How do you solve for #x# in #2x+b=w#?

Question

4446 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-02T13:54:06

#x=1/2(w-b)#

#color(red)("Using first principles")#

#color(red)("Step 1")#
Have only the terms with #x# in them on the left of the =

Subtract #color(blue)(b)# from both sides

#color(brown)(2x+b" "=" "w" "->" "2x+bcolor(blue)(-b)" "=" "w color(blue)(-b))#

#2x+0=w-b#

#2x=w-b#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(red)("Step 2")#
Have only #x# on the left hand side of the =

Divide both sides by #color(blue)(2)#

#color(brown)(2/(color(blue)(2))xx x=color(blue)(1/2)(w-b)#

#1xx x=1/2(w-b)#

#x=1/2(w-b)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(red)("Comment")#

#1/2(w-b)# is the same as #w/2-b/2#