How do you solve #2/3(3x-5)-1/2(4x+3)=1/4(2x-1)-1/12#?

1 Answer
Haunt
Nov 6, 2015

#x=-9#

Explanation:

The first step in this situation is to distribute all values outside of parenthesis

#2/3*(3x)=2x#
#2/3*(-5)=-10/3#

(The subtraction here is distributed as a negative)

#-1/2*(4x)=-2x#
#-1/2*(3)=-3/2#

#1/4*(2x)=1/2x#
#1/4*(-1)=-1/4#

This leaves us with:

#2x-10/3-2x-3/2=1/2x-1/4-1/12#

Combine all like terms (like terms have the same variables attached) (There is a lot of fractional addition but I'm assuming you can calculate those yourself c: )

#-29/6=1/2x-1/3#

Add #1/3# to both sides

#-29/6+1/3=1/2x-1/3+1/3#

#-27/6=1/2x#

Multiply both sides by 2

#-27/6*2=1/2x*2#

#-27/3=x#

Simplify

#x=-9#

You can check your answer by substituting this value for x in the equation (you need to distribute to solve anyway so I won't repeat that step)
#2(-9)-10/3-2(-9)-3/2=1/2(-9)-1/4-1/12#

Simplify
#-18-10/3+18-3/2=-9/2-1/4-1/12#

#-29/6=-19/4-1/12#

#-29/6=-58/12#

#-29/6=-29/6#

Our answer is correct!

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