sloppycanada
  • sloppycanada
How to solve rational function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sloppycanada
  • sloppycanada
|dw:1435654672339:dw|
amoodarya
  • amoodarya
multiply both side by (3*4) to make it simpler
sloppycanada
  • sloppycanada
Soo.. x-4+x = 72?

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sloppycanada
  • sloppycanada
2x = 76 x = 38
amoodarya
  • amoodarya
no this is not correct!
amoodarya
  • amoodarya
denominators are different
kropot72
  • kropot72
Not so. Multiplying both sides by 12 gives us: \[\large \frac{12(x-4)}{4}+\frac{12x}{3}=72\]
kropot72
  • kropot72
Which simplifies to: \[\large 3x-12+4x=72\]
sloppycanada
  • sloppycanada
7x = 84
kropot72
  • kropot72
Correct. Now find the value of x.
sloppycanada
  • sloppycanada
x = 14
kropot72
  • kropot72
Not so. What is the value of 84/7 ?
sloppycanada
  • sloppycanada
Sorry 12.
kropot72
  • kropot72
Correct.
sloppycanada
  • sloppycanada
|dw:1435656598968:dw|
amoodarya
  • amoodarya
\[\frac{5x+1}{x}<1 \\\frac{5x+1}{x}-1<0 \\\frac{5x+1}{x}-\frac{x}{x}<0 \]
amoodarya
  • amoodarya
now go on
sloppycanada
  • sloppycanada
Multiply both sides by "x"?
amoodarya
  • amoodarya
no no no
amoodarya
  • amoodarya
if you are sure that "x>0" you can multiply by "x" but in this case we can not be sure about x>0 if you do it by multiply ,some part of answer is lost !
amoodarya
  • amoodarya
\[5x+1
amoodarya
  • amoodarya
\[\frac{4x+1}{x}<0\] |dw:1435665521336:dw|
amoodarya
  • amoodarya
\[first \space suppose \space x>0 \\\frac{5x+1}{x}<0 \rightarrow 5x+1o ,x<\frac{-1}{4} \] for the second part note that sign will change \[x<0 \\\frac{5x+1}{x}<1 \\mutilply \space by -x\\5x+1>x\\4x>-1\\x>\frac{-1}{4}\\now\\x>\frac{-1}{4} ,x<0 \rightarrow \frac{-1}{4}
sloppycanada
  • sloppycanada
That makes no sense to me...

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