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#1 Before February 2015

32OrtonEdge32dh
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Joined: 2015-02-15
Posts: 5,166
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Do Some Math For a +rep

How much you wanna bet Jad tries to do it?

Anyway, I have a boatload of Algebra II homework, and there's one question I can't really complete.   Here it is:

My Book wrote:

Write and solve an inequality to find three consecutive whole numbers with a sum between 13 and 16.

SHOW YOUR WORK DAD GUM IT.

The use of absolute value is not necessary, but encouraged.

Last edited by 32OrtonEdge32dh (Sep 16 2011 3:34:00 pm)


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#2 Before February 2015

jakery
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Re: Do Some Math For a +rep

13 < x + (x+1) + (x+2) < 16

Did I win?

#3 Before February 2015

Greenzoid2
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Re: Do Some Math For a +rep

Hmm, seems like you guys use different terms in USA than in Canada, I don't know what an inequality is //forums.everybodyedits.com/img/smilies/tongue

#4 Before February 2015

32OrtonEdge32dh
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From: DMV
Joined: 2015-02-15
Posts: 5,166
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Re: Do Some Math For a +rep

Hmm...wanna solve it?


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#5 Before February 2015

jakery
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Re: Do Some Math For a +rep

Per IRC chat:

13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13

10/3 < x < 13/3

There. I robbed you of your education. lol

#6 Before February 2015

Gaming_Guy
Guest

Re: Do Some Math For a +rep

Algebra? No thanks.

Lemme have a go,

13 < x + (x+1) + (x+2) < 16
13 < x + (x+3) < 16
13 < (x+x)=2+3 < 16

Did I do something wrong? I fail at algebra.

#7 Before February 2015

32OrtonEdge32dh
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From: DMV
Joined: 2015-02-15
Posts: 5,166
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Re: Do Some Math For a +rep

Something went majorly wrong, there should never be an = in an inequality.


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#8 Before February 2015

Gaming_Guy
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Re: Do Some Math For a +rep

I don't know what an "inequality" is. I don't live in America therefor I don't know they're standards for math.

#9 Before February 2015

32OrtonEdge32dh
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Joined: 2015-02-15
Posts: 5,166
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Re: Do Some Math For a +rep


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#10 Before February 2015

calculusguy
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Re: Do Some Math For a +rep

jakery wrote:

Per IRC chat:

13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13

10/3 < x < 13/3

There. I robbed you of your education. lol

This.
But you need to make it whole numbers, so make 10/3 to 4 and 13/3 to 4... WHAT ARWOIHWIRUCHC
That means it has to be 4!?!?!

Last edited by calculusguy (Sep 16 2011 4:56:33 pm)

#11 Before February 2015

Baticon
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Re: Do Some Math For a +rep

This is an impossible problem IMO.
Look:

13 < (x-1) + x + (x+1) < 16        ------->        13 < 3x < 16

AND:

13 < x + (x+1) + (x+2) < 16        ------->        13 < 3x + 3 < 16        ------->        10 < 3x < 16

So:

13 < 3x < 16
10 < 3x < 13

Impossible.

#12 Before February 2015

JadElClemens
Member
From: Colorado, USA
Joined: 2015-02-15
Posts: 4,559

Re: Do Some Math For a +rep

32OrtonEdge32dh wrote:

How much you wanna bet Jad tries to do it?

Lolyeah

32OrtonEdge32dh wrote:

Algebra II

Bro, I'm only in Geometry.


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#13 Before February 2015

32OrtonEdge32dh
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From: DMV
Joined: 2015-02-15
Posts: 5,166
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Re: Do Some Math For a +rep

baticon wrote:

This is an impossible problem IMO.
Look:

13 < (x-1) + x + (x+1) < 16        ------->        13 < 3x < 16

AND:

13 < x + (x+1) + (x+2) < 16        ------->        13 < 3x + 3 < 16        ------->        10 < 3x < 16

So:

13 < 3x < 16
10 < 3x < 13

Impossible.

Well, jakery did it....   How is it impossible?
13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13
10/3 < x < 13/3
You didn't isolate the variable.


