How many ordered pairs ( x ; y ) of non-zero positive integers exists to: 6 x + 2 y = 7 0 ?
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3x+y=35, so, for y to be non-zero, positive, 3x<35, and for x to be non zero, positive, x>0 => o<x<12, thus x can be anything from 1 to 11, which accounts to 11 total values.
Because x and y are both non zero integers, they must have a value of at least one. Therefore, we can subtract 8 from 70 to create the new equation
6 x + 2 y = 6 2 , which allows for x and y to be 0.
Then, we can notice that in our new equation, for any value of x, y must be at least 1, so that the remainder will be a multiple of 6. Therefore, we can create the new equation
6 x + 2 y = 6 0
Now we need only look at the possible values of 6 (since 2 is a multiple of 6, the space that is leftover will be filled up with 2s: for every possible value of x, there exists a corresponding value for y). At most, x=10, and at its minimum, x is 0. Therefore, all values possible for x are inclusively in between 0 and 10. 10-0+1= 11. There are 11 possible values for x, therefore there are 11 ordered pairs.
Both multiples of 6 and 2 are all even. Since the ordered pairs are non-zero positive integers, we can multiply 6 by 1 up to 11 while the multiples of 2 will just complement the multiples of 6, summing up to 70. Hence, the answer is 11.
6x + 2y = 70 or, 3x + y = 35
y is alone here. So we leave that in peace, and try to match up the pairs of x. It easy to see that we can have 11 different choices, since (11)(3) = 33. This is the maximum we can reach, and then add up a y = 2 to balance up the 35.
Hence we have our answer: 11
The thing to understand here, x plays a major role here than y, y merely does the balancing act.
6x + 2y = 70 ⇒ 3x + y = 35 ⇒ y=35-3x ⇒ 35-3x>0 ⇒ 35 - 3x > 0 .(-1) ⇒ 3x - 35 < 0 ⇒ x<11,6... ∴ 0 < x ≤ 11
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The equation is actually 3 x + y = 3 5 . Divide it by 3 so that the coefficient of x is 1. Rearrange it and you can get x − 1 1 = 3 2 − y Because x and y are integers, x − 1 1 is also an integer, then 3 2 − y must be also an integer.
Suppose 3 2 − y = k where k is an integer, then y = 2 − 3 k . Because y is positive, so k < 1 .
Substitue k in 3 x + y = 3 5 you will get x = 1 1 + k . Because x > 0 , so k > − 1 1 , hence − 1 1 < k < 1 , k = − 1 1 , − 1 0 , − 9 , . . . , 0 There are only 11 solutions for k , so the answer is 11.