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To solve the equation
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{x^2 - 6x + 8} \][/tex]
we need to follow several steps. Let's go through the solution in a detailed, step-by-step manner.
### Step 1: Identify the equation and constraints
Given:
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{x^2 - 6x + 8} \][/tex]
Firstly, we factorize the quadratic expression in the denominator on the right-hand side:
[tex]\[ x^2 - 6x + 8 = (x-2)(x-4) \][/tex]
So the equation becomes:
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{(x-2)(x-4)} \][/tex]
### Step 2: Determine the common denominator
For the whole equation, the common denominator for the fractions is [tex]\((x-2)(x-4)\)[/tex]:
[tex]\[ \frac{2x(x-4) - (x+3)(x-2)}{(x-2)(x-4)} = \frac{-4x}{(x-2)(x-4)} \][/tex]
In this form, the denominators are the same, so we can equate the numerators directly:
[tex]\[ 2x(x-4) - (x+3)(x-2) = -4x \][/tex]
### Step 3: Simplify the expression
Expanding both sides:
[tex]\[ 2x(x-4) = 2x^2 - 8x \][/tex]
[tex]\[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 \][/tex]
So the equation becomes:
[tex]\[ 2x^2 - 8x - (x^2 + x - 6) = -4x \][/tex]
Simplify the left-hand side:
[tex]\[ 2x^2 - 8x - x^2 - x + 6 = -4x \][/tex]
Combine like terms:
[tex]\[ x^2 - 9x + 6 = -4x \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 - 9x + 6 + 4x = 0 \][/tex]
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To find the values of [tex]\(x\)[/tex], solve the quadratic equation:
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
Factorize:
[tex]\[ (x - 2)(x - 3) = 0 \][/tex]
Setting each factor to zero gives us the potential solutions:
[tex]\[ x - 2 = 0 \quad \rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 3 = 0 \quad \rightarrow \quad x = 3 \][/tex]
### Step 5: Verify the solutions
Check if the solutions [tex]\(x = 2\)[/tex] and [tex]\(x = 3\)[/tex] are valid by substituting them back into the original equation. We note that the original equation's denominators cannot be zero.
For [tex]\(x = 2\)[/tex]:
[tex]\[ \frac{2 \cdot 2}{2-2} - \frac{2 + 3}{2-4}\][/tex]
\]
Here, the first fraction becomes undefined because the denominator is zero. Therefore, [tex]\(x = 2\)[/tex] is not a valid solution.
For [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{2 \cdot 3}{3-2} - \frac{3+3}{3-4} = \frac{-4 \cdot 3}{3^2 - 6 \cdot 3 + 8} \][/tex]
[tex]\[ \frac{6}{1} - \frac{6}{-1} = \frac{-12}{1} \][/tex]
Simplifying:
[tex]\[ 6 + 6 = -12 \quad \text{not valid} \][/tex]
So, [tex]\(x = 3\)[/tex] is a valid solution.
### Conclusion:
Thus, the valid solution to the equation is:
[tex]\[ x = 3 \][/tex]
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{x^2 - 6x + 8} \][/tex]
we need to follow several steps. Let's go through the solution in a detailed, step-by-step manner.
### Step 1: Identify the equation and constraints
Given:
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{x^2 - 6x + 8} \][/tex]
Firstly, we factorize the quadratic expression in the denominator on the right-hand side:
[tex]\[ x^2 - 6x + 8 = (x-2)(x-4) \][/tex]
So the equation becomes:
[tex]\[ \frac{2x}{x-2} - \frac{x+3}{x-4} = \frac{-4x}{(x-2)(x-4)} \][/tex]
### Step 2: Determine the common denominator
For the whole equation, the common denominator for the fractions is [tex]\((x-2)(x-4)\)[/tex]:
[tex]\[ \frac{2x(x-4) - (x+3)(x-2)}{(x-2)(x-4)} = \frac{-4x}{(x-2)(x-4)} \][/tex]
In this form, the denominators are the same, so we can equate the numerators directly:
[tex]\[ 2x(x-4) - (x+3)(x-2) = -4x \][/tex]
### Step 3: Simplify the expression
Expanding both sides:
[tex]\[ 2x(x-4) = 2x^2 - 8x \][/tex]
[tex]\[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 \][/tex]
So the equation becomes:
[tex]\[ 2x^2 - 8x - (x^2 + x - 6) = -4x \][/tex]
Simplify the left-hand side:
[tex]\[ 2x^2 - 8x - x^2 - x + 6 = -4x \][/tex]
Combine like terms:
[tex]\[ x^2 - 9x + 6 = -4x \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 - 9x + 6 + 4x = 0 \][/tex]
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To find the values of [tex]\(x\)[/tex], solve the quadratic equation:
[tex]\[ x^2 - 5x + 6 = 0 \][/tex]
Factorize:
[tex]\[ (x - 2)(x - 3) = 0 \][/tex]
Setting each factor to zero gives us the potential solutions:
[tex]\[ x - 2 = 0 \quad \rightarrow \quad x = 2 \][/tex]
[tex]\[ x - 3 = 0 \quad \rightarrow \quad x = 3 \][/tex]
### Step 5: Verify the solutions
Check if the solutions [tex]\(x = 2\)[/tex] and [tex]\(x = 3\)[/tex] are valid by substituting them back into the original equation. We note that the original equation's denominators cannot be zero.
For [tex]\(x = 2\)[/tex]:
[tex]\[ \frac{2 \cdot 2}{2-2} - \frac{2 + 3}{2-4}\][/tex]
\]
Here, the first fraction becomes undefined because the denominator is zero. Therefore, [tex]\(x = 2\)[/tex] is not a valid solution.
For [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{2 \cdot 3}{3-2} - \frac{3+3}{3-4} = \frac{-4 \cdot 3}{3^2 - 6 \cdot 3 + 8} \][/tex]
[tex]\[ \frac{6}{1} - \frac{6}{-1} = \frac{-12}{1} \][/tex]
Simplifying:
[tex]\[ 6 + 6 = -12 \quad \text{not valid} \][/tex]
So, [tex]\(x = 3\)[/tex] is a valid solution.
### Conclusion:
Thus, the valid solution to the equation is:
[tex]\[ x = 3 \][/tex]
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