If a(x+b)=bx-c, what is x?
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"What is the oxidation state of copper in #"CuSO"_4#?"
expand the bracket
#ax+ab=bx-c#
make #x# the subject
#ax+ab+c=bx#
#ab+c=bx-ax#
#ab+c=x(b-a)#
#(ab+c)/(b-a)=x#
We have,
#color(white)(xxx)a(x + b) = bx - c#
#rArr ax + ab = bx -c# [Distributive Property]
#rArr ax +ab - bx = cancel(bx) cancel(- bx) - c# [Subtract #bx# from both sides]
#rArr ax - bx + cancel(ab) cancel(-ab) = -c - ab#
#rArr x(a- b) = (-c - ab)# [Group like terms]
#rArr x = (c + ab)/-(a - b)# [Divide both sides by #(a - b)#] [Only if #a != b#]
#rArr x = (c + ab)/(b - a)#
If #a = b#, and #ab + c != 0#, Then, The solution doesn't exist.
If #c lt 0#, #a = b# and #ab + c = 0#, Then #x in RR#
And If, #c gt 0#, #a = b# and #ab + c = 0#, Then Again, Solution is a Complex Number, so No Real Solutions.
And Gotcha.
Hope this helps.
If #a=b# and #ab+c=0# then we get all real numbers as solution.
if #a=b# and #ab+cne0# then no solution
if #ane b# then #x=(ab+c)/(b-a)#
Write your equation in the form
#x(b-a)=ab+c#
and distinguish the cases above.
