How do you solve #2x + \frac{3}{4} = 3x - \frac{3}{4}#?

2 Answers
Apr 17, 2018

#x#=#3//2#

Explanation:

We have, #2x+frac{3}4=3x-frac{3}4#
On further simplification, #2xx frac{3}4=3x-2x#
Or, #frac{3}2=x#

Apr 17, 2018

#x=3/2" "# Using First principles

A lot of detail given so that you can see where everything comes from. You would normally do this in just a few lines.

Explanation:

Given #2x+3/4=3x-3/4#

The #3x# on the right of = is greater than the #2x# on the left. So to keep the #x# terms positive we move the #2x# on the left to the right. We do this by changing the #2x# into 0.

Subtract #color(red)(2x)# from both sides

#color(green)(2x+3/4=3x-3/4color(white)("dddd")->color(white)("dddd)ubrace(2x color(red)(-2x))+3/4=ubrace(3x color(red)(-2x))-3/4 )#
#color(green)(color(white)("ddddddddddddddddddddddddddd")darrcolor(white)("dddddddddd")darr)#
#color(green)(color(white)("ddddddddddddddddddd")->color(white)("d.dddd")0color(white)("d.d")+3/4 =color(white)("dd")xcolor(white)("d.d")-3/4)#

Now we need to get the #x# on its own so we need to turn the #-3/4# onto 0.

Add #color(red)(3/4)# to both sides

#color(green)(3/4=x-3/4color(white)("ddddddddd")->color(white)("dddd")ubrace(3/4color(red)(+3/4))=xcolor(white)("d")ubrace(-3/4color(red)(+3/4)))#

#color(green)(color(white)("dddddddddddddddddd")->color(white)("dddddd") 3/2color(white)("dd")=xcolor(white)("d")+0)#

#color(green)(color(white)("dddddddddddddddddd")->color(white)("ddddddd.d")x=3/2)#