8.5. Two graphs of linear functions intersect at $x=2$. At $x=8$, the values differ by 8. At $x=20$, the value of one of the functions is 100. What could be the value of the other function
Answer: 76 or 124.

Solution 1. Let the value of the first function change by $d$ when the argument increases by $8-2=6$, then the second function changes by $d \pm 8$. From 2 to 20, the argument increases by 18, which is 3 times more than 6. Therefore, the value of the first function changes by $3d$, and the second function changes by $3(d \pm 8)=3d \pm 24$, they differ by $3d-3d \pm 24= \pm 24$. Therefore, the desired value of the second function is $100 \pm 24$, that is, 76 or 124.

Solution 2. Let $y=k_{1} x+b_{1}$ and $y=k_{2} x+b_{2}$ be the given functions. According to the condition, $2 k_{1}+b_{1}=2 k_{2}+b_{2},\left(8 k_{1}+b_{1}\right)-\left(8 k_{2}+b_{2}\right)= \pm 8,20 k_{1}+b_{1}=100$. We need to find the value of $20 k_{2}+b_{2}$.

Rewrite the first two relations as $2\left(k_{1}-k_{2}\right)+\left(b_{1}-b_{2}\right)=0$ and $8\left(k_{1}-k_{2}\right)+$ $\left(b_{1}-b_{2}\right)= \pm 8$, from which $6\left(k_{1}-k_{2}\right)= \pm 8$.

Then $20 k_{2}+b_{2}=-20\left(k_{1}-k_{2}\right)-\left(b_{1}-b_{2}\right)+\left(20 k_{1}+b_{1}\right)=$

$=-\left(2\left(k_{1}-k_{2}\right)+\left(b_{1}-b_{2}\right)\right)-3 \cdot 6\left(k_{1}-k_{2}\right)+\left(20 k_{1}+b_{1}\right)=$

$=-0-3 \cdot( \pm 8)+100=100 \pm 24$.