How do you solve #4( 2- 5x ) = 6- 3( 1- 3x )#?

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Answer 1 · 2018-07-28T03:40:00

#x=5/29#

#4(2-5x)=6-3(1-3x)#

Expand the brackets
#8-20x=6-(3-9x)#

#8-20x=6-3+9x#

Simplify
#8-20x=3+9x#

Move the #x# onto one side
#29x=5#

Divide both sides by #29#
#x=5/29#

Answer 2 · 2018-07-28T04:30:04

#x=5/29#

We can distribute the #4# on the left side, and the #-3# on the right. Doing this, we now have

#8-20x=6-3+9x#

We can combine like terms to get

#8-20x=3+9x#

Let's get our constants on one side by subtracting #8# from both sides:

#-20x=-5+9x#

Next, subtract #9x# from both sides to get

#-29x=-5#

Lastly, we can divide both sides by #-29# to get

#x=5/29#

Hope this helps!