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3(t − 2) ≥ 9(t + 2) help please
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if you need to find t, distribute the 3 and the 9 to get 3t-6 > 9t+18 then add 6 to both side so 3t-6+6 = 3t > 9t + 24 subtract 9t from both side and get -6t > 24 and now divide by -6 you get t > -4
You need to change the direction of the greater than sign when you multiply or divide by a negative number. So the answer would be \[t \le-4\]
3(t − 2) ≥ 9(t + 2) t − 2 ≥ 3(t + 2) t - 2 ≥ 3t + 6 2t <= -8 => t <= -8/2 = -4
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