1:34 Dont stop ! 5b + 3 is a factor then : multiply both sides by 5 and add 3 to both sides ---> 15a(5b + 3) + 5b + 3 = 39*5 + 3 ---> (15a + 1)*(5b+3) = 198 without using fractions !
I think we can set up the equation as b(5a+1)+3/5(5a+1-1)=39 (b+3/5)(5a+1)-3/5=39 multiply by 5 so (5b+3)(5a+1)=195+3 (5b+3)(5a+1)=198 Now if a and b are integers we can get a few pairs of solutions, as 198=2*99=2*3*33=2*3^2*11 (a,b)€{(0,39)from1*198,(-20,-1) from -99*-2,(13,0) from 66*3,(-2,-5) from -9*-22,(2,3) from 11*18,(1,6) from 6*33}
a + b = a + (39 - 3a)/(5a + 1), where a is any number, except -⅕. Nowhere in the thumbnail did it specify that a and b have to be integers. If they must be integers, then there are 5 solutions, where a equals -20, -2, 0, 1, 2 or 13, while b equals -1, -5, 39, 6, 3 or zero, respectively.
@@Taric25 -- *Thumbs-down.* Because it is a Diophantine equation, that means, in part, we are only interested in integer solutions. Also, you wrote out six solutions, not five. Do not spell out 'zero" in this context. Write it just like the others.
Nice solution especially the first method, but why you asked in the most begining to find a+b=? and not all the solutions? Can you find the a+b without finding beforehanded all the solutions?
Third Method (trial and error) 3 a + 5 a b + b = 39 a (3+5 b) = 39 - b a = (39 - b)/(3+5 b) Now set (b) value and check for resulting (a) value as integer b a ---------------------- 0 39/3 = 13 1 38/8 err 2 37/13 err 3 36/18 = 2 4 35/23 err 5 34/28 err 6 33/33 = 1 // No need for further checks, all results will be < 1 Just one only check for (b) = 39 39 0/?? = 0 Final results of (a,b) = { (0,39), (1,6), (2,3), (13,0) }
Second method, at step 1 you missed multiplying the right hand side (39) by 5 (195). Then for step 2 you add 3 to both sides to get the 198 you were looking for.
(−20,−1), (−2,−5), (0, 39), (1, 6), (2, 3), and (13, 0) are the only integral solution. I suggest rewriting as a rational function to see that you have a minimal set of values to choose from. A visual, even a rough one, often offers insight.
3a,39 are multiples of 3 So b=0,3,6,9,.. a=39,2,1, fractions So (a,b)=(39,0),(2,3),(1,6) For b=-3,-6,-9, a=fractions. There are3 pairs of integer solutions.
a + b = a + (39 - 3a)/(5a + 1), where a is any number, except -⅕. Nowhere in the thumbnail did it specify that a and b have to be integers. If they must be integers, then there are 5 solutions, where a equals -20, -2, 0, 1, 2 or 13, while b equals -1, -5, 39, 6, 3 or zero, respectively.
You're being corrected again, because you don't know what you are talking about. In the thumbnail, Diophantine equations refer to those where we are only interested in integer solutions.
1:34 Dont stop !
5b + 3 is a factor then : multiply both sides by 5 and add 3 to both sides --->
15a(5b + 3) + 5b + 3 = 39*5 + 3 ---> (15a + 1)*(5b+3) = 198 without using fractions !
You forgot:
a = 2, b = 3
I think we can set up the equation as
b(5a+1)+3/5(5a+1-1)=39
(b+3/5)(5a+1)-3/5=39 multiply by 5 so
(5b+3)(5a+1)=195+3
(5b+3)(5a+1)=198
Now if a and b are integers we can get a few pairs of solutions, as 198=2*99=2*3*33=2*3^2*11
(a,b)€{(0,39)from1*198,(-20,-1) from -99*-2,(13,0) from 66*3,(-2,-5) from -9*-22,(2,3) from 11*18,(1,6) from 6*33}
Thank God! Today he did not call the second method first 🤣🤣
You forgot (1,6); (-2,-5) and (-20,-1).
a=2, b = 3. So a+b = 5.
a+b=(39-b)/(3+5b)+b...b=6(a=1),b=3(a=2),b=0(a=13),b=-1(a=-20),b=-5(a=-2)
You forgot another solution : a=1, b=6
a + b = a + (39 - 3a)/(5a + 1), where a is any number, except -⅕. Nowhere in the thumbnail did it specify that a and b have to be integers.
