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Write 100 as the sum of two integers, one divisible by 7 and the other divisible by 11. Use your answer to find formulas giving all the solutions of the following equation where x and y are integers. So 7x+11y=100 1. First work with positive integers to see how many pairs of values you can find to satisfy the equation. Positive integers x y 7 93 14 86 21 79 28 72 35 65 42 58 49 51 54 44 63 37 70 30 The multiples of 7 are laid at the x axis stating from 7 up to 70 and the numbers at the y axis are the numbers which will add up to 100. In other words x and y = 100. And to find the satisfying condition you have to find a number in the y axis which is a multiple of 11. As for the results there is only 1 pair of positive integers which is 56 and 44 Equal to : X= 8 Y=4 7x+11y=100 7(8) +11(4) =100 56+44=100 2. Now try negative integers for x with positive integers for y and vice-versa. Negative integers x y -7 107 -14 114 -21 121 -28 128 -35 135 -42 142 -49 149 -54 154 -63 163 -70 170 -77 177 -84 184 -91 191 -98 198 -105 205 -112 212 For negative integers I did the same thing as in positive integers which is to lay in x axis the multiple of 7 but in negative numbers starting from -7 up to -112 and in y axis the number which will add up to 100. (-x+y=100). And as in positive integers I have to find a number which will satisfy both conditions in x axis a multiple of 7 and in y a multiple of 11. I found 2 pair of negative integers which are -21 & 121, -98 & 198 X= -3 7x+11y=100 Y= 11 7(-3)+11(11)=100 -21+121=100 X= -14 Y= 18 7x+11y=100 7(-14)+11(18)=100 -98+198=100 3. 7x+11y=100 solutions (positive and negative integers) x y 8 4 -3 11 -14 18 In this table at the x axis are the numbers which are multiplied to make the integers of 7 and in y axis integers of 11. 4. Do you notice a pattern? Throughout the investigation I noticed a pattern which is that in x (multiples of 7) goes subtracting 11 each time starting from 8 (7 x 8) and in y (multiple of 11) it goes adding 7 starting from 4. (11 x 4) 5. Can you make equations that describes how you get from one value of x and y to the next value? Equations: These are the equation I found for nth term of x and nth term of y X= -11n+19 Y= 7n-3 7x+11y=100 7(-11n+19) +11(7n-3) =100 6. If you can, test to see if your equations work for several other values of x and y Checking equations X= -11(7)+19 = -58 7x+11y=100 Y= 7(7)-3 = 46 7(-58)+11(46)=100 -406+506=100 X= -11(10)+19 = -91 7x+11y=100 Y= 7(10)-3 = 67 7(-91)+11(67)= 100 -637+737=100