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Mathematics 9 Online
OpenStudy (anonymous):

Solve the open sentence. 6 < 3b OR 3b + 12 < 9 A.b < –1 OR b > –6 B.b < –1 OR b < 3 C.b < 2 OR b > –3 D.b > 2 OR b < –1

OpenStudy (tapha):

hmm

OpenStudy (tapha):

ok

OpenStudy (anonymous):

can you help me

OpenStudy (tapha):

sure

pooja195 (pooja195):

solve them serperatly 6 < 3b OR 3b + 12 < 9 for 6<3b just divide both sides by 3 3b+12<9 subtract 12 from both sides and divide by 3

OpenStudy (tapha):

c i think

OpenStudy (tapha):

try it

OpenStudy (anonymous):

ok ill try it

OpenStudy (tapha):

pooja not intrested

pooja195 (pooja195):

@tapha I messaged you not to give direct answers if you wanna get in trouble keep giving out answers

OpenStudy (tapha):

ok

pooja195 (pooja195):

@WildQueen did you try?

OpenStudy (tapha):

i really gtg

OpenStudy (anonymous):

yes i still dont get it

pooja195 (pooja195):

What dont you get?

OpenStudy (anonymous):

how i divide 3 and subtract 12

pooja195 (pooja195):

Lets do this one first :) 6 < 3b in this you divide by 3 what are you left with?

OpenStudy (anonymous):

you do the other side

pooja195 (pooja195):

Right but one you divide by 3 what are you left with?

pooja195 (pooja195):

\[\huge~\rm~6 /3< 3b/3\]

OpenStudy (anonymous):

after u divide it u subtract 12

OpenStudy (ddos_dragon):

When you do the inverse operation of something, that is the same integer, it... vanishes.

OpenStudy (anonymous):

oh ok

pooja195 (pooja195):

@WildQueen we do not subtract by 12 in this one

pooja195 (pooja195):

That is for the after the OR part

pooja195 (pooja195):

What are you left with once both sides are divide by 3?

OpenStudy (anonymous):

then u subtract 12 and divide 3 again

OpenStudy (ddos_dragon):

No..no..no..no

pooja195 (pooja195):

No ignore that part. Just do the part i told you.

OpenStudy (ddos_dragon):

The division first.

pooja195 (pooja195):

Focus only on this \[\huge~\rm~6 /3< 3b/3\]

OpenStudy (ddos_dragon):

Remember that inverse trick.

pooja195 (pooja195):

What will the inequality look like if we divide both sides by 3?

OpenStudy (anonymous):

ok you will get 3 or -3

OpenStudy (ddos_dragon):

Erm... Not exactly.

pooja195 (pooja195):

You are dividing a positive by a positive there are no negatives.

OpenStudy (ddos_dragon):

We know one side has nothing but the B, because of the inverse trick. What about the other?

OpenStudy (anonymous):

oh ok ok so u get 3

OpenStudy (ddos_dragon):

6/3 = ?

OpenStudy (anonymous):

3

OpenStudy (ddos_dragon):

No.

OpenStudy (ddos_dragon):

How many times does 3 fit into 6?

pooja195 (pooja195):

Im gonna let @Ddos_Dragon finish helping u.

OpenStudy (anonymous):

ok

OpenStudy (ddos_dragon):

So, how many times does 3 go into 6?

OpenStudy (anonymous):

2

OpenStudy (ddos_dragon):

YES!

OpenStudy (ddos_dragon):

2 < B

OpenStudy (ddos_dragon):

There is your first inequality.

OpenStudy (anonymous):

ok i got you

OpenStudy (ddos_dragon):

Now for, 3b + 12 < 9

OpenStudy (anonymous):

ok ready

OpenStudy (ddos_dragon):

First, divide 3 on both sides.

OpenStudy (anonymous):

how will that look like

OpenStudy (ddos_dragon):

9/3 3/3b (Inverse trick removes both threes.)

OpenStudy (ddos_dragon):

So all you need to do is figure out how many times 3 goes into 9.

OpenStudy (anonymous):

3 times

OpenStudy (ddos_dragon):

Perfect!

OpenStudy (ddos_dragon):

Leaving us with, b + 12 < 3

OpenStudy (anonymous):

so now u subtract

OpenStudy (ddos_dragon):

b + 12 -12 (Inverse trick!!!!!) 3- 12

OpenStudy (anonymous):

is it 9

OpenStudy (ddos_dragon):

Ermm. No.

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

then how do u solve it

OpenStudy (ddos_dragon):

3 - 12 = ? If the bigger number is on the right it is a negative.

OpenStudy (ddos_dragon):

BRB

OpenStudy (anonymous):

ok

OpenStudy (ddos_dragon):

Back. And its closed XD

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