How do you solve #5 1/2x+2/3x=37#?

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Answer 1 · 2016-11-14T06:18:22

#x=6#

Treat it as a normal equation - except that you have to add fractions as coefficients rather than integers.

Improper fractions will probably be best for adding because later you can cross multiply.

#5 1/2x +2/3x = 37#

#11/2x+2/3x = 37#

#(33x+4x)/6 = 37#

#(37x)/6=37/1" "larr# cross-multiply

#37x= 6xx37#

#x= (6xx37)/37#

#x = 6#