How do you solve for x: #5(x-3)+2(3-x)=12#?

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Answer 1 · 2016-06-22T12:27:42

You first work away the parentheses

#->(5*x-5*3)+(2*3-2*x)=12#
(these brackets only used for clarification)

#->5x-15+6-2x=12->3x-9=12#

Add #9# to both sides:
#->3x-cancel9+cancel9=12+9#
#->3x=21->x=7#

Answer 2 · 2016-06-22T14:27:27

Just a different approach

#x=7#

Given:#" "5(x-3)+2(3-x)=12#

If we can change #2(3-x)" such that "(3-x)" became "(x-3)# then we have a common factor.

Note that
#-2(x-3) = -2x+6" "->" "+2(3-x)=-2x+6#

Write the given equation as:

#5(x-3)-2(x-3)=12#

Factor out #(x-3)#

#(x-3)(5-2)=12#

#3(x-3)=12#

Divide both sides by 3

#x-3=4#

Add 3 to both sides

#x=7#