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HallsofIvy Excellent! Okay, the derivative of y= 2/(3x-2)= 2(3x-2)^(-1) is y'= -2(3x-2)^(-2)(3)= -6(3x-2)^(-2). The slope of the tangent line at each point is equal to that. In order that the tangent line be parallel to y= (-3/2)x- 1, which has slope -3/2, we must have -6(3x-2)^(-2)= -3/2. That's what you have. Good! To solve that for x, first get rid of the fractions by multiplying on both sides by 2(3x-2)^2 and dividing by -3. You should get 4= (3x-2)^2. Taking the square root of both sides, $\displaystyle 3x-2= \pm 2$. The positive sign gives 3x= 4 so x= 4/3. The negative sign gives x= 0. When x= 4/3, y= 2/(3(4/3)- 2)= 2/2= 1 and when x= 0, y= 2/(3(0)-2)= -1. Thus the tangent line to the curve will be parallel to y= (-3/2)x- 1 at (4/3, 1) and (0, -1). You should be able to determine the "b" in y= (-3/2)x+ b for both of those points.