Heavy weights?

Let $B$ , and $S$ be Bob and Sue's respective weights.

• $B = 100 + \frac{S}{2}$
• $S = 100 + \frac{B}{3}$

To remove all the fractions, we multiply the first equation by 2 and the second second by 3, we get

$2B = 200 + S, \quad 3S = 300 + B$

Multiply the first equation by 3, we get $6B = 600 + 3S = 600 + (300 + B)= 700 + B$ or $5B = 900$ .

This gives $B = \dfrac{900}5 = 180$ and so, $S = 100 + \dfrac{180}3 = 100 + 60 = 160$ .

Our answer is $B+S = 180 + 160 =\boxed{340}$ .

From the first statement,

$B=100+\dfrac{1}{2}S$ $\color{#D61F06}(1)$

From the second statement,

$S=100+\dfrac{1}{3}B$ $\color{#D61F06}(2)$

Substituting $\color{#D61F06}(2)$ in $\color{#D61F06}(1)$ , we get

$B=100+\dfrac{1}{2}\left(100+\dfrac{1}{3}B \right)=100+50+\dfrac{B}{6}$

Multiplying both sides by $6$ , we get

$6B=600+300+B$ $\implies$ $5B=900$ $\implies$ $B=180$

Substitute $B=180$ in $\color{#D61F06}(2)$ , we get

$S=100\dfrac{1}{3}(180)=160$

Finally, the sum of their weights is $180+160=$ $\boxed{340}$