Mary can clean 1/4 of the classroom in one hour

# 1/2 / 2/1 # can be simplified by multiplying both the denominator and numerator by 1/2 this results in

# (1/2 xx 1/2) /( 2/1 xx 1/2) # The denominator becomes 1 leaving

# n t1/2 xx 1/2 = 1//4#

Suzzie can clean 1/6 of the classroom in one hour.

g

# 1/3 / 1/2# can be simplified by multiplying both the denominator and numerator by 1/2 this results in

# ( 1/3 xx 1/2) / (2/1 xx 1/2)# The denominator becomes 1 leaving

# 1/3 xx 1/2 = 1/6 #

The amount of the class work that can be cleaned in one hour by both working together can be expressed in this equation.

# 1/4 + 1/6 = 1/T # where T is the total amount of time needed to clean the classroom

to solve for T multiply by the LCM of all the denominators

The LCM is 12 T so

12 T( 1/4 + 1/6 = 1/T) This gives

# 3T + 2T = 12 # Adding common terms

# 5T = 12# Divide both sides by 5

# 5T/5 = 12/5# this results in

T = 2 2/5