How do you solve #1/4x-5/2=-2#?

1 Answer
May 19, 2015

Take #4# as common denominator (adapting the numerators to the new one) so you get:
#(x-5*2)/(4)=(-2*4)/(4)#
now you can cancel the denominators and get:
#(x-5*2)/cancel(4)=(-2*4)/cancel(4)#
#x-10=-8#
#x=10-8#
#x=2#

You can check this result substituting it back into your original equation. :-)