anonymous
  • anonymous
Solve the inequality (x/2) > (-8)/(x + 4) + 3 for domain
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
is the "3" in the denominator??
anonymous
  • anonymous
The 3 is added on after the fraction so I suppose its + 3/1
amistre64
  • amistre64
x -8 -- > ----- + 3 2 x+4 get common denoms...

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amistre64
  • amistre64
(x^2 +4x) > -16 + 6x +24 (x^2 +4x) > 6x + 8 x^2 +4x -6x > 8 x^2 -2x >8 x^2 -2x -8 > 0
amistre64
  • amistre64
(x-4)(x+2) >0 when x = -2 or 4; the equation = 0; so lets draw a number line: <.......-2...........4..........> - + + - - + --------------------- + - +
amistre64
  • amistre64
this tells us that when: x = -2, everything to the left is (-) and to the right is (+) x = 4, everything to the left is (-) and to the right is (+) We multiply the signs together to get the bottom stuff; our answer is everything that is (+); becasue that is greater than zero. (-inf,-2) OR (4,inf)
anonymous
  • anonymous
hmmm that doesnt appear to be one of the answers but ill double check the work
amistre64
  • amistre64
try x= -2 (-2/2) > (-8)/(-2 + 4) + 3 -1 > -8/2 + 3 -1 > -4 + 3 -1> -1... -2 is one of the "options" there :)
amistre64
  • amistre64
try x= 4 (4/2) > (-8)/(4 + 4) + 3 2 > -8/8 + 3 2 > -1 + 3 2> 2... 4 is the other "option" there :)
amistre64
  • amistre64
we know zero is between them, so lets see if we include that or exclude that area... try x= 0 (0/2) > (-8)/(0 + 4) + 3 0 > -8/4 + 3 0 > -2 + 3 0> -1 is a false statement so the area that has "0" in it is thrown out; that only leaves: (-inf,-2) U (4,inf)
anonymous
  • anonymous
These are the answers available: A. (- inf, -4) U (-2, 4) B. (- inf, -4) U [-2, 4] C. (-4, -2] U [4, inf) D. (- inf, -4] U [-2, 4) E. (-4, -2) U (4, inf)
amistre64
  • amistre64
Check the problem, did you typo a sign?
anonymous
  • anonymous
x -8 --- > ---- + 3 2 x+4
amistre64
  • amistre64
i see an error on my part :) for starters 0 IS > -1, so zero is a solution
amistre64
  • amistre64
never mind that comment, its my senility taking riit :)
amistre64
  • amistre64
root... its like my fingers are having a stroke ....
amistre64
  • amistre64
there is no "equal to" line in the equaiton, so it aint gonna have any [...] in it;
amistre64
  • amistre64
I think it narrows down to "E" -6/2 = -3 -3 > -8 + 3 -3 > -5 is true, so -6 is a solution so "E" is wrong. You got something messed up in either the answers or the questions; because the 2 dont match up.... unless you can see an error in my calculations
amistre64
  • amistre64
and -4 is not an option; becasue that zeros out the denom....
anonymous
  • anonymous
I plugged it into my calculator and im trying to eye it from the graph
amistre64
  • amistre64
no -4 and no 0 A. -------------- B. --------------- C. (-4, -2] U [4, inf) D. -------------- E. (-4, -2) U (4, inf)
amistre64
  • amistre64
there is no "greater than or EQUAL TO" sign in it so no "[4" 4/2 > -8/8 +3 2> -1 +3 2>2 ...... bad statement A. -------------- B. --------------- C. ------------- D. -------------- E. (-4, -2) U (4, inf) test for "E"?
amistre64
  • amistre64
x = -6 -6/2 = -8(-6+4) + 3 -3> -8/-2 + 3 -3 > 4+3 = 7...... -6 is a bad number and anything on that end... the answer is "E" then
anonymous
  • anonymous
I plugged it into my calculator and im trying to eye it from the graph
anonymous
  • anonymous
I keep retrying it but it keeps turning out wrong! grrr
amistre64
  • amistre64
use a pencil and paper ;)
anonymous
  • anonymous
I did, ive solved it 4 times now :/
amistre64
  • amistre64
we know that -2 and 4 are boundaries; we know that zero is not an option; we know that {4 is not an option becasue there is no ">=" in the equation; that leaves "E" as the only possible shot. the only thing we need to determine is if it ends at -4 or not; so plug in a -6 and see if it hold :)
amistre64
  • amistre64
-6 is bad value, so "E" is correct.

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