SOLUTION: Hello, I could really use help with this problem: Find all possible pairs of numbers such that the first is at least 8 more than 4 times the second, and 12 less than three ti

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Question 44339: Hello,
I could really use help with this problem:
Find all possible pairs of numbers such that the first is at least 8 more than 4 times the second, and 12 less than three times the first is at most 12 more than 8 times the second.
Thank you in advance,
Louis

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
First Number: x
Second Number: y
Find all possible pairs of numbers such that the first is at least 8 more than 4 times the second,
x+%3E=+4y+%2B+8
and 12 less than three times the first is at most 12 more than 8 times the second.
3x+-+12+%3C=+8y+%2B+12
So....
3x+-+24+%3C=+8y
%283%2F8%29x+-+3+%3C=+y shade above line
And...
x+%3E=+4y+%2B+8
x+-+8+%3E=+4y
%281%2F4%29x+-+2+%3E=+y shade below line
Red Line: above
Green Line: below
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+%283%2F8%29x+-+3%2C+%281%2F4%29x+-+2+%29+
Answer: {x|x+%3C+8}
Answer: {y|y+%3C+0}
Possible Answers: (-3,-3),(-2,-3),(-10,-5)....