The brief (4 cases) analysis is long and may be shortened. As the clip shows, the original equation, 5/a + 6/b = 7 may be written in the following way: (7a - 5) (7b - 6) = 30. Any solution for a , b ∈ Z requires that: 7a - 5 = n , 7b - 6 = 30/n , (n , 30/n ∈ Z) Therefore, n ∈ {±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30} and a , b are given by: a = (n + 5) / 7 and b = (30/n + 6) / 7 = 6 ((n + 5) / 7) / n = 6a / n For a ∈ Z+ (and even for a ∈ Z), the only two possible values for n are: n = 2 and n = 30 (n = -5 is excluded, since it gives: a = b = 0 , which is not allowed by the original equation, even if we don't restrict ourselves to positive a , b) . These values give b ∈ Z+ as well and therefore, the set solutions to this problem is: { (a , b) } = { (1 , 3) , (5 , 1) }
The general solution to this type of problem is first to multiply through by _ab_ then divide through by the coefficient of the _ab_ term and collect all terms onto the same side: _5/a + 6/b = 7_ ... ① ⇒ _5b + 6a = 7ab_ ⇒ _(⁵/₇)b + (⁶/₇)a = ab_ ⇒ _ab - (⁵/₇)b - (⁶/₇)a = 0_ ... ② Now factorise the LHS of ② as follows _(a - ⁵/₇)(b - ⁶/₇)_ which gives the LHS of ② but with an extra constant _(⁵/₇)(⁶/₇) = ³⁰/₇²_ ∴ _(a - ⁵/₇)(b - ⁶/₇) = ³⁰/₇²_ (from ②) Finally multiply through by the _7²_ denominator, one _7_ to each factor on the LHS: _(7a - 5)(7b - 6) = 30_ in which form we can consider all the factors of _30,_ not forgetting the negative ones. The following give integers for _a_ and _b_ : _7a - 5 = 30, 7b - 6 = 1 ⇒ (a, b) = (5, 1)_ _7a - 5 = 2, 7b - 6 = 15 ⇒ (a, b) = (1, 3)_ _7a - 5 = -5, 7b - 6 = -6 ⇒ (a, b) = (0, 0)_ - rejected because _a, b ∊ ℤ⁺_ and also _a_ and _b_ are denominators in ①.
@@superacademy247 I was able to give up on some intuitive solutions myself, for example, such as 1 and 3, but what a beautiful proof you got for all the other solutions.
He doesn't solve it. He just works back from one answer. I don't think there is any general solution to this type of problem, but what I did is the following: Multiply through by ab and rearrange to get: 7ab = 6a + 5b construct a small table of values 6a + 5b starting from 1,1 in the top left corner and , say 6,6 in the bottom right. This can be done quickly because you just need to count up in steps of 6 along the a axis and +5 along the b axis. Spot which ones are divisible by 7 and check whether it is equal to 7ab. You will soon discover the 1,3 and 5,1 solutions. Time needed is about 3 minutes.
The brief (4 cases) analysis is long and may be shortened.
As the clip shows, the original equation, 5/a + 6/b = 7 may be written in the following way:
(7a - 5) (7b - 6) = 30. Any solution for a , b ∈ Z requires that: 7a - 5 = n , 7b - 6 = 30/n , (n , 30/n ∈ Z)
Therefore, n ∈ {±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30} and a , b are given by:
a = (n + 5) / 7 and b = (30/n + 6) / 7 = 6 ((n + 5) / 7) / n = 6a / n
For a ∈ Z+ (and even for a ∈ Z), the only two possible values for n are: n = 2 and n = 30 (n = -5 is excluded, since it gives: a = b = 0 , which is not allowed by the original equation, even if we don't restrict ourselves to positive a , b) . These values give b ∈ Z+ as well and therefore, the set solutions to this problem is:
{ (a , b) } = { (1 , 3) , (5 , 1) }
5/a+6/b=7 a=5 b=1 final answer
(a,b)=(5,1),(1,3)
The general solution to this type of problem is first to multiply through by _ab_ then divide through by the coefficient of the _ab_ term and collect all terms onto the same side:
_5/a + 6/b = 7_ ... ①
⇒ _5b + 6a = 7ab_
⇒ _(⁵/₇)b + (⁶/₇)a = ab_
⇒ _ab - (⁵/₇)b - (⁶/₇)a = 0_ ... ②
Now factorise the LHS of ② as follows
_(a - ⁵/₇)(b - ⁶/₇)_
which gives the LHS of ② but with an extra constant _(⁵/₇)(⁶/₇) = ³⁰/₇²_
∴ _(a - ⁵/₇)(b - ⁶/₇) = ³⁰/₇²_ (from ②)
Finally multiply through by the _7²_ denominator, one _7_ to each factor on the LHS:
_(7a - 5)(7b - 6) = 30_
in which form we can consider all the factors of _30,_ not forgetting the negative ones.
The following give integers for _a_ and _b_ :
_7a - 5 = 30, 7b - 6 = 1 ⇒ (a, b) = (5, 1)_
_7a - 5 = 2, 7b - 6 = 15 ⇒ (a, b) = (1, 3)_
_7a - 5 = -5, 7b - 6 = -6 ⇒ (a, b) = (0, 0)_ - rejected because _a, b ∊ ℤ⁺_ and also _a_ and _b_ are denominators in ①.
{5a+5a ➖} /a+{6b+6 ➖}/b ={10a^2/a+12b^2/b}=22ab^4/ab=22ab^4 2^11ab^4 2^11^1ab^4 2^1^1ab^2^2 1ab^1^2 1ab^2 (ab ➖ 2ab+1).
A big like from me.
Thank you! Cheers!
@@superacademy247 I was able to give up on some intuitive solutions myself, for example, such as 1 and 3, but what a beautiful proof you got for all the other solutions.
Why is your method better than just immediately spotting the answer?
5 count only with 1 or 5
a=1 > 5/1+6/b=7 >6/b=7-5
6/b=2 > *** b=3 ****
a=5 > 5/5+6/b=7 > 1+6/b=7
6/b=7-1 6/b=6 > *** b=1 ***
END
He doesn't solve it. He just works back from one answer.
I don't think there is any general solution to this type of problem, but what I did is the following:
Multiply through by ab and rearrange to get: 7ab = 6a + 5b
construct a small table of values 6a + 5b starting from 1,1 in the top left corner and , say 6,6 in the bottom right.
This can be done quickly because you just need to count up in steps of 6 along the a axis and +5 along the b axis.
Spot which ones are divisible by 7 and check whether it is equal to 7ab. You will soon discover the 1,3 and 5,1 solutions.
Time needed is about 3 minutes.
There's a method easy and take less time
5 + 2 = 7
beautiful
Many many thanks