Question 1083260: Find all real numbers t such that 2/3t - 1 < t + 7 ≤ -2t + 15. Give your answer as an interval. Found 2 solutions by math_helper, Theo:Answer by math_helper(2461) (Show Source):
2/3 * t - 1 < t + 7
add 1 to both sides to get:
2/3 * t < t + 8
subtract t from both sides to get:
2/3 * t - t < 8
simplify to get:
-1/3 * t < 8
multiply both sides by 3 to get:
-t < 24
multiply both sides by -1 to get:
t > -24
multiplying both sides by -1 reverses the inequality.
solution for the first set of inequalities is t > -24
the second set of inequalities is:
t + 7 <= -2 * t + 15
subtract 7 from both sides to get:
t <= -2 * t + 8
add 2t to both sides to get:
3 * t <= 8
divide both sides by 3 to get:
t <= 8/3
solution for the second set of inequalities is t <= 8/3
your solution is that t > -24 and t <= 8/3
this can be written as -24 < t <= 8/3
in interval notation, this would be written as:
(-24,8/3]
you can test this out with various values of t to ensure the inequality is correct.
the inequality statement to test is:
2/3 * t - 1 < t + 7 <= -2t + 15
in order for the statement to be true, all parts of the inequality must be true.
2/3 * t - 1 must be smaller than t + 7 AND t + 7 must be smaller than or equal to -2t + 15
i checked and it all looks good, so i'm reasonably certain that the solution is good.
