How do you solve #3/4+1/2x=2+1/4x#?

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Answer 1 · 2018-05-29T19:59:12

#x=5#

We have, #frac{3}4 + x/2=2+x/4#

On further simplification,we get,

#x/2-x/4=2-frac{3}4#

Or,#x/4=5/4#

Thus, # x# is equal to 5

Answer 2 · 2018-05-29T20:03:55

5

In these cases, you always want to move terms with #x# to one side and numbers to the other. So, moving things around, we get

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

(remember to change signs when you move things to the opposite side!)

Simplifying that, we get

#2/4x-1/4x=8/4-3/4#

#1/4x=5/4#

Multiplying both sides by 4, we get

#x=5#

and voila!

Answer 3 · 2018-05-29T20:49:34

#x=5#

#3/4+1/2x=2+1/4x#

If you have an equation which has fractions, you can get rid of the denominators by multiplying by the LCM of the denominators.

(In this case it is #4#)

#(color(blue)(cancel4xx)3)/cancel4+(color(blue)(cancel4^2xx)1x)/cancel2=color(blue)(4xx)2+(color(blue)(cancel4xx)1x)/cancel4#

#3 +2x=8+x" "larr# no fractions !

#2x-x = 8-3#

#x=5#