How do you solve #3/2y-y=4+.5y#?

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Answer 1 · 2018-07-16T13:30:58

See a solution process below:

First, convert #3/2# to:

#3/2 = (2 + 1)/2 = 2/2 + 1/2 = 1 + 1/2 = 1 + 0.5 = 1.5#

We can rewrite the problem as:

#1.5y - y = 4 + 0.5y#

Next, we can combine like terms on the left side of the equation:

#1.5y - 1y = 4 + 0.5y#

#(1.5 - 1)y = 4 + 0.5y#

#0.5y = 4 + 0.5y#

Now, we can subtract #color(red)(0.5y)# from each side of the equation to show there is no solution:

#0.5y - color(red)(0.5y) = 4 + 0.5y - color(red)(0.5y)#

#0 = 4 + 0#

#0 != 4#

Because #0# is obviously not equal to #4# there is no solution for this problem

Or, the solution is the null or empty set: #{O/}#

Answer 2 · 2018-07-16T14:50:23

No solutions

#3/2# is the same thing as #1.5#. With this in mind, we can rewrite our equation as

#1.5y-y=4+0.5y#

We have two #y# terms on the left, so we can simplify them to get

#0.5y=4+0.5y#

Something already looks suspicious...let's subtract #0.5y# from both sides to get

#0=4#

This is obviously not true, which means no value of #y# satisfies this equation.

Hope this helps!