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 $c=7$

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