In fact, we have unlimited solution by make one variable as depend to another variable. Lets A is independent variable, then B=f(A). In this case, we can choose B=n×A or another equation. For exampke n=1 and B=1×A=A, then 1/A+1/B=1/A+1/A=1/13 2/A=1/13 or A=26, B=26 n=2, B=2×A=2A 1/A+1/B=1/A+1/2A=3/2A 3/2A=1/13 A=39/2, B=39 .... And so on..
If you solve the system for a you get b= 13 a/(a-13). Now you just plug in values for a in f(a) = a+ 13a /(a-13) to find natural numbers for a+b . This system has of course infinitely many solutions .
Nice Math Olympiad Algebra Equation: 1/a + 1/b = 1/13, a, b > 0; a + b =? 1/a + 1/b = (a + b)/(ab) = 1/13, 13(a + b) = ab, 13 is a prime number 1/13 > 1/a ≥ 1/b > 0, b ≥ a or 1/13 > 1/b ≥ 1/a > 0, a ≥ b b = a; 13(a + b) = 26a = a², a² - 26a = a(a - 26) = 0, a > 0, a - 26 = 0; a = 26 = b a + b = 26 + 26 = 52 b > a; a, b must be multiple integers of 13; Let: b = 13a, a ϵ ℤ 13(a + 13a) = 13(14a) = ab = 13a², a(a - 14) = 0, a - 14 = 0; a = 14, b = 13(14) = 182 a + b = 14 + 182 = 196 a > b; Let: a = 13b, b ϵ ℤ 13(13b + b) = 13(14b) = ab = 13b², b(b - 14) = 0, b - 14 = 0; b = 14, a = 13(14) = 182 a + b = 182 + 14 = 196 Answer check: a = b = 26: 1/a + 1/b = 1/26 + 1/26 = 2/26 = 1/13; Confirmed a = 14, b = 182; a = 182, b = 14: a + b = 1/14 + 1/182 = 1/182 + 1/14 = (1/14)(14/13) = 1/13; Confirmed Final answer: a + b = 52 or a + b = 196
nothing say neither a or b must be a whole number .. this is one soultion .. 1/a+1/b = 1/13 -> (a+b)/ab = 13/13^2 -> a+b=13, ab=13^2 ... √a * √b = 13 .. infinite many solutions
In fact, we have unlimited solution by make one variable as depend to another variable.
Lets A is independent variable, then B=f(A). In this case, we can choose B=n×A or another equation.
For exampke n=1 and B=1×A=A, then
1/A+1/B=1/A+1/A=1/13
2/A=1/13 or A=26, B=26
n=2, B=2×A=2A
1/A+1/B=1/A+1/2A=3/2A
3/2A=1/13 A=39/2, B=39
....
And so on..
@@sie_khoentjoeng4886 Nice Approach 👍
a,b must be positive integers
There are only 2 solutions since 13 is prime number.
For real numbers there are many solutions.
If you solve the system for a you get b= 13 a/(a-13). Now you just plug in values for a in f(a) = a+ 13a /(a-13) to find
natural numbers for a+b . This system has of course infinitely many solutions .
😮
I just understand
1/a + 1/b = 1/13
ab× (1/a + 1/b = 1/13)
b + a = ab/13
= (1/13)×(a)(b)
(1/a)+(1/b)=1/13
(a+b)/ab=1/13
[k(a+b)]]/kab=1/13; k≠0
k(a+b)=1
kab=13
when k=1,
a+b=1 =>b=1-a
ab=13
a(1-a)=13
a-a²=13
a²-a+13=0
♤=-51
♤½=i(51½)
a = (1/2)+[±i(51½)/2]
b = (1/2) -[±i(51½)/2]
ka= kr + ks*z
when k=1
r=(1/2)
s=i(51½)/2
the same for b:
kb= ku - kv*z
when, k=1
u=(1/2)
v=-i(51½)/2
u=r =1/2
s=-v = i(51½)/2
=>a= ku+kv*z
when, k=1
a+b=1
(u+vz)+(u-vz)=1
2u=1
2(1/2)=1
1=1. ok
ab=13
(u+vz)(u-vz)=13
u²-(vz)²=13;
u²=(1/2)²=(1/4)
(vz)²= v²z²= v²(-1)= -v²= -[(51½)/2]²
=> (vz)²= -51/4
ab=13
(u+vz)(u-vz)=13
u²-(vz)²=13
(1/4) -(-51/4)=13
(1/4)+(51/4)=13
(1+51)/4=13
52/4=13
26/2=13
13=13. ok
a=ku + kvz
b=ku - kvz
then,
the anwser is:
k € Z, k ≠0 and:
a=k/2 + ki√(51/4)
b=k/2 - ki√(51/4)
Where does it say they must be whole numbers?
on the screen
nowhere
Nice Math Olympiad Algebra Equation: 1/a + 1/b = 1/13, a, b > 0; a + b =?
1/a + 1/b = (a + b)/(ab) = 1/13, 13(a + b) = ab, 13 is a prime number
1/13 > 1/a ≥ 1/b > 0, b ≥ a or 1/13 > 1/b ≥ 1/a > 0, a ≥ b
b = a; 13(a + b) = 26a = a², a² - 26a = a(a - 26) = 0, a > 0, a - 26 = 0; a = 26 = b
a + b = 26 + 26 = 52
b > a; a, b must be multiple integers of 13; Let: b = 13a, a ϵ ℤ
13(a + 13a) = 13(14a) = ab = 13a², a(a - 14) = 0, a - 14 = 0; a = 14, b = 13(14) = 182
a + b = 14 + 182 = 196
a > b; Let: a = 13b, b ϵ ℤ
13(13b + b) = 13(14b) = ab = 13b², b(b - 14) = 0, b - 14 = 0; b = 14, a = 13(14) = 182
a + b = 182 + 14 = 196
Answer check:
a = b = 26: 1/a + 1/b = 1/26 + 1/26 = 2/26 = 1/13; Confirmed
a = 14, b = 182; a = 182, b = 14:
a + b = 1/14 + 1/182 = 1/182 + 1/14 = (1/14)(14/13) = 1/13; Confirmed
Final answer:
a + b = 52 or a + b = 196
nothing say neither a or b must be a whole number .. this is one soultion ..
1/a+1/b = 1/13 ->
(a+b)/ab = 13/13^2 ->
a+b=13, ab=13^2 ... √a * √b = 13 .. infinite many solutions
😮