J jaydstein New member Joined Apr 9, 2008 Messages 3 Apr 9, 2008 #1 The equation that I am given is: y = (2x+1)/(x-3) I need to rearrange the equation in terms of x. I'm stumped. Any suggestions?
The equation that I am given is: y = (2x+1)/(x-3) I need to rearrange the equation in terms of x. I'm stumped. Any suggestions?
M masters Full Member Joined Mar 30, 2007 Messages 378 Apr 9, 2008 #2 Multiply both sides of the equation by (x-3). Distribute the y and isolate the x terms.
J jaydstein New member Joined Apr 9, 2008 Messages 3 Apr 9, 2008 #3 May as well bring this up this question while I'm here: solution 1: y = (2x+1)/(x-3) y - (7/(x-3)) = (2x+1)/(x-3) - (7/(x-3)) = (2(x-3))/(x-3) = 2 y = 2 + (7/(x-3)) (y-2)/7 = (x-3)^-1 7/(y-2) = x-3 7/((y-2) + 3 = x ie x = 7/(y-2) + 3 solution 2: y = (2x+1)/(x-3) y(x-3) = 2x+1 yx-3y = 2x+1 yx -2x = 3y+1 x(y-2) = 3y+1 x = (3y+1)/(y-2) Are the two different solutions equivalent, or am I screwing up again? :mrgreen:
May as well bring this up this question while I'm here: solution 1: y = (2x+1)/(x-3) y - (7/(x-3)) = (2x+1)/(x-3) - (7/(x-3)) = (2(x-3))/(x-3) = 2 y = 2 + (7/(x-3)) (y-2)/7 = (x-3)^-1 7/(y-2) = x-3 7/((y-2) + 3 = x ie x = 7/(y-2) + 3 solution 2: y = (2x+1)/(x-3) y(x-3) = 2x+1 yx-3y = 2x+1 yx -2x = 3y+1 x(y-2) = 3y+1 x = (3y+1)/(y-2) Are the two different solutions equivalent, or am I screwing up again? :mrgreen:
M masters Full Member Joined Mar 30, 2007 Messages 378 Apr 9, 2008 #4 I believe I'd go with solution #2.