Answer :
Certainly! Let’s go through the process of simplifying the given equation:
We start with the original equation:
[tex]\[ \frac{x+2}{x+1} + \frac{3}{x} = 3 \][/tex]
First, clear the denominators by finding a common denominator, which in this case is [tex]\( x(x+1) \)[/tex]. We multiply every term by the common denominator to eliminate the fractions:
[tex]\[ x(x+1) \left( \frac{x+2}{x+1} \right) + x(x+1) \left( \frac{3}{x} \right) = 3 \cdot x(x+1) \][/tex]
Simplify each term:
[tex]\[ x(x+2) + 3(x+1) = 3x(x+1) \][/tex]
Expand each term:
[tex]\[ x^2 + 2x + 3x + 3 = 3x^2 + 3x \][/tex]
Combine like terms on the left side:
[tex]\[ x^2 + 5x + 3 = 3x^2 + 3x \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 + 5x + 3 - 3x^2 - 3x = 0 \][/tex]
Combine like terms:
[tex]\[ -2x^2 + 2x + 3 = 0 \][/tex]
Or equivalently:
[tex]\[ 2x^2 - 2x - 3 = 0 \][/tex]
Thus, we have shown that the equation [tex]\(\frac{x+2}{x+1} + \frac{3}{x} = 3\)[/tex] simplifies to:
[tex]\[ 2x^2 - 2x - 3 = 0 \][/tex]
We start with the original equation:
[tex]\[ \frac{x+2}{x+1} + \frac{3}{x} = 3 \][/tex]
First, clear the denominators by finding a common denominator, which in this case is [tex]\( x(x+1) \)[/tex]. We multiply every term by the common denominator to eliminate the fractions:
[tex]\[ x(x+1) \left( \frac{x+2}{x+1} \right) + x(x+1) \left( \frac{3}{x} \right) = 3 \cdot x(x+1) \][/tex]
Simplify each term:
[tex]\[ x(x+2) + 3(x+1) = 3x(x+1) \][/tex]
Expand each term:
[tex]\[ x^2 + 2x + 3x + 3 = 3x^2 + 3x \][/tex]
Combine like terms on the left side:
[tex]\[ x^2 + 5x + 3 = 3x^2 + 3x \][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[ x^2 + 5x + 3 - 3x^2 - 3x = 0 \][/tex]
Combine like terms:
[tex]\[ -2x^2 + 2x + 3 = 0 \][/tex]
Or equivalently:
[tex]\[ 2x^2 - 2x - 3 = 0 \][/tex]
Thus, we have shown that the equation [tex]\(\frac{x+2}{x+1} + \frac{3}{x} = 3\)[/tex] simplifies to:
[tex]\[ 2x^2 - 2x - 3 = 0 \][/tex]