Solve 1/2left(k+3right)=1/3left(k+16right)

$$\frac{ 1 }{ 2 } \left( k+3 \right) = \frac{ 1 }{ 3 } \left( k+16 \right)$$

Solve for k

$$k=23$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{1}{2} by k+3.

$$\frac{1}{2}k+\frac{1}{2}\times 3=\frac{1}{3}\left(k+16\right)$$

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

$$\frac{1}{2}k+\frac{3}{2}=\frac{1}{3}\left(k+16\right)$$

Use the distributive property to multiply \frac{1}{3} by k+16.

$$\frac{1}{2}k+\frac{3}{2}=\frac{1}{3}k+\frac{1}{3}\times 16$$

Multiply \frac{1}{3} and 16 to get \frac{16}{3}.

$$\frac{1}{2}k+\frac{3}{2}=\frac{1}{3}k+\frac{16}{3}$$

Subtract \frac{1}{3}k from both sides.

$$\frac{1}{2}k+\frac{3}{2}-\frac{1}{3}k=\frac{16}{3}$$

Combine \frac{1}{2}k and -\frac{1}{3}k to get \frac{1}{6}k.

$$\frac{1}{6}k+\frac{3}{2}=\frac{16}{3}$$

Subtract \frac{3}{2} from both sides.

$$\frac{1}{6}k=\frac{16}{3}-\frac{3}{2}$$

Least common multiple of 3 and 2 is 6. Convert \frac{16}{3} and \frac{3}{2} to fractions with denominator 6.

$$\frac{1}{6}k=\frac{32}{6}-\frac{9}{6}$$

Since \frac{32}{6} and \frac{9}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{6}k=\frac{32-9}{6}$$

Subtract 9 from 32 to get 23.

$$\frac{1}{6}k=\frac{23}{6}$$

Multiply both sides by 6, the reciprocal of \frac{1}{6}.

$$k=\frac{23}{6}\times 6$$

Cancel out 6 and 6.

$$k=23$$