Question

Determine all possible ordered pairs of positive integers (a,b) that satisfy
1/a+2/b=8/2a+b and 2a +5b greater than or equal to 54

Answer 1

\[\frac{ 1 }{ a } +\frac{ 2 }{ 2a+8 }=\frac{ 8 }{ 2a+b } and 2 +5b \le 54\]

Answer 2

Not sure if I can help you but i will try.

Answer 3

For starters lets put a restraint for b
\(2+5b \leq 54\)
\(5b \leq 52\)
\(b \leq 10.4 \)

Answer 4

@satellite73 can you help out?

Answer 5

@El_Pollo bump ur question

Answer 6

i think he meant the restraint is
\[2a + 5b \leq 54\]

Answer 7

no i think we are supposed to find the integers for that as well

Answer 8

i would probably combine the LHS and see where that takes me

\[\frac{ 1 }{ a } + \frac{ 2 }{ 2a+8 } = \frac{ 2a + 8 + 2a}{ a(2a+8) } = \frac{ 4a + 8 }{ 2a(a+4)} = \frac{ 2(a+2) }{ a(a+4) }\]

Answer 9

you wrote and 2a +5b greater than or equal to 54

Answer 10

black label interpreted the restraint as 2 + 5b <= 54

Answer 11

sorry i mean less than i put the equation under

Answer 12

let me retype the thing

Answer 13

if its \[2 + 5b \leq 54\]
then

\[5b \leq 52\]\[5b \leq 50\]since you are looking for positive integers \[b \leq 10\]

Answer 14

\[\frac{ 1 }{ a }+\frac{ 2 }{ b }=\frac{ 8 }{ 2a+b } and 2a+5b \le 54\]

Answer 15

ok im gonna go with \[2a + 5b \leq 54 \] then

Answer 16

yes

Answer 17

I am probably doing it the wrong way, im sorta stuck atm trying to figure out what to do
i show you what i have so far:

you have

\[\frac{ 2(a+2) }{ a(a+4) } = \frac{ 8 }{ 2a + b } \]
cross multiply

\[2(a+2)(2a+b) = 8a(a+4)\]\[2(2a^2 +ab + 4a + 2b) = 8a^2 + 32\]\[(4a^2 +2ab + 8a + 4b) = 8a^2 + 32\]\[-4a^2 +2ab + 8a + 4b = 32\]\[-4a^2 +2ab +18ab - 18ab+ 8a + 4b +25b^2 - 25b^2= 32\]\[-4a^2 +8a^2 - 8a^2+20ab - 18ab+ 8a + 4b +25b^2 - 25b^2= 32\]\[4a^2 -8a^2 +20ab - 18ab+ 8a + 4b +25b^2 - 25b^2= 32\]\[ -8a^2 - 18ab+ 8a + 4b - 25b^2 + (2a+5b)^2= 32\]

Answer 18

its okay, i see what you are doing. I was getting same answers

Answer 19

lol yea this is tough for me..

heres what i got so far:
starting from

\[-4a^2 + 2ab + 8a + 4b = 32\]\[-2a^2 + ab + 4a + 2b = 16\]\[ab + 4a + 2b = 2a^2 + 16\]\[4a + 2b = 2a^2 + 16 -ab\]\[2a + 2b = 2a^2 + 16 -ab-2a\]\[2a + 5b = 2a^2 + 16 -ab-2a +3b\]

and we know \[2a + 5b \leq 54\]so
\[ 2a^2 + 16 -ab-2a +3b\leq 54\]

hope that gets somewhere.. maybe someone else can help you haha

Answer 20

\[ 2a^2 -ab-2a +3b \leq 38\]

Answer 21

It's okay thank you so much!

Answer 22

lol your welcome i hope someone else can figure something out

Answer 23

Don't worry about, its supposed to be challenging. I think i can keep going, thank you for helping get a better idea

Answer 24

alright your welcome then gl figuring it out