Find equations of the tangent lines to the curve y=(x-1)/(x+1) that are parallel to the line x − 2y = 3.

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The tangent line to a curve at a given point has the slope which equals the value of the derivative of the function describing the curve at this point.

Since the tangent line has to be parallel to the line with the equation `x - 2y = 3`, it has to have the same slope. The slope of the line given by the equation in standard form (`ax + by = c` ) is determined as `m = -a/b`. So, the slope of this line is `m = -1/(-2) = 1/2`.

Now we need to determine at which point `x_0` the tangent line to the curve.

`y = (x-1)/(x + 1)` has the slope `1/2`.

The derivative of this function is as follows:

`y = (1*(x + 1) - (x-1)*1)/(x + 1)^2 = 2/(x + 1)^2`

This equals `1/2` when the following is true:

`2/(x + 1)^2 = 1/2`

Cross-multiplying yields the following:

`(x + 1)^2 = 4`

Taking the square root of both sides results in the following:

`x + 1 = 2` and `x + 1 = -2`

x = 1 and x = -3

So the...

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