anonymous
  • anonymous
Can someone please explain to me how you solve inequalities with a fraction on one side. For example: x/8+3<15. I do not know how to do this. By the way it is in Algebra II.
Algebra
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SOLVED
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schrodinger
  • schrodinger
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jdoe0001
  • jdoe0001
more or less same as you'd any linear equation, how would you solve say \(\bf \cfrac{x}{8}+3=15\)
anonymous
  • anonymous
I thought you would multiply x/8 by 8 to get rid of the denominator and then multiply 15 by 8 but I am not for sure.
jdoe0001
  • jdoe0001
\(\bf \cfrac{x}{8}+3<15\\ \textit{subtracting 3 from both sides}\\ \cfrac{x}{8}\cancel{+3-3}<15-3 \implies \cfrac{x}{8}<12\\ \textit{multiplying 8 to both sides}\\ \cfrac{x}{\cancel{8}}\times \cancel{8}<12 \times 8 \)

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anonymous
  • anonymous
Is this how you do all of them that have a fraction on one side even if the fraction is negative?
jdoe0001
  • jdoe0001
nope, that's the main difference in an inequality, if you multiply/divide/exponentialize both sides by a NEGATIVE value, then you'd need to \(\bf \text{flip the sign}\) lemme redo that for the sake of example, using a negative fraction \(\bf -\cfrac{x}{8}+3\color{blue}{<}15\\ \textit{subtracting 3 from both sides}\\ -\cfrac{x}{8}+3-3<15-3 \implies -\cfrac{x}{8}<12\\ \textit{multiplying -8 to both sides}\\ \cfrac{x}{8}\times -8<12 \times -8 \implies x \color{blue}{>} -96\)
jdoe0001
  • jdoe0001
woops, I missed the -... anyhow \(\bf -\cfrac{x}{8}\times -8<12 \times -8 \implies x \color{blue}{>} -96\)
anonymous
  • anonymous
but you would still subtract/add the whole number multiply by the denominator and then if its negative then flip the sign
jdoe0001
  • jdoe0001
the procedure is the same as in any equality, the difference is when the multiplier/divider/exponent is negative, otherwise yes
anonymous
  • anonymous
Okay and what about -1/4x-1<=-9
jdoe0001
  • jdoe0001
same thing
anonymous
  • anonymous
so itd be +1 and then multiply both sides by -1/4 and thatll give you x=> 2 right
jdoe0001
  • jdoe0001
\(\bf \cfrac{1}{4x}-1\le -9\\ \textit{adding 1 to both sides }\\ \cfrac{1}{4x}\cancel{-1+1}\le -9+1 \implies \cfrac{1}{4x}\le -8\\ \textit{multiplying both sides by 4x}\\ \cfrac{1}{\cancel{4x}}(\cancel{4x})\le -8(4x)\)
anonymous
  • anonymous
the 1/4 is negative
jdoe0001
  • jdoe0001
ohhh... ok
anonymous
  • anonymous
yeah
jdoe0001
  • jdoe0001
\(\bf -\cfrac{1}{4x}-1\le -9\\ \textit{adding 1 to both sides }\\ -\cfrac{1}{4x}\cancel{-1+1}\le -9+1 \implies -\cfrac{1}{4x}\le -8\\ \textit{multiplying both sides by -4x}\\ -\cfrac{1}{\cancel{4x}}(\cancel{-4x})\le -8(-4x)\)
anonymous
  • anonymous
Im still not getting the correct answer its coming out to -1/32 or -0.031
jdoe0001
  • jdoe0001
so that'd leave us with \(\bf 1 \ge 32x \implies \cfrac{1}{32} \ge x\) notice, on the -4x, negative, the sign flpped
jdoe0001
  • jdoe0001
well you see \(\bf -\cfrac{1}{4x}(-4x)\le -8(-4x) \implies \cfrac{1}{-4x}(-4x)\le -8(-4x)\)
anonymous
  • anonymous
Im not getting a whole number
jdoe0001
  • jdoe0001
yeap, doesn't have to be
anonymous
  • anonymous
my answers on the work sheet are x>-80, x>96, x>21, x>-54, x=>32, x=>-6, or x =>-91. I am doing a riddle work sheet called "What Happened When the King of Beasts Runs in Front of a Train?"
jdoe0001
  • jdoe0001
well, that just means your inequality is incorrect
anonymous
  • anonymous
Oh I dont know I will just have to ask my teacher then

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