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About LinkBacksI am not sure the steps to solving this problem.
1/(t-5) < 3/(t+5)
Would someone please post how to solve this problem?
Thank you in advance.
Here is one way.Originally Posted by bobbysgirl
Since inequalities are like equations except that we have to reverse the sense of the inequality when multipying or dividing both sides of the inequality by a negative quantity, then we investigate when (t-5) and/or (t+5) is/are negatives.
a) when (t-5) is negative, t -5 < 0 -----so, t < 5.
b) when (t+5) is negative, t +5 < 0 --------t < -5
c) when (t -5) is positive, t -5 > 0 -----------------t > 5
d) when (t +5) is positive, t +5 > 0 -----------------t > -5
So t = -5 and t = 5 are critical values of t. [In fact, t cannot be -5 nor 5. Division by zero is not allowed.]
Hence we investigate the intervals (t < -5), (-5 < t < 5) and (t > 5).
------------------------------------------------
i) When t < -5, or t is to the left of -5.
(t-5) is negative. -----see a) above.
(t+5) is negative also. -----see b) above.
So the Inequality goes:
1/["-"(t-5)] < 3/["-"(t+5)]
Cross multiply.
The sense reverses after one multiplication by negative, then reverses again after multiplication by the other negative, so it ends up that the newest sense is back to the original sense, which is " < ".
Thus,
(t+5) < 3(t-5)
t +5 < 3t -15
20 < 2t
10 < t
Or, t > 10 -----------not true, because t < -5 here.
So when t < -5, there is no solution for the Inequality.
----------------------------------------------
ii) When -5 < t < 5
(t-5) is negative. -----see a) above.
(t+5) is positive. -----see d) above.
So the Inequality goes:
1/["-"(t-5)] < 3/(t+5)
Cross multiply. The sense revereses, thus,
(t+5) > 3(t-5)
t +5 > 3t +15
20 > 2t
10 > t
Or, t < 10 -----------true, because t max here is only close to 5.
Therefore, the interval -5 < t < 5 is a solution of the Inequality. -------***
-----------------------------------------------------------------
iii) When t > 5, or t is to the right 5.
(t-5) is positive. -----see c) above.
(t+5) is positive also. -----see d) above.
So,
1/(t-5) < 3/(t+5)
Cross multiply.
(t+5) < 3(t-5)
t +5 < 3t -15
20 < 2t
10 < t
Or, t > 10 -----------not true, because if t = 6, t > 5, okay, but not t > 10.
So when t > 5, there is no solution for the Inequality. -----**
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We are not done yet.
We keep finding t < 10 or t > 10 in the computations. Could be that t = +,-10 are critical values of t also.
t = -10 is to the left of t = -5. We found out already that to the left of t = -5, there are no solutions for the Inequality. But is that final? What about those values of t to the left of t = -10?
Say, t = -11,
1/(t-5) < 3/(t+5)
1/(-11 -5) <? 3/(-11 +5)
1/(-16) <? 3/(-6)
1/16 <? 1/2
Yes!
Therefore, t < -10 is a solution to the Inequlity. --------------***
What about t > 10?
Say, t = 11,
1/(11 -5) <? 3/(11 +5)
1/6 <? 3/16
Cross multiply,
1*16 <? 6*3
16 <? 18
Yes!
Therefore, t > 10 is also a solution to the Inequlity. --------------***
So, the solutions of the Inequality are
t < -10, -5 < t < 5, and t > 10 ------------------answer.
Again, check when x < -10
1/(t -5) < 3/(x+5)
If t = -20,
1/(-25) <? 3/(-15)
-1/25 <? -1/3
-0.04 <? -0.3333
No.
Aha!
You are right.
My mistake was when I checked when t = -11, I automatically removed the negative signs from both sides, forgetting the rule that I should have reversed the sense....
We keep finding t < 10 or t > 10 in the computations. Could be that t = +,-10 are critical values of t also.
t = -10 is to the left of t = -5. We found out already that to the left of t = -5, there are no solutions for the Inequality. But is that final? What about those values of t to the left of t = -10?
Say, t = -11,
1/(t-5) < 3/(t+5)
1/(-11 -5) <? 3/(-11 +5)
1/(-16) <? 3/(-6)
1/16 <? 1/2
Yes!
Therefore, t < -10 is a solution to the Inequlity. --------------***
If I reviewed that correctly,
the "1/16 <? 1/2"
should have been "1/16 >? 1/2"
which is No, and therefore the t < -10 is not a solution to the Inequality as you've found in your graph.
Oh well, no detailed reviewing and no graph to check on my part.
It's okay.
It happens.
Hello, bobbysgirl!
Another approach . . .
. . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . 3
. . . 1 . . . . . . .3
. . ------ . < . ------
. . t - 5 . . . . t + 5
First, we solve the equation: . ------- . = . -------
. . . . . . . . . . . . . . . . . . . . . . t - 5 . . . . .t + 5
Note that: .t ≠ 5 and t ≠ -5 . . . These are "critical values".
Solve for t: .t + 5 .= .3(t - 5) . → . t = 10 . . . another critical value
Plot the critical values on a number line.The three points divide the number line into four intervals.Code:- - - - * - - - - - - - - * - - - * - - - - -5 5 10
Test a value from each interval in the original inequality.
. . On (-∞, -5), test x = -6: .-1/11 < -3 . . . not true
. . On (-5, 5), test x = 0: . -1/5 < 3/5 . . . . true
. . On (5, 10), test x = 6: .1 < 3/11 . . . . . .not true
. . On (10, ∞), test x = 11: .1/6 < 3/16 . . . true
Hence, the inequality is true on the intervals: .(-5, 5), (10, ∞)
Therefore: .-5 < x < 5 .and .x > 10
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