step 1: Combine the Right-Hand Side
The common denominator for the right-hand side is \((3x - 2)(x^2 - x + 3)\). Thus, we can write:

\[ \frac{A}{3x - 2} + \frac{Bx + C}{x^2 - x + 3} = \frac{A(x^2 - x + 3) + (Bx + C)(3x - 2)}{(3x - 2)(x^2 - x + 3)}. \]

### Step 2: Expand the Numerator
Now we will expand the numerator:

\[ A(x^2 - x + 3) + (Bx + C)(3x - 2). \]

Expanding \(A(x^2 - x + 3)\):

\[ Ax^2 - Ax + 3A. \]

Expanding \((Bx + C)(3x - 2)\):

\[ 3Bx^2 - 2Bx + 3Cx - 2C = 3Bx^2 + (3C - 2B)x - 2C. \]

Combining these, we have:

\[ (A + 3B)x^2 + (-A + 3C - 2B)x + (3A - 2C). \]

### Step 3: Set the Numerator Equal to the Left-Hand Side
Now we equate this to the left-hand side:

\[ 5x^2 + 8x - 2 = (A + 3B)x^2 + (-A + 3C - 2B)x + (3A - 2C). \]

### Step 4: Compare Coefficients
Now we can compare the coefficients of \(x^2\), \(x\), and the constant term:

1. For \(x^2\):
\[ A + 3B = 5 \quad \text{(1)} \]

2. For \(x\):
\[ -A + 3C - 2B = 8 \quad \text{(2)} \]

3. For the constant term:
\[ 3A - 2C = -2 \quad \text{(3)} \]

### Step 5: Solve the System of Equations
From equation (1), we can express \(A\) in terms of \(B\):

\[ A = 5 - 3B \quad \text{(4)} \]

Substituting (4) into equation (2):

\[ -(5 - 3B) + 3C - 2B = 8, \]
which simplifies to:

\[ -5 + 3B + 3C - 2B = 8 \implies B + 3C = 13 \quad \text{(5)}. \]

Now substituting (4) into equation (3):

\[ 3(5 - 3B) - 2C = -2, \]
which simplifies to:

\[ 15 - 9B - 2C = -2 \implies -9B - 2C = -17 \implies 9B + 2C = 17 \quad \text{(6)}. \]

### Step 6: Solve Equations (5) and (6)
Now we have a system of two equations:

1. \(B + 3C = 13\) (5)
2. \(9B + 2C = 17\) (6)

From equation (5), we can express \(C\) in terms of \(B\):

\[ C = \frac{13 - B}{3} \quad \text{(7)}. \]

Substituting (7) into equation (6):

\[ 9B + 2\left(\frac{13 - B}{3}\right) = 17. \]

Multiplying through by 3 to eliminate the fraction:

\[ 27B + 2(13 - B) = 51 \implies 27B + 26 - 2B = 51 \implies 25B = 25 \implies B = 1. \]

### Step 7: Find \(A\) and \(C\)
Substituting \(B = 1\) back into equation (4):

\[ A = 5 - 3(1) = 2. \]

Substituting \(B = 1\) into equation (7):

\[ C = \frac{13 - 1}{3} = \frac{12}{3} = 4. \]

### Step 8: Calculate \(A + B + C\)
Now we can find \(A + B + C\):

\[ A + B + C = 2 + 1 + 4 = 7. \]

Thus, the final answer is:

\[ \boxed{7}. \]