If #5/(a-b) = 2# and if #a=4#, what is the value of #b#?

1 Answer
Feb 25, 2017

To solve this we must substitute #color(red)(4)# for #color(red)(a)# in the expression and then solve for #b#:

#5/(color(red)(a) - b) = 2# becomes:

#5/(color(red)(4) - b) = 2#

#5/(4 - b) xx color(red)((4 - b)) = 2color(red)((4 - b))#

#5/color(red)(cancel(color(black)(4 - b))) xx cancel(color(red)((4 - b))) = (2 xx 4) - (2 xx b)#

#5 = 8 - 2b#

#5 - color(red)(8) = -color(red)(8) + 8 - 2b#

#-3 = 0 - 2b#

#-3 = -2b#

#(-3)/color(red)(-2) = (-2b)/color(red)(-2)#

#3/2 = (color(red)(cancel(color(black)(-2)))b)/cancel(color(red)(-2))#

#3/2 = b#

#b = 3/2#