Question

2x/(3x^2-x+2) - 7x/(3x^2+5x+2)=1

Answer 1

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Answer 2

To solve the equation:

2x/(3x^2-x+2) - 7x/(3x^2+5x+2) = 1

First, we need to find the common denominator for the two fractions. The denominators of the two fractions are (3x^2-x+2) and (3x^2+5x+2), respectively. To find the common denominator, we need to factor the denominators and then take their product:

3x^2-x+2 = (3x-2)(x+1) 3x^2+5x+2 = (3x+2)(x+1)

Therefore, the common denominator is (3x-2)(x+1)(3x+2).

Next, we need to convert the two fractions to have this common denominator:

2x/(3x^2-x+2) = (2x)/[(3x-2)(x+1)] -7x/(3x^2+5x+2) = (-7x)/[(3x+2)(x+1)]

We can now substitute these expressions into the original equation, using the common denominator:

(2x)/[(3x-2)(x+1)] - (-7x)/[(3x+2)(x+1)] = 1

Simplifying the left side by combining the two fractions, we get:

[2x(3x+2) + 7x(3x-2)]/[(3x-2)(x+1)(3x+2)] = 1

Expanding the numerator, we get:

[6x^2+4x+21x^2-14x]/[(3x-2)(x+1)(3x+2)] = 1

Combining like terms, we get: