anonymous
  • anonymous
Solve the following inequality. write the solution set using interval notation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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anonymous
  • anonymous
thats the question and how the answer must be.
jim_thompson5910
  • jim_thompson5910
Multiply both sides by 40 to clear out the fractions \[\Large \frac{6x+2}{8}-\frac{2x-8}{5} \le -4\] \[\Large 40*\left(\frac{6x+2}{8}-\frac{2x-8}{5}\right) \le 40*(-4)\] \[\Large 40*\left(\frac{6x+2}{8}\right)+40*\left(-\frac{2x-8}{5}\right) \le 40*(-4)\] \[\Large 5*\left(6x+2\right)-8*\left(2x-8\right) \le -160\] Do you see how to finish up?

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More answers

anonymous
  • anonymous
why is it 40? and i would distribute the 5 out into the 6X+2 then the 8 into the 2x-8 and keep going until i can take the left side and either divide the right by it or the left by the right.
jim_thompson5910
  • jim_thompson5910
because 8*5 = 40 is the LCD
jim_thompson5910
  • jim_thompson5910
if you multiply both sides by the LCD, then you clear out the fractions
anonymous
  • anonymous
ohhhhhhhhhhh ok. am i correct in the final steps?
jim_thompson5910
  • jim_thompson5910
` i would distribute the 5 out into the 6X+2 then the 8 into the 2x-8` correct `keep going until i can take the left side and either divide the right by it or the left by the right` these steps seem a bit vague to me
anonymous
  • anonymous
if the number on the left of the less than or equal to sign is less than 160 divide 160 by it. if it is more than 160 than divide that number by 160?
jim_thompson5910
  • jim_thompson5910
when you distributed, what did you get?
anonymous
  • anonymous
30x+10-16x-64≤-160
jim_thompson5910
  • jim_thompson5910
the -64 should be +64
anonymous
  • anonymous
14x-54≤-160
jim_thompson5910
  • jim_thompson5910
since -8 times -8 = 64
anonymous
  • anonymous
+54**
anonymous
  • anonymous
didnt change the sign
jim_thompson5910
  • jim_thompson5910
after distributing you should get `30x+10-16x+64≤-160` which simplifies to `14x+74≤-160` after you combine like terms
anonymous
  • anonymous
14x≤-214
anonymous
  • anonymous
oh i messed up somewhere
anonymous
  • anonymous
had to do the +64
anonymous
  • anonymous
14x≤234?
jim_thompson5910
  • jim_thompson5910
close but no
jim_thompson5910
  • jim_thompson5910
you lost a sign somewhere, it should be `14x≤-234`
anonymous
  • anonymous
ah i put it in my calculator and forgot the negative!
anonymous
  • anonymous
x=-16.4142857?
jim_thompson5910
  • jim_thompson5910
leave it as a fraction and reduce as much as possible
anonymous
  • anonymous
is it a fraction? i got a decimal when i divided
jim_thompson5910
  • jim_thompson5910
234/14 can be reduced to ???
jim_thompson5910
  • jim_thompson5910
hint: find the GCF of 234 and 14
anonymous
  • anonymous
GCF=2?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
117/7
jim_thompson5910
  • jim_thompson5910
divide each part of the fraction by 2 234/2 = 117 14/2 = 7 so 234/14 = 117/7
jim_thompson5910
  • jim_thompson5910
good
anonymous
  • anonymous
so what would the solution set be?
jim_thompson5910
  • jim_thompson5910
so -234/14 = -117/7
jim_thompson5910
  • jim_thompson5910
in the end, the solution would be \[\Large x \le -\frac{117}{7}\] what is this in interval notation?
anonymous
  • anonymous
[-117,7]?
jim_thompson5910
  • jim_thompson5910
draw a number line with -117/7 on it |dw:1443394703025:dw|
jim_thompson5910
  • jim_thompson5910
the inequality \(\LARGE x \le -\frac{117}{7}\) means that we shade to the left of -117/7 on the number line |dw:1443394759997:dw|
jim_thompson5910
  • jim_thompson5910
so the interval would start at -infinity, then stop at -117/7 giving us \[\Large \left. (-\infty, -\frac{117}{7}\right]\]
anonymous
  • anonymous
(-inf, -117/7)
jim_thompson5910
  • jim_thompson5910
|dw:1443394823713:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1443394850880:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1443394878344:dw|
anonymous
  • anonymous
is it a ] because -117/7 is included?
jim_thompson5910
  • jim_thompson5910
|dw:1443394904462:dw|
jim_thompson5910
  • jim_thompson5910
`is it a ] because -117/7 is included?` correct
jim_thompson5910
  • jim_thompson5910
and there is a closed circle at -117/7 on the number line to tell readers to include the endpoint |dw:1443394959903:dw|
anonymous
  • anonymous
ok so that should be my answer than. Thanks! i understand it all now. and i know to use the inf.
jim_thompson5910
  • jim_thompson5910
you're welcome

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