since we are dealing with integers and since all terms are positive. we can just go through increasing b and reducing a we don’t need all that algebra start with b=0 so a=5 b=1 a = 1 3/4 so that doesn’t work b=2 a=1 last possibility is a=0 b=15 but if strictly positive then only b=2, a=1
@@SALogics In mathematics, "Z+" represents the set of all positive integers, meaning the numbers 1, 2, 3, 4, and so on; essentially, the "Z" stands for the set of all integers, and the "+" indicates only positive integers are included. So, No, 0 should not be included if the domain is Z+.
It does seem like it might’ve been easier if you had multiplied by five through the whole equation and then added three. This way you would not have had to work with a fraction.
since we are dealing with integers and since all terms are positive.
we can just go through increasing b and reducing a
we don’t need all that algebra
start with b=0 so a=5
b=1 a = 1 3/4 so that doesn’t work
b=2 a=1
last possibility is a=0 b=15
but if strictly positive then only b=2, a=1
Very nice! ❤
Your problem statement said Z+, so how come you're including 0 in valid solutions?
is'nt 0 incldeded? ❤
@@SALogics In mathematics, "Z+" represents the set of all positive integers, meaning the numbers 1, 2, 3, 4, and so on; essentially, the "Z" stands for the set of all integers, and the "+" indicates only positive integers are included. So, No, 0 should not be included if the domain is Z+.
It does seem like it might’ve been easier if you had multiplied by five through the whole equation and then added three. This way you would not have had to work with a fraction.
Yes, you are right! ❤
a、bは正の整数なので解は a=１, b=２だけです。
5ab+3a+b=15. i see if a=0: 5*0*b+3*0+b=15 0+0+b=15. b=15. i see if b=0; 5a*0+3a+0=15. 0+3a+0=15. 3a=15. a=15/3=5 (a,b) (0, 15) (5, 0). 3) 5ab+(3a+b)=15. if5ab=10 yes 3a+b=5. b=5-3a. 5ab=5a(5-3a)=10. a(5-3a)=2. 5a-3aa=2. 3aa-5a+2=0. a=1. b=5-3a=5-3*1=2. (a,b) (0, 15) (5,0) (1. 2).
Very nice! ❤
0 is not positive
Yes, you are right! ❤