Since you are using trial and match, why not keep it simple? a or b cannot be lower than 1 When a=1(min), b=4(max) When b=1(min), a=7(max) When b=2 or b=3, being only other integer options in the range of 1 to 4, a has fractional values. So we can conclude that any value of a in (2,6) will not give any integer value for b. Hence a+b is (1+4) or (1+7)
Since you are using trial and match, why not keep it simple?
a or b cannot be lower than 1
When a=1(min), b=4(max)
When b=1(min), a=7(max)
When b=2 or b=3, being only other integer options in the range of 1 to 4, a has fractional values. So we can conclude that any value of a in (2,6) will not give any integer value for b.
Hence a+b is (1+4) or (1+7)
Very nice! ❤
What is the whiteboard software that you are using to write and teach ?
a+b=5
7/a + 8/b = 9
a = 1, b = 4 : OK
a = 7, b = 1 : OK
Very nice! ❤
A=1+B=4=5
Very nice! ❤
а и в - целые. а≤7. 7/а даёт конечную дробь при а=1;2;4;7. Всего 4 варианта а.
Very nice! ❤