Asked by Jones Foday
4/x+5>1/2x+3
Answers
Answered by
oobleck
assuming the usual carelessness with parentheses,
4/(x+5) > 1/(2x+3)
you can work out four separate solutions, choosing which expressions are positive or negative, or take a shortcut via
16/(x+5)^2 > 1/(2x+3)^2
16(2x+3)^2 > (x+5)^2
64x^2 + 192x + 144 > x^2 + 10x + 25
63x^2 + 182x + 119 > 0
9x^2 + 26x + 13 > 0
(9x+17)(x+1) > 0
x is in (-∞,-17/9) U (-1.∞)
But this does not take into account the fact that x ≠ -5 , -3/2
So the real solution is
(-∞,-5) U (-5,-17/9) U (-1.∞)
4/(x+5) > 1/(2x+3)
you can work out four separate solutions, choosing which expressions are positive or negative, or take a shortcut via
16/(x+5)^2 > 1/(2x+3)^2
16(2x+3)^2 > (x+5)^2
64x^2 + 192x + 144 > x^2 + 10x + 25
63x^2 + 182x + 119 > 0
9x^2 + 26x + 13 > 0
(9x+17)(x+1) > 0
x is in (-∞,-17/9) U (-1.∞)
But this does not take into account the fact that x ≠ -5 , -3/2
So the real solution is
(-∞,-5) U (-5,-17/9) U (-1.∞)
Answered by
mathhelper
I usually do these the following way:
consider the actual equation 4/(x+5) = 1/(2x+3)
easy to solve and x = -1
also x ≠ -5, -3/2
So we have the following critical parts of the domain for x
x < -5
-5 < x < -3/2
-3/2 < x < -1 , and
x > -1
pick a value in each section
for x < -5, let x = -6 .... is 4/-1 > 1/-9 ?? , NO
-5 < x < -3/2, let x = -3 ... is 4/2 > 1/-3 ?? , YES
-3/2 < x < -1, let x = -1.1 ... is 4/3.9 > 1/.8 ??, NO
finally, x > -1, let x = 10, ..... is 4/15 > 1/23 ?? YES
so we have:
-5 < x < -3/2 OR x > -1
graphing y = 4/(x+5) and y= 1/(2x+3) using DESMOS shows this is correct.
By squaring both sides, sometimes solutions are introduced which do
not satisfy the original relation. That is why all solutions must be verified after squaring.
The value of x = -17/9 does not show up when graphing
consider the actual equation 4/(x+5) = 1/(2x+3)
easy to solve and x = -1
also x ≠ -5, -3/2
So we have the following critical parts of the domain for x
x < -5
-5 < x < -3/2
-3/2 < x < -1 , and
x > -1
pick a value in each section
for x < -5, let x = -6 .... is 4/-1 > 1/-9 ?? , NO
-5 < x < -3/2, let x = -3 ... is 4/2 > 1/-3 ?? , YES
-3/2 < x < -1, let x = -1.1 ... is 4/3.9 > 1/.8 ??, NO
finally, x > -1, let x = 10, ..... is 4/15 > 1/23 ?? YES
so we have:
-5 < x < -3/2 OR x > -1
graphing y = 4/(x+5) and y= 1/(2x+3) using DESMOS shows this is correct.
By squaring both sides, sometimes solutions are introduced which do
not satisfy the original relation. That is why all solutions must be verified after squaring.
The value of x = -17/9 does not show up when graphing
Answered by
oobleck
well said; always check your answers against the original equations.
Unfettered cleverness will get you in trouble.
Unfettered cleverness will get you in trouble.
Answered by
DrBob222
Can't beat being unfettered no matter what !
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.