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 tangent lines have to touch the curve at the points with the following coordinates:
`x_0 = 1` and `x_0 = 3`. At these points, the y-coordinates are as follows:
`y_0 = y(x_0) = (1- 1)/(1 + 1) = 0` and `y_0 - y(x_0) = (-3 - 1)/(-3 + 1) = 2`.
This means the first tangent line passes through the point (1, 0) and the second tangent line passes through the point (-3, 2).
The equation of a line with the given slope (`m = 1/2)` passing through the given point can be found using the point-slope form:
`y -y_0 = m(x - x_0)`.
For the first line, this gives us the following:
`y - 0 = 1/2(x - 1)` , or `y = 1/2x - 1/2`.
For the second line, this gives us the following:
`y-2 = 1/2(x + 3)` , or `y = 1/2x + 7/2`.
These equations can be converted to standard form by multiplying both sides by 2 and bringing the terms containing x and y to the same side:
`x-2y = 1 ` and `x - 2y = -7` .
The equations of the lines tangent to the given curve and parallel to x - 2y = 3 are x - 2y = 1 and x - 2y = 7.