Mathematics
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OpenStudy (anonymous):
Solve each inequality 8<2v-4<16
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jimthompson5910 (jim_thompson5910):
the idea is to isolate v
jimthompson5910 (jim_thompson5910):
so we first add 4 to all sides
then we divide all sides by 2
OpenStudy (anonymous):
Wait ill tey to answer
jimthompson5910 (jim_thompson5910):
alright
OpenStudy (anonymous):
6<v<10?
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jimthompson5910 (jim_thompson5910):
good, here are the steps if you're curious
\[\large 8 < 2v-4 < 16\]
\[\large 8+4 < 2v-4+4 < 16+4\]
\[\large 12 < 2v+0 < 20\]
\[\large 12 < 2v < 20\]
\[\large \frac{12}{2} < \frac{2v}{2} < \frac{20}{2}\]
\[\large 6 < \frac{1v}{1} < 10\]
\[\large 6 < 1v < 10\]
\[\large 6 < v < 10\]
jimthompson5910 (jim_thompson5910):
I'm sure you did all that
OpenStudy (anonymous):
Yeah im right. But i did not that 1 i just cancel it out
jimthompson5910 (jim_thompson5910):
that's ok, I put more steps than needed really to be really technical
OpenStudy (anonymous):
-7d <_4d-3<_-1
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jimthompson5910 (jim_thompson5910):
hint: use the idea that
x < y < z
breaks up into
x < y and y < z
OpenStudy (anonymous):
How can i use that in this equation whats the y?
jimthompson5910 (jim_thompson5910):
-7d <_4d-3<_-1
breaks up into
-7d <_4d-3 and 4d-3<_-1
jimthompson5910 (jim_thompson5910):
solve each for d
OpenStudy (anonymous):
I dnt know how to solve this one its complicated
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jimthompson5910 (jim_thompson5910):
give it a shot
jimthompson5910 (jim_thompson5910):
tackle each one on its own
OpenStudy (anonymous):
Ill try
OpenStudy (anonymous):
d>_d<_-4?
jimthompson5910 (jim_thompson5910):
hmm that seems off
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OpenStudy (anonymous):
Dnt know pls explain thnx :)
jimthompson5910 (jim_thompson5910):
what do you get when you solve -7d <_4d-3
OpenStudy (anonymous):
-d<d
jimthompson5910 (jim_thompson5910):
hint:
-7d <_4d-3
-7d -4d <_ -3
OpenStudy (anonymous):
d<_-3?
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jimthompson5910 (jim_thompson5910):
you're missing something on the left side
OpenStudy (anonymous):
-3?
jimthompson5910 (jim_thompson5910):
no -7d - 4d = ??
OpenStudy (anonymous):
-11
OpenStudy (anonymous):
How about the other side?
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jimthompson5910 (jim_thompson5910):
so we have -11d <_ -3
now divide both sides by -11 to isolate d, this will flip the inequality sign
OpenStudy (anonymous):
<_ 3.66
OpenStudy (anonymous):
>_ imean
jimthompson5910 (jim_thompson5910):
leave it as a fraction
OpenStudy (anonymous):
I got 3.6
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jimthompson5910 (jim_thompson5910):
leave it as a fraction to get
d >= 3/11
jimthompson5910 (jim_thompson5910):
now solve the other piece
OpenStudy (anonymous):
Ow i got >_ 2d
jimthompson5910 (jim_thompson5910):
solve 4d-3<_-1 for d
OpenStudy (anonymous):
Yeah i put 3 in both sides
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OpenStudy (anonymous):
Then the answer was divided into 4 so i got 2
jimthompson5910 (jim_thompson5910):
2/4 is not 2 though
OpenStudy (anonymous):
Why
jimthompson5910 (jim_thompson5910):
4/2 = 2
not the other way around
jimthompson5910 (jim_thompson5910):
4/2 = 2
2/4 = ???
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OpenStudy (anonymous):
.5
jimthompson5910 (jim_thompson5910):
or 1/2
jimthompson5910 (jim_thompson5910):
so d <= 1/2
jimthompson5910 (jim_thompson5910):
so all together, we know this
d >= 3/11 and d <= 1/2
put together, the final answer is
\[\large \frac{3}{11} \le d \le \frac{1}{2}\]
OpenStudy (anonymous):
So ill combine and thats the final asnwer
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jimthompson5910 (jim_thompson5910):
yes
OpenStudy (anonymous):
This is the equation i cant understand
jimthompson5910 (jim_thompson5910):
ok last one
OpenStudy (anonymous):
6y-3<27 or -4y+2<-26 what ir means
jimthompson5910 (jim_thompson5910):
you solve each inequality for y
so what do you get when you solve 6y-3<27
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OpenStudy (anonymous):
Hold on a sec
OpenStudy (anonymous):
Y<5
jimthompson5910 (jim_thompson5910):
good
jimthompson5910 (jim_thompson5910):
solve -4y+2<-26 for y
OpenStudy (anonymous):
Y<-6
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jimthompson5910 (jim_thompson5910):
no that's incorrect
OpenStudy (anonymous):
7?
OpenStudy (anonymous):
-7?
OpenStudy (anonymous):
Is that right?
jimthompson5910 (jim_thompson5910):
it should be y > 7
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jimthompson5910 (jim_thompson5910):
so after solving 6y-3<27 or -4y+2<-26 , we get this
y < 5 or y > 7
OpenStudy (anonymous):
So thats the final answer?
jimthompson5910 (jim_thompson5910):
yes
OpenStudy (anonymous):
Thank you so much
jimthompson5910 (jim_thompson5910):
you're welcome