How do you solve #3/2y-y=4+.5y#?
See a solution process below:
We can rewrite the problem as:
Next, we can combine like terms on the left side of the equation:
Now, we can subtract
Or, the solution is the null or empty set:
We have two
Something already looks suspicious...let's subtract
This is obviously not true, which means no value of
Hope this helps!