Math Question

@PTX187 said in #8:
> I think we need to solve the system of inequalities here:
>
> B+D<200
> B+D>122
> D-B>18
>
> In my opinion, there are a lot of children on this forum. They should know this better than me)

Method: Replace B&D by x&y. Then replace "<" by "=". These are 3 lines in the xy plane. each inequality is one side of the line. Shade that side for each. Look at the intersection of all 3 shaded areas. That is the solution if it exists.

@natbee56 said in #1:
> A+52=B
> C+70=D
> If C=2A, A>0 and B+D<200
> Find the number of possible values for B
> Show us your working.
Infinite. There are infinite numbers between 1 and 2 for A, and they all work. 1+52=53=B 2*1=2=C 2+70=72=D 53+72=125<200

Possible values for A is {1 - 43}
Possible values for B is {53 - 95}
Possible values for C is {2 - 86}
Possible values for D is {72 - 156}
& here is proof
youtu.be/e6TWeOBl9qE
EDIT:
Lol i calculated it wrong because i plugged (a+d<200)
It should be plugged (b+d<200) as mentioned by you.
So the solution is
Possible values for A {1-25}
Possible values for B {53-77}
Possible values for C {2-50}
Possible values for D {72-120}
youtu.be/0eS9b5O8OAE

I'm late to the party but I wanted to share my solution. Of course it is the same method of #14 and #18, except I thought it would be more direct if I started from D to obtain A and therefore B.

B = A + 52 and C = 2A
D = C + 70 ---> D = 2A + 70
D + B = <200 = 199 (max) ---> 2A + 70 + A + 52 = 199 ---> 3A + 122 = 199 ---> A = (199 - 122)/3 = 77/3 = 25 ca. (max)
A>0 ---> 1 ≤ A ≤ 25
B = A + 52 ---> 53 ≤ B ≤ 77 which means that B has 25 possible values.