Find all real numbers t such that (2/3)t- 1 < t + 7 is < or = to -2t + 15. Give your answer as an interval.
(2/3)t- 1 < t + 7 ≤ -2t + 15
(2/3)t - 1 < t+7 AND t + 7 ≤ -2t + 15
2t - 3 < 3t + 21 AND 3t ≤ 8
-t < 24 AND t ≤ 8/3
t > -24 AND t ≤ 8/3
-24 < t ≤ 8/3
change this to whatever notation asked for.
Thank you!
To solve this inequality, we'll work step by step:
Step 1: Let's simplify the inequality by combining like terms:
(2/3)t - 1 < t + 7 ≤ -2t + 15
Step 2: To eliminate the fraction and make it easier to work with, let's multiply all terms by 3 to clear the denominator:
3 * [(2/3)t - 1] < 3 * (t + 7) ≤ 3 * (-2t + 15)
This simplifies to:
2t - 3 < 3t + 21 ≤ -6t + 45
Step 3: Now, let's collect all the terms with "t" on one side of the inequality and the constants on the other side:
2t - 3 - 3t < 3t + 21 - 3t ≤ -6t + 45 + 3t
Simplifying further:
-t - 3 < 21 ≤ -3t + 45
Step 4: Now, let's isolate "t" in each inequality separately:
-t < 21 + 3
-t < 24 (adding 3 to both sides)
To eliminate the negative sign, we'll multiply both sides by -1, which reverses the inequality:
t > -24 (changing the direction of the inequality)
Next:
21 ≤ -3t + 45
Subtracting 45 from both sides:
21 - 45 ≤ -3t
-24 ≤ -3t
Now, dividing both sides by -3 (remembering to reverse the inequality since we're dividing by a negative number):
8 ≥ t
So, we have found two different inequalities:
t > -24
8 ≥ t
Step 5: Combining the two solutions:
Since we are looking for common solutions, we need to find the intersection between both intervals. From the inequalities, we know that t has to be greater than -24 and less than or equal to 8. In interval notation, this can be written as:
(-24, 8]
Therefore, the interval representing all real numbers t that satisfy the given inequality is (-24, 8].