Andrew M. answered • 07/04/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
1/a + 2/b = 1/c + 1/2
1/a + 2/b -1/2 = 1/c
Multiply both sides by c...
c(1/a + 2/b - 1/2) = 1
Get a common denominator with our 3 fractions
c (2b/2ab + 4a/2ab - ab/2ab) = 1
c((2b+4a-ab)/2ab) = 1
c = 1/[((2b+4a-ab)/2ab)]
c = (2ab)/(2b+4a-ab)
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If a = 3 we have c = 6b/(12-b)
multiplying both sides by (12-b)...
6 b = 12c - bc
I tried plugging 1, 2, and 3 in for b to see if I got an integer for c
Using b = 3 I got...
18 = 12c-3c
18 =9c
c = 2
Thus I find that a=3, b=3, c=2 will solve this equation
Check: 1/a + 2/b = 1/c + 1/2
1/3 + 2/3 = 1/2 + 1/2
3/3 = 2/2
1=1
This checks
Michael J.
In my solution, I obtained the equation
3b(2 + c) = 2c(b + 6)
From this equation, we can create two separate equations with like variables. One for b and one for c.
2 + c = 2c and 3b = b + 6
We can do this because each side of the original equation is a product. Solve both of these equations,
c = 2 and b = 3
Plug these values into the equation that I obtained.
3(3)(2 + 2) = 2(2)(3 + 6)
9(4) = 4(9)
36 = 36
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07/04/15
Marietta H.
But that only gives you one possible pair of values. And my teacher said I'm supposed to have 7. (Wished she'd given me a more helpful hint). I have no idea how to get that.
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07/04/15
Andrew M.
From the point which Michael and I both obtained, as well as yourself:
3b(2+c)=2c(b+6)
We can rearrange also to
3b = 2c eqn 1
2+c = b+6 eqn 2
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From eqn 1 we have b = 2/3 c
Plug that into eqn 2 and we have
2+c=2/3 c +6
1/3 c = 4
c = 4(3)
c = 12
3b = 2c so... 3b = 24 .. b = 8
So another solution is a=3, b = 8, c =12
I'm not certain about how to graph out all the possibilities but we know we have
two solutions for a,b,c ...
3,3,2 and 3,8,12
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07/04/15
Andrew M.
Looking at this again I went back to the original equation 1/a + 2/b = 1/c + 1/2
Put 3 in for a:
1/3 + 2/b = 1/c + 1/2
2/b-1/c = 1/2 - 1/3
Setting 2/b = 1/2 and 1/c = 1/3 gives b = 4 and c = 3
We now have 3 solutions: a,b,c = 3,3,2 3,4,3, 3,8,12
Lets work with even numbers for b;
Let b = 6
1/3 + 2/6 = 1/c + 1/2
4/6 - 1/2 = 1/c
1/6 = 1/c
c = 6
a,b,c = 3,3,2 3,4,3, 3,8,12 3,6,6
Continuing in the vein let b = 10.... We come up with c = 30
So a 5th solution is 3,10,30
We now have a,b,c = {3,3,2} {3,4,3} {3,8,12} {3,6,6} {3,10,30}
When I tried b = 12 it does not work because I get 0 = 1/c ....
Let's try even values for c now: we have so far c = 2,3,6,12,30
Let c = 18 (continuing in the vein of multiples of 6)
1/3 + 2/b = 1/c + 1/2
1/3 + 2/b = 1/18 + 1/2 This ends up with b = 9 giving solution 3,9,18
We are now up to a,b,c = {3,3,2} {3,4,3} {3,6,6} {3,8,12} {3,9,18} {3,10,30}
I'm having difficulty finding that last one. There has to be a pattern to these answers....
but I'm not quite seeing it. But we do have 6 out of 7 answers.
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07/04/15
Andrew M.
For a final solution, since we have b = 3,4,6,8,9,10... I put in b = 11
1/3 + 2/11 = 1/c + 1/2
22/66 + 12/66 = 1/c + 33/66
34/66 = 1/c + 33/66
c = 66
OUR solution set is
{a,b,c} = {3,3,2} {3,4,3} {3,6,6} {3,8,12} {3,9,18} {3,10,30}{3,11,66}
It no longer worked once I got up to b = 12
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07/04/15
Marietta H.
I got the answer before you! From c=(2ab)/(4a-ab+2b) it can be simplified to get 12c-bc=6b.
I put in all the numbers 1-12 for b. If b is any greater than 12, e.g. 12c-13c, the c would be negative which is not possible since the question said they were all positive integers.
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07/04/15
Andrew M.
Fantastic. lol. However, in my defense, I'm at work and doing this between busy spells. :-). Great job.
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07/04/15
Marietta H.
07/04/15