# How do you solve #(2x)/(x-2)+(x^2+3x)/((x+1)(x-2))=2/((x+1)(x-2))#?

x=

mutiply throughout by (x+1)(x-2) to remove the fractions

(3x-1)(x+2)=0

3x-1=0 or x+2=0

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To solve the equation (2x)/(x-2)+(x^2+3x)/((x+1)(x-2))=2/((x+1)(x-2)), we can start by finding a common denominator for the fractions on both sides of the equation, which is (x+1)(x-2).

Multiplying both sides of the equation by (x+1)(x-2), we get:

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

Expanding and simplifying the equation, we have:

2x^2 + 2x + x^2 + 3x = 2

Combining like terms, we get:

3x^2 + 5x = 2

Rearranging the equation to bring all terms to one side, we have:

3x^2 + 5x - 2 = 0

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use factoring.

Factoring the quadratic equation, we have:

(3x - 1)(x + 2) = 0

Setting each factor equal to zero, we get:

3x - 1 = 0 or x + 2 = 0

Solving these equations, we find:

3x = 1 or x = -2

Dividing both sides of the first equation by 3, we have:

x = 1/3

Therefore, the solutions to the given equation are x = 1/3 and x = -2.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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