sloppycanada
  • sloppycanada
Five rational function equations to help me with, if you'd be oh so kind. (5x+1)/x < 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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sloppycanada
  • sloppycanada
|dw:1435669658962:dw|
anonymous
  • anonymous
Ok i have an idea but i might be wrong so hopefully this might help. https://www.khanacademy.org/math/algebra2/rational-expressions/solving-rational-equations/v/solving-rational-equations-2
anonymous
  • anonymous
Im sorry ; ~;

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More answers

UnkleRhaukus
  • UnkleRhaukus
\[\frac{5x+1}{x}<1\\ \frac{5x}x+\frac1x<1\\ \quad\dots\quad<1\]
UnkleRhaukus
  • UnkleRhaukus
the first term simplifies . . .
sloppycanada
  • sloppycanada
5x^2 + 1/x <1?
UnkleRhaukus
  • UnkleRhaukus
not quite, the x's in the first term dont multiply, they cancel
sloppycanada
  • sloppycanada
Oh. Okay so 5 + 1/x <1?
UnkleRhaukus
  • UnkleRhaukus
good, now the next step is to take away 5 form both sides
UnkleRhaukus
  • UnkleRhaukus
*from
sloppycanada
  • sloppycanada
1/x < -4
UnkleRhaukus
  • UnkleRhaukus
yes
UnkleRhaukus
  • UnkleRhaukus
now if we multiply both sides by x . . .
sloppycanada
  • sloppycanada
1 = -4x So x = -1/4
UnkleRhaukus
  • UnkleRhaukus
we get -1/4 < x but this is because we assumed that x was not negative (we didn't swap the direction of inequality)
UnkleRhaukus
  • UnkleRhaukus
if we had assumed that x was negative, we would have switched the direction of inequality, when we multiplied both sides by x, we would have gotten -1/4 > x
UnkleRhaukus
  • UnkleRhaukus
hmmm,
sloppycanada
  • sloppycanada
Okay thanks. For this next one (assuming we're good here), wouldn't I just want to find the quadratic parts?
sloppycanada
  • sloppycanada
|dw:1435671020481:dw|
UnkleRhaukus
  • UnkleRhaukus
im not sure what our final answer should be for the first one
sloppycanada
  • sloppycanada
If we used -1/4 as our answer, I get the 1 on my calculator.
UnkleRhaukus
  • UnkleRhaukus
but is x less or greater? than -1/4
sloppycanada
  • sloppycanada
According to the original problem, the answer (once you plug in the value for x) should be less than one.
UnkleRhaukus
  • UnkleRhaukus
oh, ok
sloppycanada
  • sloppycanada
So for the first question, I think the answer would just be -1/4 < x Which would mean anything smaller than -1/4 right?
UnkleRhaukus
  • UnkleRhaukus
x < -1/4 says all values of x that are less than -1/4
Loser66
  • Loser66
If \(\dfrac{a}{b}<1\) , that is b>a hence, in your case, x > 5x+1 that gives us \(x<-\dfrac{1}{4}\). Dat sit
sloppycanada
  • sloppycanada
So I think that would be right... let me plug in -1/5.. that equals "0" which means x < -1/4 as our final answer right?
UnkleRhaukus
  • UnkleRhaukus
that works!
sloppycanada
  • sloppycanada
|dw:1435671952655:dw| Next question, if you have time
sloppycanada
  • sloppycanada
|dw:1435672689916:dw|
Loser66
  • Loser66
a/b <0 iff a, b are different sign. That is if a >0, then b <0 and vice versa solve for them.
Loser66
  • Loser66
got it?
sloppycanada
  • sloppycanada
Yeah but it's =<
Loser66
  • Loser66
|dw:1435673140046:dw| for this, not for the last one

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