First note that the values #x = -8/3# and #t = -1# cause division by #0#, so are excluded values.
Multiply both sides of the equation by #(3x + 8)# to get:
#(3x+8)/(t+1) = 2x-6#
Multiply both sides of this equation by #(t+1)# to get:
#3x+8 = (2x-6)(t+1)#
#=(2x-6)t + (2x - 6)#
Subtract #(2x-6)# from both sides to get:
#(2x-6)t = (3x+8)-(2x-6)#
#=3x + 8 - 2x + 6#
#=3x - 2x + 8 + 6#
#=(3-2)x+14#
#=x+14#
Divide both ends by #(2x-6)# to get:
#t = (x+14)/(2x-6)#
#=(x-3+17)/(2(x-3))#
#=(x-3)/(2(x-3))+17/(2(x-3))#
#=1/2+17/(2(x-3))#
#=1/2+17/(2x-6)#
So:
#color(red)(t = 1/2 + 17/(2x-6))#
To find #x# in terms of #t#, first subtract #1/2# from both sides to get:
#t - 1/2 = 17/(2x-6)#
Multiply both sides by #2# to get:
#2t-1 = 17/(x-3)#
Multiply both sides by #(x-3)# and divide both sides by #(2t-1)# to get:
#x-3 = 17/(2t-1)#
Add #3# to both sides to get:
#color(red)(x = 17/(2t-1)+3)#
