An assembly worker is making 3-legged and 5-legged stools. She has 67 legs to use in all for the day's work. What is the largest number of 3- legged stools she can make and still use all the legs to make 5-legged stools with the rest of the legs?

Step-by-step explanation:

let number of 5 legs=x

so 67-5x =multiple of 3

x=1,67-5=62 not divisible by 3

x=2,67-5×2=57 ÷3=19

so he can make 19 three -legged table and 2 five -legged table.

true? is that what you're asking or do you mean something else?