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Mathematics 20 Online
OpenStudy (anonymous):

I need help! 12x-5/12 + 3-x/4 < 4/3 *** supposed to be a less than or equal to, but I don't have that sign.

OpenStudy (phi):

you can use <= or less than or equal

OpenStudy (anonymous):

Okay

OpenStudy (phi):

is it \[ \frac{12x-5}{12} + \frac{3-x}{4} \le \frac{4}{3} \] ? I would multiply both sides by 12 (that means *every* term on both sides) as a first step. can you do that ?

OpenStudy (anonymous):

Yes that's correct

OpenStudy (anonymous):

So like distribute a 12 in?

OpenStudy (phi):

yes, but on both sides

OpenStudy (anonymous):

12x-5+3 (3-x) <= 16?

OpenStudy (phi):

yes, now distribute the 3

OpenStudy (anonymous):

12x-5+9-3x <=16

OpenStudy (phi):

now combine "like terms"

OpenStudy (phi):

12 x's take away 3 x's and -5+9

OpenStudy (anonymous):

9x+4 <=16

OpenStudy (phi):

now add -4 to both sides

OpenStudy (anonymous):

9x <=12?

OpenStudy (phi):

yes. now divide both sides by 9 (remember it's ok to divide by positive numbers)

OpenStudy (anonymous):

Ok

OpenStudy (phi):

\[ \frac{9x}{9} \le \frac{12}{9} \] and 9/9 on the left side simplifies to 1

OpenStudy (phi):

and we can simplify 12/9 (divide top and bottom by 3)

OpenStudy (anonymous):

Uhm but I think the answer is supposed to be x greater than or equal to -12

OpenStudy (phi):

Are you sure we started with the correct problem? for the problem I posted we get x <= 4/3

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

I'm confused hsha

OpenStudy (phi):

if the answer is really x>= -12 then x= 2 should work (but won't because we know x<= 4/3 (1 and 1/3) ) using x=2 in the problem \[ \frac{12x-5}{12} + \frac{3-x}{4} \le \frac{4}{3} \\ \frac{24-5}{12} + \frac{3-2}{4} \le \frac{4}{3} \\ \frac{19}{12} +\frac{3}{12} \le \frac{4}{3} \\ \frac{22}{12} \le \frac{4}{3} \\ 1.833... \le 1.333... \] which is not true (as we know)

OpenStudy (phi):

Here are the possible problems: 1) typo in the problem 2) typo in answer key

OpenStudy (anonymous):

Wait... it's actually a problem that some one did and I have to explain where they went wrong

OpenStudy (phi):

maybe if you post a copy of the page...

OpenStudy (phi):

take a screen shot or a picture

OpenStudy (anonymous):

How do I post the picture tho?

OpenStudy (phi):

click on "attach file"

OpenStudy (anonymous):

It doesn't say that for me

OpenStudy (phi):

are you at a computer or using the iPad app?

OpenStudy (anonymous):

I'm on my phone

OpenStudy (phi):

do you see a paper clip icon ?

OpenStudy (anonymous):

just got on laptop gimme a sec

OpenStudy (anonymous):

OpenStudy (phi):

do you see what they did wrong?

OpenStudy (anonymous):

they didnt change the 4/3?

OpenStudy (phi):

yes. as for *our problem* we started with (12x-5)/12 and it should be (2x-5)/12

OpenStudy (anonymous):

oh. orry

OpenStudy (anonymous):

sorry*

OpenStudy (phi):

of course that changes everything.

OpenStudy (anonymous):

yeah

OpenStudy (phi):

can you do the work ? this one is more tricky because we will get a negative x

OpenStudy (anonymous):

let me try

OpenStudy (anonymous):

im stuck at -x+4<=16

OpenStudy (phi):

add -4 to both sides

OpenStudy (anonymous):

i subtract 4 and then divide by -1 right?

OpenStudy (anonymous):

then the sign flips

OpenStudy (phi):

with relations, if you divide by a negative number you have to remember to flip the relation once you get to -x <= 12 you can divide by -1 and flip to get x >= -12 or , remember it's always safe to add or subtract. so we could add +x to both sides: -x + x <= 12 + x 0 <= 12 + x now add -12 to both sides -12 <= x which can also be written x>= -12

OpenStudy (anonymous):

okay :)

OpenStudy (anonymous):

thanks so much!

OpenStudy (phi):

yw

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