How do you solve #8- 3r \leq 5- 2r#?

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Answer 1 · 2016-12-14T14:08:39

#r >= 3#

First, do the necessary arithmetic to isolate the #r# term on one side of the inequality and the constants on the other side of the inequality:

#8 - 3r + 3r - 5 <= 5 - 5 - 2r + 3r#

#8 - 0 - 5 <= 0 - 2r + 3r#

#8 - 5<= r#

#3 <= r#

To get this solution in terms of #r# we need to reverse or "flip" the inequality:

#r >= 3#