How do you solve # (y-1)/(y-2)=-y/(y+1)#?

Answer 1

Refer to explanation

We multiply both sides with #(y-2)*(y+1)# hence we get

#(y-1)/(y-2)=-y/(y+1)=>(y+1)*(y-1)=-(y-2)*y=>y^2-1=-y^2+2y=> 2y^2+2y-1=0=>y_1=-1/2-sqrt3/2 or y_2=-1/2+sqrt3/2#

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Answer 2

To solve the equation (y-1)/(y-2)=-y/(y+1), we can start by cross-multiplying to eliminate the fractions. This gives us (y-1)(y+1) = -y(y-2). Expanding both sides of the equation, we get y^2 - 1 = -y^2 + 2y. Rearranging the terms, we have 2y^2 - 2y - 1 = 0. To solve this quadratic equation, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac))/(2a), where a = 2, b = -2, and c = -1. Plugging in these values, we get y = (2 ± √((-2)^2 - 4(2)(-1)))/(2(2)). Simplifying further, we have y = (2 ± √(4 + 8))/4, which becomes y = (2 ± √12)/4. Simplifying the square root, we have y = (2 ± 2√3)/4. Factoring out a 2 from the numerator, we get y = (1 ± √3)/2. Therefore, the solutions to the equation are y = (1 + √3)/2 and y = (1 - √3)/2.

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Answer from HIX Tutor

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.

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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