How do you solve #3x+4=5x+2 #?

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Answer 1 · 2015-11-07T13:20:25

Subtract #3x+2# from both sides, then divide by #2# to find #x=1#

Isolate #x# on one side of the equation, by subtracting #3x+2# from both sides to get:

#2 = 2x#

(Since #3x+4-(3x+2) = 3x-3x+4-2=2# and #5x+2-(3x+2) = 5x-3x+2-2 = 2x#)

Then divide both sides by #2# to get #1 = x#. That is #x = 1#.

Alternatively, just notice that #3+4 = 7 = 5 + 2#, so #x=1# will work.

Answer 2 · 2015-11-07T13:25:01

#3x + 4 = 5x + 2#

Subtract 5x from both sides so you remove the variable on the right:
#3x - 5x + 4 = 5x - 5x + 2#

Add variables together:
#-2x + 4 = 2#

Subtract 4 from both sides so you remove the constant on the left:
#-2x + 4 - 4= 2 -4#

Add constants together:
#-2x = -2#

To only get one #x#, divide both sides by #-2#:
#(-2x)/-2 = (-2)/-2 to (cancel(-2)x)/(cancel(-2)) = (cancel(-2))/(cancel(-2)) to x = 1#