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How do you solve for x in #a(x-2)=b(4-x)#?

1 Answer

Mar 20, 2016

We first get rid of the parentheses:

Explanation:

#->ax-2a=4b-bx#

Then all #x#'s to one side by adding #bx# to both sides:
#->ax+bx-2a=4b-cancel(bx)+cancel(bx)#

Then add #2a# to both sides to keep only #x#'s on the left:
#->ax+bx-cancel(2a)+cancel(2a)=2a+4b#

Get the #x#'s together:
#->(a+b)x=2a+4b#

Now divide by #a+b#:
#->x=(2a+4b)/(a+b)or2xx(a+2b)/(a+b)#

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