How do you solve $\frac{1}{3} (c + 1) = \frac{1}{6} (3 c - 5)$ $1/3(c+1)=1/6(3c-5)$ ?

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Answer 1 · 2017-05-24T11:12:20

$c = 7$ $c=7$

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

$6 \times \frac{1}{3} (c + 1) = 6 \times \frac{1}{6} (3 c - 5)$ $6xx1/3(c+1)=6xx1/6(3c-5)$

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

Step 2. Multiply 2 through on left hand side

$2 c + 2 = 3 c - 5$ $2c+2=3c-5$

Step 3. Add 5 to both sides

$2 c + 7 = 3 c$ $2c+7=3c$

Step 4. Subtract $2 c$ $2c$ from both sides

$7 = c$ $7=c$ or $c = 7$ $c=7$

How do you solve 13(c+1)=16(3c−5)1/3(c+1)=1/6(3c-5)?