Bunuel wrote:
How many integer solutions exist for the equation 8x – 5y = 221 such that x*y < 0
A. 4
B. 5
C. 6
D. 7
E. 8
Project PS Butler
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Subscribe via RSS - RSS Are You Up For the Challenge: 700 Level QuestionsThis equation is of the form Linear Diophantine Equations - courtesy
AnthonyRitz -
https://www.youtube.com/watch?v=SCf-HrtYUMgFind the first pair for the equation 8x – 5y = 221 such that xy < 0
x has to be such that 8x-221 is divisible by 5 for y to be integer => Let's try x=2 => Will give us y=-41
Now we just need to add a value to x and subtract from y or vice versa (In this case we subtract value to x, thus add 5 i.e. -(-5) and add in y - +8)
1) x=2; y=-41
2) x=2+5 =7 (Because of -5 in y); y=-41+8 = -33 (Because of 8 in x)
3) x=7+5 =12 (Because of -5 in y); y=-33+8 = -25 (Because of 8 in x)
4) x=12+5 =17 (Because of -5 in y); y=-25+8 = -17 (Because of 8 in x)
5) x=17+5 =22 (Because of -5 in y); y=-17+8 = -9 (Because of 8 in x)
6) x=22+5 =27 (Because of -5 in y); y=-9+8 = -1 (Because of 8 in x)
If we add any further value to y => y will become positive and xy > 0 which is a constraint in the question stem
Hence 6 integer solutions
Answer - C