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#14 Before February 2015

musuki
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Re: Do Some Math For a +rep

baticon wrote:

13 < 3x < 16
10 < 3x < 13

Impossible.

x isn't the same thing in those two though. In the top one x is the middle number, whereas in the bottom one x is the first number. If you're consistent with calling x the first number, the top one is really 13 < 3(x+1) < 16, which is clearly the same as the second inequality.

#15 Before February 2015

jakery
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Re: Do Some Math For a +rep

Baticon, your first inequality will work as well, but you have to remember that you subtracted 1 from each of the three consecutive numbers, so you must subtract three from the other two sides of the inequality:

13 < x + (x+1) + (x+2) < 16
10 < (x-1) + x + (x+1) < 13

And from there, you still get lead to:
10 < 3x < 13

#16 Before February 2015

catking00
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Re: Do Some Math For a +rep

look it up!
x
x + 1
x + 2

Equation: 13 < x + (x + 1) + (x + 2) < 16

Simplify the middle part first:
13 < x + x + 1 + x + 2 < 16
13 < 3x + 3 < 16

Now subtract 3 from all three sections
13 - 3 < 3x + 3 - 3 < 16 - 3

10 < 3x < 13

Divde through by 3
10/3 < x < 13/3
3.3333 < x < 4.33333

The only whole number between 3.333 and 4.3333 is 4, so the first number will be 4

x = 4
x + 1 = 5
x + 2 = 6

CHECK:
4 + 5 + 6 = 15......this is between 13 and 16

Answer: 4, 5, 6
Source(s):
Math teacher for many years

found this on yahoo answers.

#17 Before February 2015

calculusguy
Guest

Re: Do Some Math For a +rep

calculusguy wrote:
jakery wrote:

Per IRC chat:

13 < x + (x+1) + (x+2) < 16
13 < 3x + 3 < 16
10 < 3x < 13

10/3 < x < 13/3

There. I robbed you of your education. lol

This.
But you need to make it whole numbers, so make 10/3 to 4 and 13/3 to 4... WHAT ARWOIHWIRUCHC
That means it has to be 4!?!?!

Y u no read?

#18 Before February 2015

tailschao
Guest

Re: Do Some Math For a +rep

ok, I love math, but jeeze. I am only 11. I did'nt know half of the words in that problem XD

#19 Before February 2015

Betaguy
Guest

Re: Do Some Math For a +rep

Ok I wish i could solve this, but the problem is I am Norwegian (dont know what the some of the words mean), 13 years, and I havent had algebra at school yet >.<

#20 Before February 2015

Ratburntro44
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Joined: 1970-01-01
Posts: 1,383
Website

Re: Do Some Math For a +rep

baticon wrote:

This is an impossible problem IMO.
Look:

13 < (x-1) + x + (x+1) < 16        ------->        13 < 3x < 16

AND:

13 < x + (x+1) + (x+2) < 16        ------->        13 < 3x + 3 < 16        ------->        10 < 3x < 16

So:

13 < 3x < 16
10 < 3x < 13

Impossible.

Uhh... not really... one of your x values is different from the other.

The first x is one more than the second x. So your math is off...

And this has already been solved here, easily.

13<x+x+1+x+2<16
13<3x+3<16
10<3x<13
10/3<x<13/3
3.33<x<4.33
x must be a whole number, so x is 4. 4+5+6=15. Yay. Easy. Not to mention it already was solved before this...

Well, that's the best a 12 year old in Calc 1 can do...

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#21 Before February 2015

RhazzleFrazzle
Member
Joined: 2015-11-10
Posts: 4,260

Re: Do Some Math For a +rep

damn. why an inequality. thats the only thing i cant do. i can tell you that the consectutive whole numbers are 4 5 and 6, but i cant do the inequality.

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#22 Before February 2015

32OrtonEdge32dh
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Joined: 2015-02-15
Posts: 5,166
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Re: Do Some Math For a +rep

It's already been answered, though.


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#23 Before February 2015

RhazzleFrazzle
Member
Joined: 2015-11-10
Posts: 4,260

Re: Do Some Math For a +rep

oh. well. i took algebra 2 last year, and that class was torture. i know how it feels to have to use the internet to get work done.

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#24 Before February 2015

Chewy
Banned

Re: Do Some Math For a +rep

Locked on request of OP.

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