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Mathematics 15 Online
OpenStudy (anonymous):

Solve the inequality. Show your work. |r + 3| ≥ 7

OpenStudy (anonymous):

Ok so this is greater that, do you know how to start?

OpenStudy (anonymous):

No

OpenStudy (anonymous):

well first, writeit two ways, greater than and less than

OpenStudy (anonymous):

What does that mean?

OpenStudy (anonymous):

ok well just to show you\[(r+3\ge7) or (r+3\le-7)\]

OpenStudy (anonymous):

solve each of those

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

4 or -4

OpenStudy (anonymous):

close remember to subtract 3 from -7 is to add -3 to -7

OpenStudy (anonymous):

10 or -10

OpenStudy (anonymous):

4 or -10, you subtract 3 from both sides, do you understand or should I go back?

OpenStudy (anonymous):

Please go back.

OpenStudy (anonymous):

ok well first do you understand why you write it the two ways?

OpenStudy (anonymous):

is it because its absolute value

OpenStudy (anonymous):

right. |-5|=|5| they equal the same thing. so let's look at the first one the original. what does that look like now?

OpenStudy (anonymous):

[|r+3|=|r+-3|

OpenStudy (anonymous):

hold on slow down. let's look at that. I understand how you got there but let's look at it. if we plug in 5 for r, then |5+3|=|5-3| this gets us |8|=|2|. so try again. remember what the original problem is.

OpenStudy (anonymous):

|r+3|=|r-3|

OpenStudy (anonymous):

you're thinking about the wrong part. do you know what this will look like on a graph?|dw:1363025883735:dw| kind of like that

OpenStudy (anonymous):

what was inside the absolute value bars stays the same.

OpenStudy (anonymous):

OK so it should look like this |r + 3|

OpenStudy (anonymous):

ok let's think about it like this. we are trying to find what r is right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so we are told that the absolute value of r plus 3 is greater that or equal to 7.

OpenStudy (anonymous):

Yes

OpenStudy (anonymous):

so because r is inside the absolute value bars it can be negative or positive

OpenStudy (anonymous):

Would R=4?

OpenStudy (anonymous):

r will be 2 things because it is in the absolute value bars. here to make it easier make the greater than or equal to an equal sign for now. |r+3|=7 what do you do first.

OpenStudy (anonymous):

Minus 3 from both sides |r+3|=7 -3 -3 r=4

OpenStudy (anonymous):

wait. slow down. let's just take 3 away for now. |r|=7 what is r

OpenStudy (anonymous):

-7

OpenStudy (anonymous):

or...

OpenStudy (anonymous):

4

OpenStudy (anonymous):

ok if you have |r|=7 then r= 7 and r= -7 do you understand?

OpenStudy (anonymous):

Yes, I understand

OpenStudy (anonymous):

so if we write it out |r|=7 and |r|=-7 so if we have |r+3|=7 how do we write it out

OpenStudy (anonymous):

r ≤ -10 and r ≥ 4

OpenStudy (anonymous):

just to make sure you understand so you can apply it later how do you get to that?

OpenStudy (anonymous):

First I subtracted 3 from both sides |r+3|=7 -3 -3 r=4

OpenStudy (anonymous):

Then I added -3+-7 And got -10 which is greater than 7

OpenStudy (anonymous):

-10 which is less than 7

OpenStudy (anonymous):

ok let me show you the work I did |r+3| ≥7 r+3 ≥7 r+3≤-7 r ≥4 r≤-10 plug both in to the original equation |4+3| ≥7 |-10+3| ≥7 |7| ≥7 |-7| ≥7 7 ≥7 7 ≥7 does that make sence

OpenStudy (anonymous):

Yes it does make sense. Thank you so much for everything, you have been a huge help to me.

OpenStudy (anonymous):

No problem any time

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