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(Int. Algebra) I need help with an inequality problem

Hello All,

So here's the problem. I got it right but I'm still not clear on flipping the sign. As I understand it, you flip the sign when you multiply or divide by a negative number. In the operation below, why is the sign "not" flipped (from -1/3)? That's the only I got wrong and admittedly, I still don't understand it.

Attachment 38912

Thanks.

ALex

Re: (Int. Algebra) I need help with an inequality problem

it's not flipped because you divide by 3, not negative 3.

Re: (Int. Algebra) I need help with an inequality problem

Ok... I understand your answer. I'm still a little confused because I thought that -1/3, 1/-3, and -(1/3) were all the same. Oh I think I see now... The entire inequality is divided by "3." The 3 doesn't suddenly change to "-3." The way I have it written makes it look like there are individual fractions. Right? Or am I still not getting it. Wow! you have to be really careful.

Re: (Int. Algebra) I need help with an inequality problem

Solve (3x + 1)/4 = 2, then solve then solve (3x + 1)/4 = -3 : get my drift?

Re: (Int. Algebra) I need help with an inequality problem

Quote:

Originally Posted by

**alexcordero** Ok... I understand your answer. I'm still a little confused because I thought that -1/3, 1/-3, and -(1/3) were all the same. Oh I think I see now... The entire inequality is divided by "3." The 3 doesn't suddenly change to "-3." The way I have it written makes it look like there are individual fractions. Right? Or am I still not getting it. Wow! you have to be really careful.

let's just work through it

$-3 \leq \dfrac{3x +1}{4} \leq 3$

we first multiply everything by 4. inequality signs do not flip

$-12 \leq 3x+1 \leq 12$

we now subtract 1 from all sides.

$-13 \leq 3x \leq 11$

we now divide by 3, inequality signs do not flip

$-\dfrac{13}{3} \leq x \leq \dfrac{11}{3}$

Re: (Int. Algebra) I need help with an inequality problem

Quote:

Originally Posted by

**romsek** let's just work through it

$-3 \leq \dfrac{3x +1}{4} \leq 3$

we first multiply everything by 4. inequality signs do not flip

$-12 \leq 3x+1 \leq 12$

we now subtract 1 from all sides.

$-13 \leq 3x \leq 11$

we now divide by 3, inequality signs do not flip

$-\dfrac{13}{3} \leq x \leq \dfrac{11}{3}$

Ok, thank you very much for the explanation. On another question: How did you get the inequality and fractions to display as such?

Again, thanks for help.

Re: (Int. Algebra) I need help with an inequality problem

Quote:

Originally Posted by

**alexcordero** Ok, thank you very much for the explanation. On another question: How did you get the inequality and fractions to display as such?

If you click the tab *reply with quote* you will see the LaTeX code.

Here is some: -3 \leq \dfrac{3x +1}{4} \leq 3

-12 \leq 3x+1 \leq 12

You can warp the code \$ -12 \leq 3x+1 \leq 12\$ in dollar signs you get $ -12 \leq 3x+1 \leq 12$

Re: (Int. Algebra) I need help with an inequality problem

Quote:

Originally Posted by

**Plato** If you click the tab *reply with quote* you will see the LaTeX code.

Nice!! I'm going to find a LaTeX syntax cheat sheet. Thanks again.