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# Suppose a and b are integers...

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Suppose a and b are integers. How many solutions are there to the equation ab = 2a + 3b?

Guest Sep 27, 2017
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#1
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a*b=2a + 3b

a = -3 and b = 1

a = 0 and b = 0

a = 1 and b = -1

a = 2 and b = -4

a = 4 and b = 8

a = 5 and b = 5

a = 6 and b = 4

a = 9 and b = 3

Guest Sep 27, 2017
#2
+77092
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ab  = 2a + 3b

Square both sides

a^2b^2  = 4a^2 + 12ab + 9b^2   rearrange as

[ 4-b^2]a^2 + [ 12b]a + 9b^2  = 0

Setting this up as a quadratic and solving for  a, we have that

a  =  -12b ±√ [  (-12b)^2 - 4 (4 - b^2)( 9b^2) ]

________________________________

2 [ 4 - b^2]

a  =  -12b ±√ [  144b^2  - 144b^2  + 36b^4 ]

_______________________________

2 [ 4 - b^2 ]

a =    -12b  ± 6b^2

___________

2 [ 4 - b^2 ]

a  =     6b [  -2  ± b ]

___________

2 [ 4 - b^2 ]

a  =   3b [ -2  ± b ]

_____________

(2 + b)  (2 - b)

Choosing        a  =       3b [ -2 - b]                           -3b [ 2 + b]                     -3b

__________     =                _____________      =        ___

(2 + b) (2 -b)                         (2 + b) (2 -b)                    2 -b

Will  produce all the possible integer answers

Simplifying, we have that

a  =  -3b               3b

_____   =     ____

2 - b             b - 2

Dividing this fraction produces :

a =   3  +    6

____

b - 2

And when   b  =     8      5      4     3    1    0    -1   -4

a =      4      5      6     9   -3    0     1    2

So.........( a, b)  =  (2, -4), (1, -1), (0, 0), (-3, 1), (9, 3), (6,4), (5,5)  and (4,8)

a  =

CPhill  Sep 27, 2017
edited by CPhill  Sep 27, 2017

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