# 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 curve y=(x-1)/(x+1). At any point on the curve, the slope of the tangent to the curve is equal to the value of y' at that point.

As equations of the tangents have to be determined that are parallel to the line x - 2y = 3, first determine the slope of x - 2y = 3

=> y = (1/2)*x - 3/2

The slope of the line is (1/2).

y' = ((x + 1) - (x - 1))/(x + 1)^2 = 2/(x + 1)^2 = 1/2

=> (x + 1)^2 = 4

=> (x + 1) = 2 and x + 1 = -2

=> x = 1 and x = -3

At x = 1, y = (x - 1)/(x + 1) = 0

The equation of the tangent is y/(x - 1) = (1/2)

=> 2y = x - 1

=> x - 2y - 1 = 0

At x = -3, y = (x - 1)/(x + 1) = -4/-2 = 2

The equation of the tangent is (y - 2)/(x + 3) = (1/2)

=> 2y - 4 = x + 3

=> x - 2y + 7 = 0

**The equation of the tangents to y = (x - 1)/(x + 1) that are parallel to x - 2y = 3 are x - 2y - 1 = 0 and x - 2y + 7 = 0**