Work problem

let $x$ be the number of days it takes Friz to do the job alone and $y$ be the number of days it takes Jones to do the job alone

the fractional part of the work which Friz does in one day is $\dfrac{1}{x}$ , and for Jones is $\dfrac{1}{y}$ . Since they finish the job in each case, we have

$5\left(\dfrac{1}{x}\right)+6\left(\dfrac{1}{y}\right)=1$ $\implies$ $\dfrac{5}{x}+\dfrac{6}{y}=1$

let $a=\dfrac{1}{x}$ and $b=\dfrac{1}{y}$ then

$5a+6b=1$ $\color{#D61F06}(1)$

if they work, they could finish it in $5\frac{1}{2}$ days, so

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5\frac{1}{2}}=\dfrac{2}{11}$

$a+b=\dfrac{2}{11}$ $\implies$ $a=\dfrac{2}{11}-b$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(2)$ in $\color{#D61F06}(1)$ .

$5\left(\dfrac{2}{11}-b\right)+6b=1$ $\implies$ $\dfrac{10}{11}-5b+6b=1$ $\implies$ $b=1-\dfrac{10}{11}$ $\implies$ $b=\dfrac{1}{11}$

It follows that $y=11$ .

Solve for $a$ in $\color{#D61F06}(2)$ .

$a=\dfrac{2}{11}-\dfrac{1}{11}=\dfrac{1}{11}$

It follows that $x=11$ .

Finally,

$x+y=11+11=$ $\boxed{22}$