If they must be integers, then there are 5 solutions, where a equals -20, -2, 0, 1, 2 or 13, while b equals -1, -5, 39, 6, 3 or zero, respectively.
@@Taric25 -- *Thumbs-down.* Because it is a Diophantine equation, that means, in part, we are only interested in integer solutions. Also, you wrote out six solutions, not five. Do not spell out 'zero" in this context. Write it just like the others.
@@robertveith6383 If I apologize, would you hug me?
1,6
I liked it, and I liked the use of mod to examine factors of 198. Thank you!
You're welcome! 😊
Nice solution especially the first method, but why you asked in the most begining to find a+b=? and not all the solutions? Can you find the a+b without finding beforehanded all the solutions?
Third Method (trial and error)
3 a + 5 a b + b = 39
a (3+5 b) = 39 - b
a = (39 - b)/(3+5 b)
Now set (b) value and check for resulting (a) value as integer
b a
----------------------
0 39/3 = 13
1 38/8 err
2 37/13 err
3 36/18 = 2
4 35/23 err
5 34/28 err
6 33/33 = 1 // No need for further checks, all results will be < 1
Just one only check for (b) = 39
39 0/?? = 0
Final results of (a,b) = { (0,39), (1,6), (2,3), (13,0) }
Second method, at step 1 you missed multiplying the right hand side (39) by 5 (195). Then for step 2 you add 3 to both sides to get the 198 you were looking for.
Yep! You're right! 😅
3a•5+5ab•5+b•5=39•5
(5a+1)(5b+3)=195+3=198=2•99
=2•3•3•11=11•18=6•33
=1•198=66•3
5a+1=11,5b+3=18
5a+1=6,5b+3=33
5a+1=1,5b+3=198
5a+1=66,5b+3=3
(−20,−1), (−2,−5), (0, 39), (1, 6), (2, 3), and (13, 0) are the only integral solution. I suggest rewriting as a rational function to see that you have a minimal set of values to choose from. A visual, even a rough one, often offers insight.
You said integral. You meant integer.
Very cool!
@@Taric25 Integral solutions refer to solutions of equations that are integers.
@@ronbannon Integral solutions are solutions of definite or indefinite integrals. Claiming otherwise introduces ambiguity.
3a,39 are multiples of 3
So b=0,3,6,9,..
a=39,2,1, fractions
So (a,b)=(39,0),(2,3),(1,6)
For b=-3,-6,-9,
a=fractions.
There are3 pairs of integer solutions.
If you're going to "solve" it by trying all the possibilities, you could just write a short piece of program code.
Can you do that for me? 😁
5, 7, 13, 39. What a headache!
problem
3a + 5ab + b = 39
a+b = ?
Factored into
(5 a + 1) (5 b + 3) = 198
= 11 • 18
= 6 • 33
Got
(a, b) ∈ { (2, 3), (1, 6) }
answer
a+ b ∈ { 5, 7 }
"put a little 1 here"
I pretty sure that it was a normal 1 😅
😄
RHS is 195, not 198.
Nice
Should have been pos. integers only
(1, 6), (2, 3)
a + b = a + (39 - 3a)/(5a + 1), where a is any number, except -⅕. Nowhere in the thumbnail did it specify that a and b have to be integers.
If they must be integers, then there are 5 solutions, where a equals -20, -2, 0, 1, 2 or 13, while b equals -1, -5, 39, 6, 3 or zero, respectively.
chk the title
@SyberMath Fantastic, did I say thumbnail or title? I said thumbnail, thank you very much.
You're being corrected again, because you don't know what you are talking about. In the thumbnail, Diophantine equations refer to those where we are only interested in integer solutions.
@robertveith6383 It didn't say that in the thumbnail. It said that in the title.
RHS x5