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nick1816
Another way is

ac+bd = 34
ad+bc=43

Add both equations

ac+ad+bd+bc = 77

a(c+d)+b(c+d) = 77

(a+b)(c+d) = 1*77 or 7*11

1*77 case is not possible, since all of them are positive integers.

(a+b) and (c+d) are equal to 7 and 11 in any order.

a+b+c+d = 7+11=18

nick1816 does it mean 1st equation i.e. ad+cd=38 makes no sense.
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There are multiple ways to find the value of a+b+c+d. There are solutions in which all 3 equations are required (if i find the values of a, b, c and d individually and then add them) . But the way i solved this problem, equation 1 is certainly redundant.

ammuseeru
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Another way is

ac+bd = 34
ad+bc=43

Add both equations

ac+ad+bd+bc = 77

a(c+d)+b(c+d) = 77

(a+b)(c+d) = 1*77 or 7*11

1*77 case is not possible, since all of them are positive integers.

(a+b) and (c+d) are equal to 7 and 11 in any order.

a+b+c+d = 7+11=18

nick1816 does it mean 1st equation i.e. ad+cd=38 makes no sense.
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Using expression (1) and (2):

ab+cd+ac+bd = 38+34
a(b+c)+d(b+c)=72
(a+d)(b+c)=72

72=2^3 . 3^2

Then we have 4 solutions:

2 and 2^2 . 3^2 = 36 --> sum = 38
2^2 = 4 and 2 . 3^2 = 18 --> sum = 22
2^3 = 8 and 3^2 = 9 --> sum=17
2^3 . 3 = 24 and 3 --> sum=27

Differ from the solution provided using equation (3).

is it wrong?
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anrocha

ac + bd = 34
ab + cd = 38
ad + bc = 43­

If a, b, c and d positive integers such that they satisfy the system of equations above, find a+b+c+d.

A. 12
B. 14
C. 16
D. 18
E. 21­

Using expression (1) and (2):

ab+cd+ac+bd = 38+34
a(b+c)+d(b+c)=72
(a+d)(b+c)=72

72=2^3 . 3^2

Then we have 4 solutions:

2 and 2^2 . 3^2 = 36 --> sum = 38
2^2 = 4 and 2 . 3^2 = 18 --> sum = 22
2^3 = 8 and 3^2 = 9 --> sum=17
2^3 . 3 = 24 and 3 --> sum=27

Differ from the solution provided using equation (3).

is it wrong?
ac + bd = 34
ab + cd = 38
ad + bc = 43­

Here are the serts satisfying the equations:

a = 2 and b = 5 and c = 7 and d = 4
a = 4 and b = 7 and c = 5 and d = 2
a = 5 and b = 2 and c = 4 and d = 7
a = 7 and b = 4 and c = 2 and d = 5­
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