ALGEBRA

Integer solutions to equations of the type Ax ± By = C:

• If A, B and C have a common factor divide the entire equation by this common factor. The following matter considers that you have done this step and A and B are considered after the division.

• Now if A and B are not co-prime, the equation will not have any integer solution. Think why!!!

• Consecutive integer solutions to the equation will have x values differing by B and y values differing by A. For equation of type Ax + By = C, a increase (decrease) in x will cause a decrease (increase) in y and vice-versa. For equation of type Ax – By = C, a increase (decrease) in x will cause a increase (decrease) in y and vice-versa.

• Thus in any set of A consecutive integers, y can assume one and only one value for which it will be a integer solution to the equation. And in any set of B consecutive integers, x can assume one and one one value for which it will be a integer solution to the eqn. (To understand this point, i.e. if you have’nt, refer to the concept explained in this article under the heading of LCM and HCF)

• E.g. For the equation 17x + 5y = 513:
There has to be a value of x from 1 to 5 (or from 0 to 4) that will satisfy this equation. For an integral solution 513 – 17x has to be divisible by 5 i.e. 513 – 17x has to end with a 0 or 5. Thus from 0 to 4, x can only take a value 4. Thus one can easily find that (4, 97) is a solution to this equation.

Further if the total number of natural number solution is asked, as x increases in steps of 5 starting with 4 (there is no question of x decreasing as it will not be a natural number then), y will decrease in steps of 17, starting from 97. Since 17 × 5 = 85, there can be 5 such steps of decrease and the number of natural number solutions is 6
If the question was asked as to how many integer solutions exists for the equation such that -43 = x = 67, from -43 to 67 there exists a total of 110 integers. Since x can take one and only one value in any given 5 consecutive integers, the number of solutions will be 22.

If the question was asked as to how many integer solutions exists for the equation such that 0 < x < 168, from 1 to 165 there exists a total of 165 integers. Since x can take one and only one value in any given 5 consecutive integers, the number of solutions will be for 0 < x = 165 will be 33. For x = 166 and 167 would not be a solution as x has to be of the type 5n – 1 to be a solution.

• Consider this question, how many non-negative integral solution exists for the equation 17x – 4y = 1 such that y = 1000.
y can take one and only one value out of 17 consecutive integers to be a solution to the equation. 17 × 60 = 1020. Thus 17 × 58 = 986. Thus from 1 to 986, y will take 58 values. Now all that remains to be checked is that from 987 to 1000 will there be a value that y can take to be a solution to the equation. One solution x = 1 and y = 4 is evident. Thus if we consider sets of 17 consecutive integers as 1 to 17, 18 to 34, 35 to 51, the fourth number in the set will be a solution. Thus from 987 to 1003, y = 991 will be the 59th solution.

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