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

Question

2718 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T23:23:21

See the entire solution process below:

First, subtract #color(red)(3/2)# and #color(blue)(1/4a)# from each side of the equation to isolate the #a# term while keeping the equation balanced:

#3/2 + 3/4a - color(red)(3/2) - color(blue)(1/4a) = 1/4a - 1/2 - color(red)(3/2) - color(blue)(1/4a)#

#3/2 - color(red)(3/2) + 3/4a - color(blue)(1/4a) = 1/4a - color(blue)(1/4a) - 1/2 - color(red)(3/2)#

#0 + (3/4 - color(blue)(1/4))a = 0 - 1/2 - color(red)(3/2)#

#2/4a = -4/2#

Now, multiply each side of the equation by #color(red)(4)/color(blue)(2)# to solve for #a# while keeping the equation balanced:

#color(red)(4)/color(blue)(2) xx 2/4a = color(red)(4)/color(blue)(2) xx -4/2#

#8/8a = -16/4#

#1a = -4#

#a = -4#