Arithmetic

=Arthmetic=

Aim: 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.

7x+11y=100

Working out:

Here I listed the multiples of 7 plus the number they need to make a 100. I highlighted 56+44=100 because from all, 44 is the only number which is multiple of 11

7+93=100 14+86=100 21+79=100 28+72=100 35+65=100 42+58=100 49+51=100 56+44=100 63+37=100 70+30=100 77+23=100 84+16=100 91+9=100 98+2=100

7x8=56 and 11x4=44 so 7(8)+11(4)

Now I'm gonna try it, this time with negative numbers -7+107=100 -14+114=100 -21+121=100 -28+128=100 -35+135=100 -42+142=100 -49+149=100 -56+156=100 -63+163=100 -70+170=100 -77+177=100 -84+184=100 -91+191=100 -98+198=100 -105+205=100 -112+212=100 -119+219=100 -126+226=100 -133+233=100 -147+247=100 -154+254=100 -161+261=100 -168+268=100 -175+275=100

X Y 30 -10 19 -3 8 4 -3 11 -14 18 -25 25

the difference between the numbers in X is 11 (eg. 30-19=11, 19-8=11...) And the difference between the numbers in Y is 7(eg. -10- -3= 7, -3 - 4=7...) Now Im Using 8 and 4 because, from all the numbers, it's the only time that both numbers are positive. I'ts easier

Rule: X= -11n+19 Y= 7n-3

Checking:

1. X: -11(3)+19=-14 Y: 7(3)-3=18 7(-14)=-98 11(18)=198 -98+198=100

2. X:-11(7)+19=-58 Y: 7(7)-3=46 7(-58)=-406 11(46)=506 -406+506=100

as you can see, IT WORKS! :)