How do you solve #d - 2 1/12 = 3 1/12#?

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Answer 1 · 2018-03-15T17:16:22

See a solution process below:

Add #color(red)(2 1/12)# to each side of the equation to solve for #d# while keeping the equation balanced:

#d - 2 1/12 + color(red)(2 1/12) = 3 1/12 + color(red)(2 1/12)#

#d - 0 = 3 + 1/12 + color(red)(2) + color(red)(1/12)#

#d = 3 + color(red)(2) + 1/12 + color(red)(1/12)#

#d = 5 + 2/12#

#d = 5 + 1/6#

#d = 5 1/6#