How do you solve for u in #y= (u+1)/(u+2)#?

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Answer 1 · 2016-03-18T14:50:27

#color(blue)(u = (1 -2y)/(y-1) #

#y= (u+1) / color(green)((u+2)#

#y *color(green)( (u+2)) = (u+1) #

#y * color(green)( (u)) + y * color(green)((2)) = (u+1) #

#yu + 2y = u+1 #

Isolating #u# on the L.H.S

#ycolor(blue)(u) -color(blue)( u) = 1 -2y #

#u# is common to both terms of the L.H.S:

#color(blue)(u) (y-1) = 1 -2y #

#color(blue)(u = (1 -2y)/(y-1) #

Answer 2 · 2016-03-18T15:00:15

#u=(-2y+1)/(y-1)#

Another (more difficult) method:

Rewrite #u+1# as #u+2-1#.

#y=(u+2-1)/(u+2)#

Split up the numerator.

#y=(u+2)/(u+2)-1/(u+2)#

#y=1-1/(u+2)#

Rearrange the terms.

#y-1=-1/(u+2)#

Multiply both sides by #(u+2)#.

#(y-1)(u+2)=-1#

Distribute the #(y-1)# into #u# and #2#:

#u(y-1)+2(y-1)=-1#

#u(y-1)+2y-2=-1#

#u(y-1)=-2y+1#

After this algebraic rearrangement, divide both sides by #(y-1)# to isolate #u#.

#u=(-2y+1)/(y-1)#