How do you solve #1/3(c+1)=1/6(3c-5)#?

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Answer 1 · 2017-05-24T11:12:20

#c=7#

Step 1. Multiply both sides of this equation by 6, to get rid of the fractions

#6xx1/3(c+1)=6xx1/6(3c-5)#

#2(c+1)=3c-5#

Step 2. Multiply 2 through on left hand side

#2c+2=3c-5#

Step 3. Add 5 to both sides

#2c+7=3c#

Step 4. Subtract #2c# from both sides

#7=c# or #c=7#