Question

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

Answer 1

in 8/2a, is a in the denominator?

Answer 2

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

Answer 3

oic

Answer 4

No the equation is above

Answer 5

Well re-write LHS as:\[\bf \frac{1}{a}+\frac{2}{b}=\frac{2a+b}{ab}=\frac{ 8 }{ 2a+b }\]Cross-multiply both sides to get:\[\bf (2a+b)^2=8ab \implies 4a^2-4ab+b^2=0 \implies 4a^2-2ab-2ab+b^2=0\]\[\bf \implies 2a(2a-b)-b(2a-b)=0 \implies (2a-b)^2=0 \implies 2a-b = 0 \]\[\bf \implies b = 2a\]We know that \(\bf 2a+5b \le 54 \implies 12a \le 54 \implies a \le \frac{9}{2} \) Because \(\bf a \) is a positive integer, we have: \(\bf \implies a \le 4\). This means that \(\bf a=1,2,3,4\). Since we know that \(\bf b=2a\) then we get the following 4 ordered pairs:\[\bf (a,b)=(1,2),(2,4),(3,6),(4,8)\]And we are done.

Answer 6

@El_Pollo

Answer 7

Do you understand? @El_Pollo