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graph this problem . 5s – 3 + 1 < 8
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Alright, we can simplify the expression down to 5s<10, s<2, through a bit of algebraic manipulation. s<2 can be represented on a line graph by a hollow circle drawn over the point s=2, and with a line extending in the negative direction.
does a hollow circle look like this ... |dw:1330106693703:dw|
Yes. The reason why you draw a hollow circle is because s is LESS THAN two, but not LESS THAN AND EQUAL TO two. Therefore you must specify that s cannot exist at that point.
5s – 3 + 1 < 8 -> 5s < 8 +2 => s < 10/5 = 2
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